2nd PUC Statistics Question Bank Chapter 6 Statistical Inference

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Karnataka 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference

Section – A

2nd PUC Statistics Statistical Inference One Mark Questions and Answers

Question 1.
What is simple random sample?
Answer:
A simple random sample is a sample that is chosen in such a way where every unit has a possibility to be selected.

Question 2.
What is parameter?
Answer:
A parameter is a statistical constant of the population or It is population constant.

Question 3.
What is statistic?
Answer:
A statistic is the function of a sample value or it is sample constant.

Question 4.
What is parameter space?
Answer:
The parameter space is the set of all admissible values of the parameter.

Question 5.
What is sample space?
Answer:
Sample space is the set of the samples that can be drawn from the population.

Question 6.
What is sampling distribution of a statistic?
Answer:
Sampling distribution of a statistic refers to the different values of the sample size.

Question 7.
What is standard error?
Answer:
Standard error is the standard deviation of the sampling distribution of a statistic.

Question 8.
Write the formula of S.E (x̄)
Answer:
S.E(x̄) = \(\frac{\sigma}{\sqrt{n}}\)

Question 9.
Given σ² = 9 cm² and n = 36 calculate standard error of sample mean.
Answer:
σ² = 9, σ = √9 = 3 n = 36
S.E.(x̄) = \(\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{36}}=\frac{3}{6}\) = 0.5

Question 10.
Write the formula of S.E.(x̄₁ – x̄₂)
Answer:
S.E.(x̄₁ – x̄₂) = \(\sqrt{\frac{\sigma_{1}^{2}}{\mathrm{n}_{1}}+\frac{\sigma_{2}^{2}}{\mathrm{n}_{2}}}\)

Question 11.
Sizes of two samples are 50 and 100 population standard deviations are 20 and 10. Compute S.E.(x̄₁ – x̄₂)
Answer:
n₁ = 50, n₂ = 100, σ₁ = 20 and σ₂ = 10
S.E.(x̄₁ – x̄₂) = \(\sqrt{\frac{\sigma_{1}^{2}}{\mathrm{n}_{1}}+\frac{\sigma^{2}}{\mathrm{n}_{2}}}\) = \(\sqrt{\frac{20^{2}}{50}+\frac{10^{2}}{100}}=\sqrt{8+1}=\sqrt{9}=3\)

Question 12.
Write the formula of S.E.(P)
Answer:
S.E.(P) = \(\sqrt{\frac{P Q}{n}}\)

Question 13.
If P = 0.02 and n = 64 then find S.E (P)
Answer:
P = 0.02 n = 64 Q = 1 – P ⇒ 1 – 0.02 ⇒ 0.98
S.E = \(\sqrt{\frac{P Q}{n}}\)= \(\sqrt{\frac{(0.02)(0.98)}{64}}=\sqrt{\frac{0.0196}{64}}=0.0175\)

Question 14.
If P = 0.5 and n = 100 then find S.E(P)
Answer:
P = 0.5; n = 100 Q = 1 – P ⇒ 1 – 0.5 ⇒ 0.5
S.E(P) = \(\sqrt{\frac{P Q}{n}}\) = \(\sqrt{\frac{(0.5)(0.5)}{100}}=\sqrt{\frac{0.25}{100}}=0.05\)

Question 15.
Write the formula of S.E(P₁ – P₂) when P₁ ≠ P₂
Answer:
S.E. (P₁ – P₂) when P₁ ≠ P₂ = \(\sqrt{\frac{\mathrm{P}_{1} \mathrm{Q}_{1}}{\mathrm{n}_{1}}+\frac{\mathrm{P}_{2} \mathrm{Q}_{2}}{\mathrm{n}_{2}}}\)

Question 16.
A lot contains 2% defective items 40 items are chosen from it. Another lot contains 1% defective items. 60 items are chosen from it. Find E(P₁ – P₂) and S.E (P₁ – P₂)
Answer:
P₁ = 2% P₂ = l% n,₁ = 40 n₂ = 60
Q₁ = 1 – P₁ = 1 – 0.02 = 0.98
Q₂ = 1 – P₂ = 1 – 0.01 = 0.99
S.E(P₁ – P₂) = \(\sqrt{\frac{P_{1} Q_{1}}{n_{1}}+\frac{P_{2} Q}{n_{2}}}\) ⇒ \(\sqrt{\frac{(0.02)(0.98)}{40}+\frac{(0.01)(0.99)}{60}}\)
\(=\sqrt{0.00049+0.000165}=\sqrt{0.000655}=0.0256\)
⇒ E(P₁ – P₂) (0.02 – 0.01) = 0.01

Question 17.
Write the formula of S.E.(P₁ – P₂) when P₁ = P₂ = P
Answer:
S.E.(P₁ – P₂) when P₁ = P₂= P is \(\sqrt{P Q\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}\)

Question 18.
Write the utility of standard error.
Answer:

• Determine the efficiency and consistency of the statistic as an estimator.
• Obtain the confidence intervals of an estimate
• Standardise the distribution of test statistic in testing of hypothesis.

Question 19.
What is statistical inference?
Answer:
Statistical inference is the theory of making decisions about the population parameters by utilising sampling and concept of probability.

Question 20.
Mention two branches of statistical inference.
Answer:

1. Estimation
2. Testing of hypothesis.

Question 21.
What is meant by estimation?
Answer:
Estimation is the method of obtaining the most likely value of the population parameter using statistic.

Question 22.
What is an estimator?
Answer:
Any statistic which is used to estimate an unknown parameter is called as estimator.

Question 23.
What is an estimate?
Answer:
Estimate is the numerical value of the unknown parameter.

Question 24.
What is point estimation?
Answer:
If a single value is proposed to estimate an unknown parameter, then it is point estimation.

Question 25.
What is interval estimation?
Answer:
If an interval is proposed to estimate an unknown parameter, then it is interval estimation.

Question 26.
What is confidence interval?
Answer:
An interval (T₁,T₂) which contains an unknown parameter is called as confidence interval.

Question 27.
What are confidence limits?
Answer:
The boundary values of confidence interval are confidence limits.

In this activity, students will discover the relationship between the degree and leading coefficient calculator of a polynomial.

Question 28.
What is confidence coefficient?
Answer:
The probability that a confidence interval includes an unknown parameter is named as confidence coefficient.

Question 29.
What is statistical hypothesis?
Answer:
A statistical hypothesis is a statement about the parameters of the population.
Ex.: H : μ = μ₀

Question 30.
What is null hypothesis? Give an example?
Answer:
The hypothesis is being tested for possible rejection is called as Null hypothesis.
Ex.: H₀ : μ = μ₀

Question 31.
What is alternative hypothesis? Give an example.
Answer:
The hypothesis which is accepted when null hypothesis is rejected is called Alternative hypothesis.
E.g., If H₀ : μ = μ₀, then the alternative hypothesis could be H₁ = population mean differs from a specified mean μ ≠ μ₀ (two tailed).

Question 32.
What is type I error?
Answer:
Type I error is the error which occurs by rejecting null hypothesis when it is actually true

Question 33.
What is type II error?
Answer:
Type II error is the error which occurs by accepting null hypothesis when it is actually not true

Question 34.
What is size of the test?
Answer:
The probability of rejecting H₀, when it is true is called as the size of the test.

Question 35.
What is level of significance?
Answer:
Maximum size of the test is called level of significance.

Question 36.
What is power of a test?
Answer:
The probability of rejecting H₀, when it is not true is called as power of a test. It is denoted by (1 – β).

Question 37.
What is critical region?
Answer:
Critical region is the set of those values of the test statistic, which leads to the rejection of the null hypothesis.

Question 38.
What is critical value?
Answer:
Critical value is the value which separates the critical region and acceptance region.

Question 39.
What is two tailed test?
Answer:
It is a test of statistical hypothesis, where rejection region is located at both the tails of the probability curve.

Question 40.
What is one tailed test?
Answer:
It is a test of statistical hypothesis, where the rejection region will be marked at only one tail of the probability curve. The rejection region will be either left of right tail of the curve depending upon the alternative hypothesis.

In testing H₀ : μ = μ₀ against H₁ : μ < μ₀ (left tailed or lower tailed) and H₁: μ > μ₀ (right tailed or upper tailed) are examples of one tailed tests.

Question 41.
What is null distribution
Answer:
The statistical distribution of the test statistic under the null hypothesis is named null distribution.

Question 42.
What is test statistic?
Answer:
A test statistic is based on the distribution where the test of hypothesis is conducted

2nd PUC Statistics Large Sample Test Exercise Problems

Question 1.
Write the testing procedure of large sample tests?
Answer:
The test procedure contains following steps

• Setting up of the null hypothesis (H₀)
• Setting up of the alternative hypothesis (H₁)
• Identification of the test statistic and its null distribution and computation of test statistic.
• Identification of the critical region.
• Drawing a random sample and actually conducting the test
• Making the decision.

Question 2.
Write the testing procedure of population mean?
Answer:
Test for population mean has following steps
1. H₀: The population mean is μ₀, μ = μ0

2. H₁: The population mean two tailed test, μ ≠ μ₀ (two tailed test)
(or) μ < μ₀ (Left tailed test)
(or) μ > μ₀ (right tailed test)

3. Calculation of test statistic Z = \(\frac{\bar{x}-\mu_{0}}{\frac{\sigma}{\sqrt{n}}}\)
Here x̄ = sample mean 7n
σ = population standard deviation
If it is unknown, then it should be replaced by sample standard deviation ‘s’ and ‘n’ is sample size.
Under H₀ Z ~ N(0,1)

4. Depending on H₁ and α the critical value ‘k’ is selected

5. If the calculated value of the test statistic (Z_cal) is in acceptance region, then H₀ is accepted, otherwise its rejected.

6. Prepare conclusion and make a decision.

Question 3.
A sample of 100 students is taken from a college. The mean and SD of their weights are 51 kg and 5 kg respectively, test at 1% level of significance that the average weight of college students is 50 kg?
Answer:
n = 100, x̄ = 51, s = 5, μ₀= 50, α = 1% .
H₀: let Average weight of college students is 50 i.e. μ = 50
H₁ : let Average weight of college student is not equal to 50 i.e. μ ≠ 50 .

The test is two tailed test

Level of significance is α = 0.01(1%) the critical values are -2.58 and +2.58 since Z_cal value lies between the critical values (-2.58 and +2.58),
-2.58 < Z _cal < + 2.58
∴ H₀ is accepted
Conclusion: We can assume that average weight of a student is 50.

Question 4.
A machine is designed to fill 500 ml of milk to polythene bags. A randomly selected 100 milk bags filled by this machine are inspected. The mean milk is found to be 498 ml and SD is 10 ml. Is machine is functioning properly a 5% level of significance?
Answer:
n = 100, x̄ = 498, s = 10, α = 5%, μ = 500
H₀ : let machine works properly i.e., it fills 500 ml of milk to polythene bags i.e. μ = 500
H1₁ : let machine does not function properly i.e., it doesn’t fill 500 ml of milk to polythene bags i.e. μ ≠ 500

The test is two tailed test

Level of significance is α = 0.05 (i.e., 5%)
The critical values are -1.96 and 1.96
Since Z_cal value lies in the rejection region
-1.96 < Z_cal >+1.96
H₀ is rejected
Conclusion: We can assume that Machine does not function properly

Question 5.
A random sample of 64 children is taken from a school. The average weight of the children is 29 kg SD is 5 kg. Can we assume that the average weight of the school children is less than 30 kg? (use α = 0.05)
Answer:
n = 64, x̄ = 29 kg, s = 5, μ = 30, α = 0.05
H₀ : let average weight of the school children is 30 i.e. μ = 30
H₁ : let average weight of the school children is not equal to 30 i.e. μ < 30 (left tailed)


The test is one tailed
Level of significance α = 5%
Here
The critical value is -1.65
Z_cal = -1.6
Z_cal > critical value
H₀ is accepted
Conclusion: We can assume that the Average weight of the school children is 30 kg.

Question 6.
A company manufactures car tyres their average life is 40,000 kms and SD 5000 kms. A change in the production process is believed to result in a better product. A test sample of 100 new tires has mean life of 41,000 kms, can you conclude at 5% LOS that the new product gives better result.
Answer:
n = 100; x̄ = 41,000 ; σ = 5000 μ = 40,000; α = 5%
H₀ : let average life of car tires is 40,000 kms i.e. μ = 40,000 kms .
H₁ : let average life of car tires is not equal to 40,000 kms i.e. μ ≠ 40,000 kms


It is two tailed test
Level of significance is α = 0.05 (5%)
The critical value are -1.96 and + 1.96
Z _cal value does not lie between the critical values.
-1.96 < Z_cal > +1.96
H₀ is rejected
Conclusion: We can assume that the Average life of car tires is not equal to 40,000 kms

Question 7.
A specified brand of automobiles tire is known to average life of 10,000 km with a SD of 500 km. A random sample of 36 tires of this brand, when tested resulted in the average life of 9800 km. Regarding quality what is your conclusion at 1% level of significance
Answer:
x̄ = 9800 ; n = 36; σ = 500 km μ = 10,000 α = 1% = 0.01
H₀ : let automobile tire has average life of 10,000 km i.e. μ = 10,000 km
H₁ : let automobile tire does not have average life of 10,000 km i.e. μ ≠ 10,000 km


It’s a two tailed test
The level of significance is α = 0.01(1%)
The critical values are -2.58 and +2.58
Z_cal lies between the critical values
-2.58< Z_cal < +2.58. H0 is accepted
Conclusion: We can assume that the Automobile tire has an average life of 10,000 kms

Question 8.
Given x̄ = 203gm μ = 200 gm, σ = 10 gm and n = 64 calculate test statistic Z.
Answer:
x̄ = 203 ; μ = 200 σ = 10 n = 64

Question 9.
Write the testing procedure of equality of population means.
Answer:
1. H₀ : population means are equal μ₁ = μ₂

2. H₁: population means are not equal, μ₁ ≠ μ₂
If the mean of the first population is less than the second population i.e. μ₁ < μ₂
Its a left tailed test
If the mean of the first population is more than the second population i.e. μ₁ > μ₂
It’s a right tailed test

3. Calculate of test statistic = z = \(\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\)
Here x̄₁, x̄₂ = sample means
σ₁, σ₂ = standard deviations of populations.
If SD are unknown these values are replaced by s₁ and s₂ sample standard deviations
n₁ n₂= sample size

4. Critical value ‘k’ will be chosen depending on H₁ and α through the table.

5. If the calculated value Z_cal lies in acceptance region, then H₀ is accepted otherwise H₀ is rejected.

6. Find conclusion and make a decision.

Question 10.
For the following data, test whether means differ significantly.

Answer:
n₁ = 90; x̄ = 52 s₁ = 9; n₂ = 40 x̄ = 54 ; s₂ = 2
H₀ : means of two population are equal i.e. μ₁ = μ₂
H₁ : means of two population are not equal, i.e. μ₁ ≠ μ₂


It is two tailed test
At 5% significance
Critical values = -1.96 and +1.96
Z_cal value lies in rejection region
-1.96 > Z _cal <+1.96
H₀ is rejected
Conclusion: means of populations are not equal

Question 11.
450 boys and 350 girls are appeared for II PUC examination. The mean and SD of marks obtained by boys are 53 and 18 respectively. The mean and SD of marks obtained by girls are 50 and 14 respectively. Is there any significant difference between mean marks obtained by boys and girls?
Answer:
n₁ = 450, x̄₁ = 53 σ₁ = 18, n₂ = 350, x̄₂ = 50, σ₂ = 14

H₀ : mean marks obtained by boys and girls are same i.e. μ₁ = μ₂
H₁ : mean marks obtained by boys and girls are not same i.e. μ₁ ≠ μ₂

Its two tailed test
Level of significance α = 5% = 0.05
Critical values = -1.96, +1.96
-1.96 < Z_cal > +1.96
Z_cal value does not lie in the acceptance region
H₁ is rejected
Conclusion: mean marks obtained by boys and girls are not same

Question 12.
A random sample of 100 workers from South India shows that their mean wages are 146 per day with SD 20. A random sample of 150 workers from North India shows that their mean wages are 150 per day with SD 30. Test at 1% level of significance that the mean wages of south Indian is less than mean wages of North India.
Answer:
n₁ = 100, x̄₁ = 146, σ₁ = 20, n₂ = 150, x̄₂ = 150, σ₂ = 30
H₀ : mean wages of south Indian and North Indian is same μ₁ = μ₂
H₀ : mean wages of South Indian is less than North Indian μ₁ < μ₂


Its lower tailed test
Level of significance α = 1% = 0.01
The critical value is -2.33
Z_cal > -2.33, H₀ is accepted
Conclusion: Mean wages of south Indian and North Indian are same

Question 13.
As examination was conducted to two sections A and B consisting of 50 & 40 students respectively. Mean marks obtained by section A is 74 with a SD of 8 and that of B is 78 with a SD of 7. Is there a significant difference between the performances of the two sections at 1% level of significance?
Answer:
n₁ = 50, x̄₁ = 74, σ₁ = 8, n₂ =40, x̄₂ = 78, σ₂ = 7
H₀ : mean marks of Section-A and Section-B is same i.e. μ₁ = μ₂
H₀ : mean marks of Section-A and Section B are not same i.e. μ₁ ≠ μ₂


Its two tailed test
Level of significance α = 1% = 0.01
Z_cal value -2.5273 lies between the critical values
-2.58 < Z_cal < +2.58
H₀ is accepted
Conclusion: Mean marks obtained by Section A and Section B are same

Question 14.
Intelligent test given to two groups boys and girls gave the following information

difference in the mean scores of boys and significant? Use 5% LOS
Answer:
n₁ = 100, x̄₁ = 70, σ₁ = 12, n₂ = 50, x̄₂ = 74, σ₂ = 10
H₀: mean scores of boys and girls are same = μ₁ = μ₂
H₀: mean scores of boys and girls are not same= μ₁ ≠ μ₂


Its two tailed test
Level of significance α = 5 % = 0.05
Critical values = -1.96 & +1.96
Z_cal = -2.1567
Z_cal does not lie within the acceptance region
-1.96> Z_cal <+1.96
H₀ is rejected
Conclusion: Mean scores of boys and girls are not same

Question 15.
Write the testing procedure of population proportion
Answer:
The test has following steps:
(a) H₀ : P = P₀ (A given value for population)

(b) H₁: P ≠ P₀ (two tailed test) OR
P < P₀ (left tailed test) OR
P > P₀ (Right tailed test)

(c) Calculation of test statistic

x = number of items processing
n = number of items in a sample

(d) Critical value ‘k’ is chosen depending on H₁ and α

(e) If the computed value of the test statistic Z_cal lies in the acceptance region then H₀ is accepted if not H₀ is rejected.

Question 16.
In a random sample of 1000 persons from large populations 470 are females. Can it be said that males and females are in the equal ration in the populations? Use α = 0.05
Answer:
n = 1000, x = 470, P₀ = 0.5, Q₀ = 0.5
p = \(\frac{x}{n}=\frac{470}{1000}=0.47\)
H_o : males and females are in the equal ratio in the population P_o = 0.5
H₁ : males and females are not in the equal ratio in the population P_o ≠ 0.5

It is two tailed test
Level of significance α = 0.05 = 5%
The critical values = -1.96 & +1.96

Z_cal lie between the critical values
-1.96 < Z_cal <+1.96
H₀ is accepted
Conclusion: males and females are in the equal ratio in the population

Question 17.
In an election the leaders of a party contend that they would secure more than 36% of votes. A pre-pole survey of 400 voters revealed that the percentage is 42. Does the survey supports the leaders claim? use 1% LOS
Answer:
n₁ = 400, P₀ = 36% = 0.36, p = 42% = 0.42,
α= 1% = 0.01, Q₀ = 1 – P₀ = 1 – 0.36 = 0.64
H₀ : mean leaders contention in that they can secure exactly 36% of them votes P = 36%
H₁ : mean leaders contention is that they secure more than 36% of votes P > 36%

It is right tailed test
Level of significance, α = 0.01 = 1%
The critical value is 2.33
Z_cal doesn’t lie between the critical values
Z_cal > 2.33
H₀ is rejected
Conclusion: mean leaders contention is that they secure more than 36% of votes

Question 18.
In a random sample of 100 PUC statistics students 9 are distinction holders, at the 5% level of significance can we conclude that 10% of II PUC statistics students are distinction holders?
Answer:
n = 100, x = 9, P_o = 10% = 0.1, p = \(\frac{x}{n}=\frac{9}{100}\) = 0.09, Q₀ = 1 – P₀ = 1 – 0.1 = 0.9
H₀: 10% of II PUC statistics students are distinction P₀= 10%
H₁ : 10% of II PUC statistics students are not distinction P₀ ≠ 10%


It is two tailed test
Level of significance α = 0.05 = 5%
The critical values = -1.96 & +1.96
Z_cal lie between the critical values
-1.96 < Z_cal < +1.96. H₀ is accepted
Conclusion: 10% of II PUC statistics students are distinction holders.

Question 19.
A stock broker claims that he can predict with 80%accuracy whether a stock market value will rise or fall during the coming month. In a sample of 40 predictions he is correct in 28. Does this evidence supports broker’s claim at 1% level of significance
Answer:
n = 40, x = 28,p = \(\frac{x}{n}=\frac{28}{40}\) = 0.7, P₀ = 80% = 0.8, Q₀ = 1 – P₀ = 1 – 0.8

H₀ : we can support brokers claim that can predict with 80% accuracy whether a stock’s market will rise or fall
P₀ = 80%
H₁: we cannot support P₀ ≠ 80%

It is two tailed test
Level of significance α = 0.01 = 1%
The critical values = -2.58 & +2.58
Z_cal lies between the critical values
-2.58 < Z_cal < +2.58. H₀ is accepted .
Conclusion: We can support broker’s claim that he can predict with 80% accuracy whether a stock market value will rise or fall.

Question 20.
The manufacturer of surgical instruments claims that less than 2% of the instruments he supplied to a certain hospital are faulty. A sample of 400 instruments revealed that 12 were faulty. Test his claim at 5% level of significance
Answer:
n = 400, x = 12, p = \(\frac{x}{n}=\frac{12}{400}\) = 0.03, p₀ = 2% = 0.02 Q₀ = 1 – p₀ = 1 – 0.02 = 0.98

H₀ : Surgical instruments supplied to hospital has 2% faulty P₀ = 0.02
H₁ : Surgical instruments supplied to hospital has less than 2% faulty P₀ < 0.02

It is left tailed test. LOS α = 0.05 = 5%
The critical values are – 1.65
Z_cal value is greater than critical value 0
Z_cal > -1.65
H₀ is accepted
Conclusion: Surgical instruments supplied to hospital has 2% fault

Question 21.
Write the testing procedure of equality of population proportions
Answer:
The test has following steps:
(a) H₀: (The population proportions are equal) P₁ = P₂
(b) H₁ : (The population proportions are not equal) P₁ ≠ P₂
(1^st population proportion is less than 2nd population proportion) P₁ < P₂ (left tailed test)
(1^st population proportion is greater than 2nd population proportion) P₁ > P₂(right tailed test)
(c) Calculation of test statistic

(d) The critical value ‘k’ is chosen depending on H₁ and α
(e) If Z_cal lies in acceptance region then H₀ is accepted if not H₀ is rejected.

Question 22.
A machine produced 26 defective articles among 250. Another machine produced 4 defective articles among 50. Test whether there is a significant difference between population proportions at 5% level of significance
Answer:
n₁ = 250, X₁ = 26, n₂ = 50, x₂ = 4
p₁ = \(\frac{x_{1}}{n_{1}}=\frac{26}{250}\) = 0.104, P = \(\frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{26+4}{250+50}\) = 0.1
Q = 1 – P = 1 – 0.1 = 0.9 p₂ = \(\frac{x_{2}}{n_{2}}=\frac{4}{50}\) = 0.08
H₀: There is significant difference between population proportions P₁ = P₂
H₁: There is no significant difference between population properties P₁ ≠ P₂


\(=\frac{0.024}{(0.09)(0.024)}=0.5164\)
It is a two tailed test
Level of significance α = 5%
The critical values = -1.96 and +1.96
Z_cal = 0.5164 lies in between the critical values
-1.96 < Z_cal > +1.96
H₀ is accepted
Conclusion: There is significant difference between population proportions

Question 23.
Out of 400 PUC students of college A, 72% of students were passed, out of 200 PUC students of college B 66% of students were passed. Can it be concluded that performance of college A is better than performance of college B? Use 5% LOS.
Answer:
n₁ = 400, p₁ = 72% = 0.72 n₂ = 200, p₂ = 66% = 0.66
H₀ : the performance of college A is same as B, P₁ = P₂
H₁ : the performance of college A better than B, P₁ > P₂
p = \(\frac{\mathrm{n}_{1} \mathrm{p}_{1}+\mathrm{n}_{2} \mathrm{p}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}\) = \(\frac{(400) 0.72+(200) 0.66}{400+200}=\frac{288+132}{600}=\frac{420}{600}=0.7\)

Q = 1 – P = 1 – 0.7 = 0.3

It is upper tailed test
Level of significance α = 5% = 0.05
The critical value = 1.65
Z_cal = 1.512 < 1.65
H₀ is accepted
Conclusion: Performance of college A and B are same.

Question 24.
In a random sample of 100 people from a city in the year 2011 revealed that 65 were cricket match viewers. In another random sample of 100 peoples from same city in the year 2013 revealed that 75 were cricket match viewers examine whether there is a significant increase in cricket match viewers. Use 1% level of significance
Answer:
n₁ = 100; x₁ = 65 P₁ = \(\frac{x_{1}}{n_{1}}=\frac{65}{100}=0.65\)
n₂ = 100; x₂ = 75 p₂ = \(\frac{x_{2}}{n_{2}}=\frac{75}{100}=0.75\)
\(P=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{65+75}{100+100}=\frac{140}{200}=0.7\)
Q = 1 – p = 1 – 0.7 = 0.3
H₀ : there is no significant increase in cricket match viewers from 2011 to 2013 p₁ = p₂
H₁ : there is significant increase in cricket match viewers from 2011 to 2013 p₁ < p₂


It is lower tailed test
Level of significance α = 1% = 0.01.
The critical value = -2.33. Z_cal = -1.543 > -2.33
H₀ is accepted
Conclusion: There is no significant increase in cricket match viewers from 2010 to 2013.

Question 25.
From the following data, test whether the difference between the proportions in the samples is significant use 5% level of significance.

Sample	I	II
Size	100	400
Proportion	0.02	0.01

Answer:
n₁ = 100, n₂ = 400, p₁ = 0.02, p₂ = 0.01
p = \(\frac{n_{1} p_{1}+n_{2} p_{2}}{n_{1}+n_{2}}\) = \(\frac{(100) 0.02+(400) 0.01}{100+400}=0.012\)
Q = 1 – P = 1 – 0.012 = 0.988
H_o: There is no difference between the proportions in the sample P₁ = P₂
H₁ : There is difference between the proportions on the sample P₁ ≠ P₂


It is a two tailed test. Level of significance α = 5% 0.05,
The critical values = -1.96 and +1.96. Z_cal = 0.82141 lies in between the critical values
-1.96 < Z_cal > +1.96
H₀ is accepted
Conclusion: There is no difference between the proportions in the sample. P₁ = p₂

2nd PUC Statistics T- Test Exercise Problems

Question 1.
Write two applications of t-Test?
Answer:

1. To test the population mean.
2. To test the equality of two population means.

Question 2.
Write down the t-test statistics and degrees of freedom in case of test for mean?
Answer:
T-test statistic

Degrees of freedom means, number of independent observations if there are ‘n’ observations then, Degrees of freedom = (n – c) where ‘c’ is number of independent constraints.

Question 3.
Write down the t-test statistic & degrees of freedom in case of test for for equality of means of two independent samples.
Answer:
T-test statistic,

Question 4.
Write down the t-test statistic & degrees of freedom in case of test for for equality of means of praised observations.
Answer:
T – test statistic,

Question 5.
In paired t-test if n = 5 d̄ = 3 and s_d = 1.5. Then what would be the value of test statistic?
Answer:
n = 5, d̄ =3, s = 1.5

Question 6.
A random sample of size 20 taken from normal population gives a sample mean of 42 and standard deviation of 6. Test the hypothesis that the population mean is 44.
Answer:
n = 20, x̄ = 42, s = 6, μ = 44
H₀ : The population mean is 44 μ = 44
H₁ : The population mean is not equal to 44 μ ≠ 44


It is two-tailed test
The d.f is n – 1 = 20 – 1 = 19
Critical values = -2.09 and +2.09
t_cal = 1.4529 is in the accepted region
-2.09 < 1.4529 < 2.09.
H₀ accepted
Conclusion: The population mean is 44

Question 7.
A company has been producing steel tubes with mean inner diameter of 2.00 cms. A sample of 10 tubes gives a mean inner diameter of 2.01 cms and a variance of 0.004 cms². Is the difference in the values of mean is significant?
Answer:
μ = 2.00 cm, n = 10, x̄ = 2.01, variance = s² = 0.004
s = √0.004 = 0.0632
H₀ : there is no difference between population mean and sample mean μ = 2.00
H₁ : there is a difference between population mean and sample mean μ ≠ 2.00


It is two tailed test. Los α = 0.05=5%
The d.f is n – 1 = 10 – 1 = 9.
Critical values = – 2.26 and +2.26.
-2.26 < 0.4748 < +2.26
t_cal value 0.4748 lies in between the critical values
H₀ is accepted.
Conclusion: There is no difference between population mean and sample mean.

Question 8.
The mean weekly sale of ice-cream bars was 140 bars. After an advertising campaign the mean weekly sale in 26 shops for a typical week increased to 150 and showed a SD 20. Is this evidence indicates that the advertising was successful?
Answer:
x̄ =150, μ₀ = 140, s = 20, n = 26
H₀: μ = 140 Mean weekly sale of ice-cream bars after advertisement has not increased
H₁: μ > 140 Mean weekly sale of ice-cream bars after advertisement has increased.


It is upper tailed test
α = 0.05 = 5%
The .d.f is n – 1 = 26 – 1 = 25
Critical value = 2.49
t_cal value 1.970 > 1.72.
H₀ is rejected.
Conclusion: The mean weekly sale of ice – cream bars after advertising has increased.

Question 9.
A fertilizer mixing Machine is set to give 12 kg of nitrate for every bag of fertilizer. Then 10 such bags are examined. The weight of nitrate in each bag (in kg) are given below:
11 14 13 12 13 12 13 14 11 12
Is there reason to believe that the machine is defective?
Answer:
μ = 12, n = 10,
Calculation of mean and standard deviation

Σx = 125, Σx² = 1573
x̄ = \(\frac{\Sigma x}{n}\) = \(\frac{11+14+13+12+13+12+13+14+11+12}{10}=\frac{125}{10}=12.5\)


The test

The test is two-tailed
α = 5% = 0.05
The d.f is n – 1 = 10 – 1 = 9
Critical values = -2.26 and 2.26
t_cal value 1.464 lies between the critical value in acceptance region
H₀ is accepted
Conclusion: The machine is not defective

Question 10.
A random sample of size 16 has mean 53. The sum of the squared deviations taken from mean is 150. Can this sample be regarded as taken from the population having mean 56 (Use α = 0.01)
Answer:
n = 16, x̄ = 53, μ = 56, α = 0.01 = 1%, Σ(x – x̄)² = 150
H₀ : population having mean 56, μ = 56
H₁ : population is not having mean 56, μ ≠ 56


It is two tailed test
The d.f. is n – 1 = 16 – 1
Critical values = -2.95 and 2.95
χ²_cal value is in accepted region
H₀ is accepted

Question 11.
The marks obtained by two groups of students in a statistics test are given below

On the basis of this data, can it be concluded that there is a significant difference in the mean marks obtained by the two groups?
Answer:
n₁ = 15, n₂ = 11, x̄₁ = 42, x̄₂ = 38, s₁ = 10, s₂ = 15
H₀ : Mean marks of students of group A & B are same, μ₁ = μ₂
H₁ : mean marks of group A and B are not same μ₁ ≠ μ₂

Under H₀


Level of significance = α = 0.05
d.f = 15 + 11 – 2 = 24
It is two tailed test
Critical value= – 2.06 and + 2.06
t_cal = 0.7829 lies between critical values
– 2.06 < 0.7829 <+ 2.06
H₀ is accepted

Question 12.
For the following data examine if the means of two samples differ significantly.

Answer:
n₁ = 12; n₂ = 7; x̄₁ = 57.2; s₁ = 3.41; x̄₂ = 52.3; S₂ = 3.62
H₀ : μ₁ = μ₂ means are same
H₁: μ₁ ≠ μ₂ means are not same


It is a two tailed test
d.f. n₁ + n₂ – 2 = 12 + 7 – 2 = 17
Critical values = – 2.11 and + 2.11. LOS = 5% = 0.05
t_cal value 2.793 lies between the critical values
= – 2.11 < 2.793 > + 2.11
H₀ is accepted.

Question 13.
A group of 5 patients treated with medicine A weighs 42, 39, 48, 60 and 41 kg, second group of 7 patients from the same hospital treated with medicine B weighs 38, 42, 56, 61, 69, 68 and 67 kg. Do you agree with the claim that medicine B increases the weight significantly?
Answer:
n₁ = 5 n₂ = 7


H₀ : Medicine B does not increase the weight, μ₁ = μ₂
H₁ : Medicine B increase the weights significantly, μ₁< μ₂


It is lower tailed test
Level of significance = α = 0.05 = 5%
d.f. = n₁ + n₂ – 2 = 5 + 7 – 2 = 10
Critical value= – 1.81
t_cal = – 1.695 > -1.81
H₀ is accepted
Conclusion: Medicine B does not increase weight

Question 14.
Two new types of rations are fed to pigs. A sample of 11 pigs is fed with type A ration and another sample of 11 pigs is fed with type B ration, the gains in weight are recorded be low (in Pounds):

At 1% level, test whether type A ration is better than type B ration.
Answer:
n₁ = 11 n₂ = 11




\(=\frac{3.2}{\sqrt{9.94(0.09+0.09)}}=\frac{3.2}{\sqrt{9.94 \times 0.1809}}\)
\(=\frac{3.2}{\sqrt{1.7982}}=\frac{3.2}{1.34}=2.38\)
At 5% los (n₁ + n₂ – 2) = 11 + 11 – 20 df.
T_tab = CV = 2.53
t_cal = 2.38 < CV_tab = 2.53
H₀ is accepted.

Question 15.
Two laboratories A and B carry out independent estimates of fat content in ice-cream made by a firm. A sample is taken from each batch, halved and the separate halves sent to the two laboratories. The fat content obtained by the aboratories is recorded below.

Is there a significant difference between the means of fat content obtained by the two laboratories A and B?
Answer:
H₀: μ₁ = μ₂ means are same
H₀: μ₁ ≠ μ₂ means are not same



It is a two tailed test
α = 0.05=5% d.f = n₁+ n₂ – 2 ⇒ 10 + 10 – 2 = 18
Critical value = -2.10 and 2.10.
-2.10 < -0.386 <+2.10
t_cal = -0.386 this in accepted region
H₀ is accepted.
Conclusion: means are same.

Question 16.
Twelve students were given intensive coaching and 2 tests were conducted before and after coaching. The score of 2 tests are given below. Do the scores before and after coaching show an improvement?

Answer:
H₀: Coaching is not useful, μ₁ = μ₂
H₁: Coaching is better, μ1₁ < μ₂



The test statistics is

It is left tailed test
Critical value =- k = -1.80
LOS = α = 0.05 = 5%
d.f = n – 1 = 12 – 1 = 11
t_cal = – 4.891 < critical value,
-1.80 <-4.891
H₀ is rejected
Conclusion: Coaching is not useful.

Question 17.
A certain stimulus administered to each of the 12 patients resulted in the following changes in blood pressure
5 2 8 -1 3 0 -2 1 5 0 4 6
Can it be concluded that the stimulus will be in general accompanied by change in blood pressure
Answer:
H₀: stimulus accompanied doesn’t increase BP, μ₁ = μ₂
H₁ stimulus accompanied can increase BP, μ₁ < μ₂
d = 5 2 8 -1 3 0 -2 1 5 0 4 6
n = 12


\(=\sqrt{15.417-(2.583)^{2}}\)
\(=\sqrt{15.417-6.673}\)
\(=\sqrt{8.744}=2.9570\)


It is left tailed test
LOS = α = 0.05 = 5%
d.f. = n – 1 = 12 – 1 = 11
Critical value = k = 1.80
t_cal values is greater than critical value,
2.896 > 1.80
H₀ is rejected
Conclusion: stimulus accompanied can increase BP.

Question 18.
Nine patients, to whom a certain drug was administrated, registered the following increments in blood pressure.
7 3 -1 4 -3 5 6 -4 1
Show that the data do not indicate that the drug was responsible for these increments.
Answer:
n = 9.



H₀ : The drug is not responsible for increments in B.P.
H₁ : The drug is responsible for increments in B.P.

It is two tailed test
LOS = α = 0.05 = 5%, d.f = (n – 1) = 9 – 1 = 8
K critical values = 2.31, -2.31
T_cal lies in acceptance region
H₀ is accepted
Conclusion: The drug is responsible for increrment in b.p.

2nd PUC Statistics Chi – Square (χ²) Tests Exercise Problems

Question 1.
Mention 2 applications of χ² test.
Answer:

1. To test if the population has a given variance
2. To test goodness of fit in a given data of a theoretical distribution.

Question 2.
Write χ² test statistic with degree of freedom in testing of population variance?
Answer:

Question 3.
Write χ² – test statistic with degrees of freedom in testing of goodness of fit
Answer:
χ² test statistic,

Question 4.
Mention two conditions for applicability of χ² test of goodness of fit
Answer:
Condition:

1. The total frequency ‘N’ should be large
2. The theoretical frequencies E₁ ≥ 5. If any E₁ < 5, it should be pooled with the adjacent frequency.

Question 5.
When is the pooling of the frequencies done in testing of goodness of fit?
Answer:
The theoretical frequencies E₁ ≥ 5 . If any E₁ < 5, it should be pooled with the adjacent frequency.

Question 6.
In a χ² test for goodness of fit, if there are 6 classes and one parameter is estimated, then find the value of degrees of freedom of test statistic?
Answer:
D.f. in testing of goodness of fit is n – c; then d.f 6 – 1 = 5

Question 7.
Write the formula of χ² – test with degrees of freedom in testing of independence of attributes in 2 × 2 contingency table.
Answer:

Question 8.
What is the value of d.f. is 2 × 2 contingency table?
Answer:
The value is 1df.

Question 9.
What are the conditions applicable while testing independent of attribute in 2 × 2 contingency table?
Answer:
Conditions.

1. The total frequency ‘N’ should be large (> 50)
2. The observations should be independent.

Question 10.
For the χ²-test, what is the condition for expected cell frequency?
Answer:
Each of the expected cell frequencies E₁ should be 5 are more.

Question 11.
A random sample of size 25 taken from a population gives the sample standard deviation as 8.5. Test the hypothesis that the population standard deviation (a) is 10.
Answer:
n = 25, s = 8.5, σ = 10 i.e., σ² = 10² = 100
H₀: σ² = 10² Population SD is 10
H₁ : σ² ≠ 10² Population SD is not 10

It is a two-tailed test
d.f. = (n – 1) = 25 – 1 = 24
α = 0.05

Critical values K₁ = 12.4, k₂ = 39.36
χ²_cal value lies with in the critical values
H₀ is accepted
Conclusion: Population SD is 10

Question 12.
Weights in kg. of 10 students are given below:
38 40 45 53 47 43 55 48 52 49
Can we say that variance of the distribution of weights is equal to 20 kg² ?
Answer:

n = 10 σ² = 20
x̄ = \(\frac{\Sigma x}{n}\) = \(\frac{38+40+45+53+47+43+55+48+52+49}{10}\)
x̄ = \(\frac{470}{10}\) = 47
The test statistic is
χ² = \(\frac{\sum(x-\bar{x})^{2}}{\sigma^{2}}=\frac{280}{20}=14\)

H₀ : Variance of the distribution of weights is 20kg σ² = 20
H₁ : Variance of the distribution of weights is not 20kg σ² ≠ 20
It is a two-tailed test
Degree of freedom = 10 – 1 = 9
Level of significance = 0.05
Critical values are k₁ = 2.70 and k₂=19.02
k₁ < 14 < k₂
H₀ is accepted
χ²_cal value lies in acceptance region
Conclusion: variance of the distribution of weights of students is equal to 20 kgs²

Question 13.
In 120 throws of a single die, the following distribution of faces were obtained:

Test at 5% level of significance that the die in unbiased
Answer:
H₀ : The die is unbiased
H₁ : The die is biased
Under H₀ all the sides of die are equally probable.
Then frequencies should be equal and so, E₁ = \(\frac{N}{n}=\frac{120}{6}\) = 20


The test = χ² = \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 12.9
THe degree of freedom is (n – c) = 6 – 1 = 5
α = 0.05
The critical value 11.07
χ²_cal > c.v
H₀ is rejected
Conclusion: The die is biased

Question 14.
Fit a binomial distribution to the following data and test for goodness of fit?

Answer:
H₀ : Binomial distribution is a good fit.
H₁ : Binomial distribution is not a good fit.

x̄ = np = \(\frac{\Sigma f x}{n}=\frac{439}{100}\) = 4.39
np = 4.39, 9p = 4.39 p = \(\frac{4.39}{9}\) = 0.48
q = 1 – p = 1 – 0.48 = 0.52
The p.m.f is p(x) = ⁹C_x × (0.48)^xim × (0.52)^{9 – x} x = 0,1, 2 9
The frequency function is
p(0) = ⁹C₀ × (0.48)⁰ – (0.52)^{9 – 0} = 1 × 1 × 0.002779
E(0) = N. p(0) = 100 × 0.002779 = 0.2779
By using recurrence relation for expected frequencies
E(x) = \(\frac{n-x+1}{x \cdot q} E(x-1)\)
E(1) = \(\frac{9-1+1}{1} \times \frac{0.48}{0.52} \times 0.2779\) = 2.62 = 9 × 0.9231 × 0.2779 = 2.31
E(2) = \(\frac{9-1+1}{1} \times \frac{0.48}{0.52} \times 0.2779\) = 4 × 0.9231 × 2.31 = 8.529
E(3) = \(\frac{9-3+1}{3} \times \frac{0.48}{0.52} \times 8.52\) = 2.33 × 0.9231 × 8.529 = 18.344
E(4) = \(\frac{9-4+1}{4} \times \frac{0.48}{0.52} \times 18.344\) = 18.344 = 1.5 × 0.923 l × 18.344 = 25.40
E(5) = \(\frac{9-5+1}{5} \times \frac{0.48}{0.52} \times 25.40\) = 1 × 0.9231 × 25.40 = 23.45
E(6) = \(\frac{9-6+1}{6} \times \frac{0.48}{0.52} \times 23.45\) = 0.67 × 0.9231 × 23.45 = 14.5
E(7) = \(\frac{9-7+1}{7} \times \frac{0.48}{0.52} \times 14.5\) = 0.43 × 0.9231 × 14.5 = 5.76
E(8) = \(\frac{9-8+1}{8} \times \frac{0.48}{0.52} \times 5.76\) = 0.25 × 0.9231 × 5.76 = 1.33
E(9) = \(\frac{9-9+1}{9} \times \frac{0.48}{0.52} \times 1.33\) = 0.11 × 0.9231 × 1.33 = 0.14
Applying χ² test
The theoretical frequencies are

Test statistic under H₀ is
χ²_cal \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 47.47
At 5% LOS (α = 0.05) for (n – c) = 6 – 2 = 4 df
The critical value is k₂ = 9.49
since; χ²_cal > k₂
∴ H₀ is rejected
Conclusion: BD is not a good fit

Question 15.
Following is the data regarding number of mistakes per page found in a book. Fit a Poison distribution test at 5%. L.O.S. that it is good fit.

Answer:

x̄ = λ = \(\frac{\Sigma f x}{N}=\frac{120}{100}\) = 1.2, pmf,
p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) , x = 0,1,2,3,4
put x = 0 p(0) = \(\frac{\mathrm{e}^{-1.2}(1.2)^{0}}{0 !}\) = 0.3102
F(0) = NP(x) = 100 × \(\frac{\mathrm{e}^{-1.2} \times(1.2)^{0}}{0 !}\) =100 × 0.3012 = 30.12
Calculation of theoretical frequency using recurrence relation
T(x) = \(\frac{\lambda}{x} \cdot E(x-1)\)
T(1) = \(\frac{1.2}{1} \times 30.12\) = 36.144
T(2) = \(\frac{1.2}{2} \times 36.14\) = 21.68
T(3) = \(\frac{1.2}{3} \times 21.68\) = 8.67
T(4) = \(\) = 2.6


H₀ : Poisson distribution is good fit.
H₁ : Poisson distribution is not a good fit.
It is upper tail test χ²_cal = \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 0.5210
d.F = n – c
=4 – 2 = 2
LOS = α = 0.5
Critical value is 5.99
χ²_cal < C.V
H₀ is accepted
Conclusion: The P.D. is a good fit
Note: e^-1.2 = e^-1 × e^-0.2 = 0.3679 × 0.8187 = 0.30119

Question 16.
The following data shows the suicides of 1096 women in 8 Punjab cities during 14 years

Fit a Poisson distribution to the data and show that the distribution is not good fit.
Answer:
H₀ : Poisson distribution is a good fit
H₁ Poisson distribution is not a good fit

x = λ = \(\frac{\Sigma f x}{N}=\frac{1295}{1096}\) = 1.18
P(X) = \(\frac{\mathrm{e}^{-1.18} \cdot(1.18)^{\mathrm{x}}}{\mathrm{x} !}(\mathrm{x}=0,1, \ldots 7)\)
E(0) = N . P(0) = 1096 . \(\frac{\mathrm{e}^{-1.18} \cdot(1.18)^{0}}{0 !}\) = 1096 × 0.3072 = 336.69
The expected frequencies can be calculated by using recurrence relation E(x) = \(\left(\frac{\lambda}{x} E(x-1)\right)\)
E(l) = \(\frac{1.18}{1} \times 336.69\) = 397.296 118
E(2) = \(\frac{1.18}{2} \times 397.29\) = 234.405
E(3) = \(\frac{1.18}{3} \times 234.405\) = 92.19 118
E(4) = \(\frac{1.18}{4} \times 92.19\) = 27.196 118
E(5) = \(\frac{1.18}{5} \times 27.196 \) = 6.418 118
E(6) = \(\frac{1.18}{6} \times 6.418\) = 1.2627
E(7) = \(\frac{1.18}{7} \times 1.2627\) = 0.213

Test statistic under H₀ is

χ² = \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 13.935
At 5% LOS (α = 0.05)
(n – c) = 6 – 2 = 4 df
Critical value = 9.49
since, χ²_cal value does not lies in AR.
χ² > K₂,
H₀ rejected
Conclusion: PD is not a good fit.
Note:
e^-1.18 = e^-1 × e^-0.1 × e^-0.08 = 0.3679 × 0.9048 × 0.9231 = 0.3072

Question 17.
Consider the following data.

Fit a poisson distribution to the data and test the goodness of fit (use 5% Los.)
Answer:
H₀ : Poisson distribution is a good fit
H₁ : Poisson distribution is not a good fit

Mean x̄ = λ = \(\frac{\Sigma f x}{n} \Rightarrow \frac{99}{100}\) = 0.09
The P.M.F
p(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) ; n = 0,1,2…..
E(0) = N.P(0)
E(0) = p(x = 0) = 100 × \(\frac{\mathrm{e}^{-0.99} \times(0.99)^{0}}{0 !}\) ⇒(0.3176)100 = 37.16
Using recurrence relation E(x) = \(\frac{\lambda}{x} E(x-1)\)



χ²_cal = 11.09
α = 5% LOS n – 2 = 4 – 2 = 2
Critical value k₂ = 5.99
Here χ²_cal does not lie in AR.
H₀ is rejected
Conclusion: Poission distribution is not a good fit.
Note: e^-0.09 = e^-0.9 × e⁰⁰⁹
= 0.4066 × 0.9139 = 0.37159.

Question 18.
Among 64 off spring of a certain cross breeds of guinea pigs, 34 were red, 10 were black and 20 were white. According to genetic model this numbers should be in the ratio of 9 : 3 : 4. Are the data consistent with model at the 5% level?
Answer:
H₀ : The data is consistent :
H₁ : The data is not consistent
Given
The theoretical frequency ratio is 9 : 3 : 4
E₁ = 64 × \(\frac{9}{16}\)= 36
E₂ = 64 × \(\frac{3}{16}\) = 12
E₃ = 64 × \(\frac{4}{16}\) = 16


The test χ²_cal = \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 1.44
df = n – C = 3 – 1 = 2
Critical value = 5.99, α = 5%
χ²_cal < C.V
H₀ is accepted
Conclusion: The data is consistent

Question 19.
The following information was obtained in a sample of 50 small general shops.

Can it be said that there are relatively more women owners of small general shops in villages than in towns?
Answer:
H₀ : Women owner and small general shops in villages are independent.
H₁ : Women owner and small general shops in village are not independent


It is upper tailed test
α = 5% = 0.05 For 1 d.f.
The critical value k₂ = 3.84
χ²_cal < C.V
H₀ is accepted
Conclusion: Women owners of small general shops are more in villages than in towns.

Question 20.
The opinion poll was conducted to find the reaction to a proposed civic reform in 100 members of each of the two political parties.

Test whether political parties and the reaction to a proposed civic reform are independent
Answer:
H₀ : political parties and the reaction to a proposed civic reform are independent.
H₁ : political parties and the reaction to a proposed civic reform are not independent.


It is upper tailed test
LOS = 0.05 with 1 df
Critical value 3.84
χ²_cal< C.V
H₀ is accepted
Conclusion: Political parties and the reaction to a proposed civic reform are independent.

Question 21.
From the following data, test whether education and employment are independent at 5% LOS

Answer:
H₀ : Education and employment are independent
H₁ : Education and employment are dependent



α = 0.05 for 1 df
k₂ = 3.84
χ²_cal value lies in acceptance region
χ²_cal < K₂
H is not accepted.
Conclusion: Education and employment are independent.

Question 22.
Of the 500 workers in a factory exposed to an epidemic 350 in all were attacked, 200 had been inoculated and of these 100 were attacked. Test whether inoculation and attack are independent.
Answer:
H₀ : Inoculation and attack are independent.
H₁ : Inoculation and attack are not independent.




It is upper tailed test
α = 0.05 for 1 d.f.
C.V. is k₂ = 3.84
χ²_cal > c.v
H₀ is rejected
Inoculation and attack are not independent.

2nd PUC Statistics Large Sample Tests Pratical Assignments

Question 1.
A machine is designed to fill 1 litre milk to polythene packets. A randomly selected 100 milk packets filled by this machine are inspected. The mean milk is found to be 998 ml and S.D. is 10 ml. Is the machine functioning properly at 1% level of significance?
Answer:
Given: n = 100
x̄ = 998ml, s = 10ml, µ = 1 Litre = 100ml
H₀ : Machine functioning properly (ie., machine fills 1 Ltr. Milk) i.e., µ = l Litre / 1000 ml.
H₁ : Machine is not functioning properly [i.e., machine does’t fills 1 litre milk) i.e., µ ≠ 1 Litre / 100 ml (Two tail
test) under H₀ the test statistic is : z = \(\frac{\bar{x}-\mu}{s \sqrt{n}}\) ~ N(0,1)
z_cal = \(\frac{998-1000}{10 \sqrt{100}}=\frac{-2}{10 / 10}=-2\)
At α = 1% the two critical values are ± k = ± 2.58
Here Z_cal lies acceptance region (A.R)
i.e., k ≤ Z_cal ≤ k
∴ H₀ is accepted.

Conclusion: Machine is functioning properly; µ = 1 Ltr.

Question 2.
A random sample of 225 cans containing baby food has mean weight 998 gm and S.D. 15gm. Test whether the mean contents of the cans be considered as 2kg. (Use 5% L.O.S.)
Answer:
Given: n = 225, x̄ = 998, s = 15, µ = lkg = 1000 grams
H₀: Mean content of the can is 1 kg i.e., µ = 1000 grams
H₁: Mean content of the can differs from 1 kg i.e., µ ≠ 1000 grams. (Two tail test) under H₀, the test statistic is:

At α = 5% the two critical values are ± k = ± 1.96
Here Z_cal lies in rejection region (R.R)
∴ H₀ is rejected.

Conclusion: Mean content of the can differs from 1 kg. i.e., µ ≠ 1000 grams.

Question 3.
On 60 different days the numbers of passengers in a bus were noted. The mean and S.D. of the number of passengers was found to be 40 and 5 respectively. At 5% level of significance, test the hypothesis that the average number of passengers in the bus is more than 38.
Answer:
Given: n = 60, x̄ = 40, s = 5, α = 5% µ = 38
H₀ : The average number of passengers is 38. i.e., µ = 38
H₁ : The average number of passengers is more than 38. i.e., µ > 38(upper tail test)
Under H₀, the test statistic is:

At α = 5% the upper tail critical value k = 1.65
Here Z_cal lies in rejection region
∴ H₀ is rejected.

Conclusion: The average number of passengers is more than 38.

Question 4.
A company manufactures car tyres. Their average life is 40,000 kilometers and standard deviation 5,000 kilometers. A change in the production process is believed to result in a better product. A test sample of 100 new tyres has mean life of 41,000 kilometers. Can you conclude at 1% L.O.S. that the new product gives better result?
Answer:
Given: µ = 40,000, σ = 5,000, n = 100,
x̄ = 41,000, α = 1%
H₀ : Mean life of tyres is 40,000 km. i..e, µ = 40,000 kms
H₁: Mean life of tyres is more than 40,000 kms.
(i.e., New product of tyres gives better result] i.e., µ > 40,000 kms
Under H the test statistic is z = \(\frac{\bar{x}-\mu}{\sigma \sqrt{n}}\) ~ N(0,1)
z_cal = \(\frac{41,000-40,000}{5000 / \sqrt{100}}=2\)
At α = 1% the upper tail critical value k = 2.33
Here Z_cal lies in acceptance region i.e., Z_cal < k
∴ H₀ is rejected.

Conclusion: Mean life of tyres is 40,000 Kms
i.e., µ = 40,000 Kms.

Question 5.
A certain brand of soap is said to weigh 125 gm on the average. A consumer agency received complaints that the weight is much less. To test the claim the agency selects 100 pieces and finds mean is 123 gm with S.D. 5 gm. complete the test at 5% level of significance?
Answer:
Given: µ = 125gm, n = 100, x̄ = 123 gm, s = 5gm
H₀ : Average weight is 125 gmi..e, µ = 125 kms
H₁ : Average weight is less than 125 gm i.e., µ < 125 gm (Lower tail test)
under H₀ the test statistic is

Z_cal = \(\frac{123-125}{5 \sqrt{100}}\) = -4
At α = 5% the lower tail critical value – k = -1.65
Here Z_cal lies in rejection region (R. R)
i.e., Z_cal < -k
∴ H₀ is rejected.

Conclusion: Average weight is less than 125 gms
i.e., µ < 125 gms.

Question 6.
The mean of weight of 50 boys of a college is 58 kg and that of 40 girls of the same college is 54 kg. The variances of weights of boys and girls are 64 and 49 respectively. Can we conclude that boys and girls studying in the college have same weight? Use 5% level of significance.
Answer:
Given: n₁ = 50, x̄₁ = 58, n₂ = 40; x̄₂ = 54, s²₁ = 64, s²₂ = 49
H₀ : Boys and girls have same weight i.e., µ₁ = µ₂
H₁ Boys and girls do not have same weight i.e., µ₁≠ µ₂ (two tail test)

At α = 5% the lower tail critical values are ± k = ± 1.96
Here Z_cal lies in rejection region (R. R)
∴ H₀ is rejected.

Conclusion: Boys and girls do not have same weight i.e., µ₁≠ µ₂

Question 7.
400 women shoppers are chosen at random in market A. Their average weekly expenditure on food to be ?500 with S.D. of ?40. The figures are ?492 and ?50 respectively, in the market B, where 500 women shoppers are chosen at random from this place. Test at 5% level of significance whether the average weekly food expenditure of population of shoppers are equal.
Answer:
Given: n₁ = 400, x̄₁ = 500, s₁ = 40,
x̄₂ = 492, s₂ = 50, n₂ = 500, α = 5%
H₀ : Average weekly expenditure of women shopers in market A and B are same i.e., µ₁= µ₂
H₁ : Average weekly expenditure of women shopers in market A and B are not same i.e., µ₁≠ µ₂ (two tail test)

At α = 5% the two tail critical values are ± k = ± 1.96
Here Z_cal lies in rejection region (R. R)
∴ H₀ is rejected.

Conclusion: Average weekly expenditure of women shoppers in market A and B are not same µ₁≠ µ₂

Question 8.
For the following data, test whether means differ significantly, (Use α = 0.01)

Answer:
Given: n₁ = 40, x̄₁ = 70, s₁ = 8, n₂ = 60, x̄₂ = 66, s₂ = 6
H₀: Means does’t differ significantly i.e., µ₁ = µ₂
H₁ : Means differ significantly i.e., µ₁≠ µ₂ (two tail test)

At α = 0.01 the two tail critical values are ±k = ±2.58
Here Z_cal lies in rejection region (R. R)
∴ H₀ is rejected.

Conclusion: Means differ significantly.

Question 9.
Test at 1% level of significance, that average life of bulbs manufactured by Firm – A is less than Firm – B.

Answer:
Given:
Let n₁ =32, x̄₁ =1300, s²₁ = 64,
n₂ = 50, x̄₂ = 1305, s²₁ = 100
H₀ : Average life of bulbs of firm A and B is same, (i.e., µ₁= µ₂)
H₁ : Average life of bulbs of firm A is less than Firm B i.e., µ₁ < µ₂ (Lower tail test)

At α = 1% the lower tail critical value – k = -2.33
Here Z_cal lies in rejection region (R. R)
i.e., Z_cal < – k
∴ H₀ is rejected.

Conclusion: Average life of bulbs of firm A is less than firm B. i.e., µ₁ < µ₂.

Question 10.
Following is data regarding mean weights of randomly selected boys and girls of P.U.C. students. Test whether, mean weight of boys are greater than mean weight of girls. (Use α = 0.05.)

Answer:
Given: n₁ = 64, x̄₁ = 63, s₁ = 8, x̄₂ = 60, s₂ = 12
H₀ : Mean weight of Boys and girls is same, (i.e., µ₁= µ₂)
H₁ : Mean weight of Boys is greater than girls i.e., µ₁> µ₂ (upper tail test)

At α = 0.05 the upper tail critical value +k = +1.65
Here Z_cal lies in acceptance region (A. R)
i.e., Z_cal < k
∴ H₀ is accepted

Conclusion: Mean weight of Boys and girls is same i.e., 0 µ₁> µ₂

Question 11.
Intelligence test given to two groups of boys and girls gave the following information:

Is the difference in the mean score of boys and girls statistically significant Use 1% L.O.S.
Answer:
Given: n₁ – 100, x̄₁ = 74, s₁ = 12,
n₂ = 50, x̄₂ = 70, s₂ = 10
H₀ : The mean score of boys and girls does’t differs significantly i.e., µ₁= µ₂
H₁ : The mean score of boys and girls differ significantly i.e., µ₁ ¹µ₂ (two tail test)

At α = 1% the tail critical values are ± k = ± 2.58
Here Z_cal lies in acceptance region (A. R)
∴ H₀ is accepted

Conclusion: Mean scores of boys and girls does’t differ significantly i.e., µ₁= µ₂

Question 12.
A coin is tossed 400 times and head turns up 220 times. Can we conclude that the coin is unbiased ? (Use α = 0.05)
Answer:
Given: n = 400, x = 200, P₀ = 0.5 (unbiased)
H₀ : The coin is unbiased i.e., P₀ = 0.5
H₁ : The coin is biased i.e., P₀ ≠ 0.5 (Two tailed test)
Under H0, the test statistic is : z = \(\frac{p-p_{0}}{\sqrt{\frac{P_{0} Q_{0}}{n}}}\)
Here sample proportion :
p = \(\frac{x}{n}=\frac{220}{400}\) = 0.55 and Q₀ = 1 – P₀ = 1 – 0.5
∴ z_cal = \(\frac{0.55-0.5}{\sqrt{\frac{0.5 \times 0.5}{400}}}=\frac{0.05}{0.025}=2\)
At α = 0.05 the two tail critical values are ±k = ±1.96
Here Z_cal lies in rejection region (R. R) i.e., Z_cal > k
∴ H₀ is rejected

Conclusion: The coin is biased i.e., p₀ ≠ 0.5

Question 13.
In a city, out of 900 men 486 were smokers. Does this information indicate that the majority of men in the city are smokers?
Answer:
Given: n = 900, x = 486, P₀ = 0.5 (majority)
H₀ : Men smokers in the city are50% i.e., P₀ = 0.5
H₁ : Majority of men in the city are smoker i.e., P₀ > 0.5 (upper tailed test)
Under H ₀, the test statistic is : z = \(\frac{p-P_{0}}{\sqrt{\frac{P_{0} Q_{0}}{n}}}\) ~ N(0,1);
Here sample proportion :
P₀ = \(\frac{x}{n}=\frac{486}{900}\) = 0.54 and Q₀ = 1 – P ₀ = 1 – 0.5 = 0.5
∴ Z_cal = \(\frac{0.54-0.5}{\sqrt{\frac{0.5 \times 0.5}{900}}}=\frac{0.04}{0.0167}=2.3952\)
At α = 1% the upper tail critical value k = 2.33
Here Z_cal lies in rejection region (R R)
i.e., Z_cal > k
∴ H₀ is rejected

Conclusion: Majority of men in the city are smokers i.e., p₀ > 0.5.

Question 14.
A manufacturer claims that less than 2% of his products are defective. A retailer buys a batch of 400 articles from the manufacturer and finds that 12 are defectives. Test at 1% level of significance that, whether the manufacturer claim is justifiable.
Answer:
Given p₀ = 2% = \(\frac{2}{100}\) = 0.02, n = 1000; x = 10
∴ p = \(\frac{x}{n}=\frac{10}{100}=0.01\) and
Q₀ = 1 – P₀ = 1 – 0.02 = 0.98
H₀ : 2% of the products are defectives i.e., P₀ = 0.02
H₁ : Less than 2% of the products are defectives i.e., P₀ < 0.02 (Lower tail test)

At α = 5% the lower tail critical value = k = – 1.65
Here Z_cal lies in rejection region (R R)
i.e., Z_cal < – k
∴ H₀ is rejected

Conclusion: Less than 2% of the products are defectives i.e., P₀ < 0.02.

Question 15.
A stock broker claims that he can predict with 75% accuracy whether a stock’s market value will rise or fall during the coming month. In a sample of 50 predictions he is correct in 35. Does this evidence supports broker’s claim at 5% level of significance.
Answer:
Given: P₀ = 75% = 0.75.
∴ Q₀ = 1 – p₀ = 1 – 0.75 = 0.25 n = 50, x = 35,
∴ Sample proportion p = \(\frac{x}{n}=\frac{35}{50}\) = 0.7
H₀ : Stock broker can predict 75% accurately i.e., P₀ = 0.75
H₁ : Stock broker cannot predict 75% accurately i.e., P₀ ≠ 0.75 (two tail test)

At α = 5% the two tail critical value ±k = ± 1.96
Here Z_cal lies in acceptance region (A. R) – k ≤ k
∴ H₀ is accepted

Conclusion: Stock broker can predict 75% accurately i.e., P₀ = 0.75.

Question 16.
The manufacturer of surgical instruments claims that less than 1% of the instruments he supplied to a certain hospital are faulty. A sample of 300 instruments revealed that 6 were faulty. Test his claim at 1% level of significance.
Answer:
Given: P₀ = 1% = 0.01 and
Q₀ = 1 – P₀ = 1 – 0.01 = 0.99; n = 300, x = 6,
P = \(\frac{x}{n}=\frac{6}{300}\) = 0.02
H₀ : 1% of instruments are faulty i.e., P₀ = 0.01
H₁ : Less than 1% instruments are faulty i.e., P₀ < 0.01 (Lower tail test)

At α = 1% lower tail critical values – k = – 2.33
Here Z_cal lies in acceptance region (A. R)
i.e., Z_cal > k
∴ H₀ is accepted

Conclusion: 1% of the instruments are faulty i.e., p₀ = 0.01

Question 17.
Among randomly selected 100 students of college A, 66 were passed. Among randomly selected 200 students of college B, 114 were passed. Test whether passing proportion is same in both the colleges. Use 5% L.O.S.
Answer:
Given: n₁ =100, x₁ =66, ∴ p₁ = \(\frac{x_{1}}{n_{1}}=\frac{66}{100}\) = 0.66,
n₂ = 200, x₂ = 114 ∴ p₂ = \(\frac{x_{2}}{n_{2}}=\frac{114}{200}\) = 0.57
H₀ : Passing proportion is same in both colleges i.e., P₁ = P₂
H₁ : Passing proportion is not same in both colleges – i.e., P₁ = P₂ (Two tail test)

At α = 5% the two tail critical values are ±k = ± 1.96
Here Z_cal lies in acceptance region (A. R)
i.e., H₀ is accepted

Conclusion: passing proportion is same both colleges, i.e., P₁ = P₂.

Question 18.
Among 80 randomly selected persons from district A, 36 are interested in viewing hockey match. Among 40 randomly selected persons from district B, 12 are interested in viewing hockey match. Test at 5% level of significance that, the proportion of viewers in district A is more than district B.
Answer:
Given: n₁ = 80, x₁= 36, n₂ = 40, x₂ = 12
The sample proportions:
p₁ = \(\frac{x_{1}}{n_{1}}=\frac{36}{80}\) = 0.45
p₂ = \(\frac{x_{2}}{n_{2}}=\frac{12}{40}\) = 0.3
H₀ : Proportion of hockey viewers in district A and B is same i.e., P₁ = P₂
H₁ : Proportion of Hockey viewers in district A is more than district B i.e., P₁ > P₂ (upper tail test)

At α = 5% the upper tail critical values are + k = + 1.65
Here Z_cal lies in acceptance region (A. R)
i.e., -k ≤ Z_cal ≤ k
∴ H₀ is accepted

Conclusion: The proportion of Hockey viewers in district A and B is same (P₁ = P₂)

Question 19.
In a random sample of 200 people from a city in the year 2011 revealed that 150 were cricket match viewers. In another random sample of 100 people from same city in the year 2013 revealed that 90 were cricket match viewers. Examine whether there is a significant increase in cricket match viewers. Use 5% level of significance.
Answer:
Given: n₁ = 200, x₁ = 150, ∴ P₁ = \(\frac{x_{1}}{n_{1}}=\frac{150}{200}\) = 0.75
n₂ = 100, x₂ = 90, ∴ P₂ = \(\frac{\mathrm{x}_{2}}{\mathrm{n}_{2}}=\frac{90}{100}\) = 0.9
H₀ : Proportion of cricket viewers in 2011 and 2013 is same i.e., P₁ = P₂
H₁ : Proportion of cricket viewers in 2011 is less than 2013 P₁ < P₂
(i.e., increases means more in 2013 than 2011)
under H₀ the test stastistic is :

At α = 5% the lower tail critical values – k = -1.65
Here Z_cal lies in rejection region (R. R)
∴ H₀ is rejected

Conclusion: proportion of cricket viewers in 2011 is less than 2013 i.e., ₁ < P₂

Question 20.
From the following data, test whether the difference between the proportions in the samples is insignificant. Use 5% level of significant.

Answer:
Given: n₁ = 200, p₁ = 0.12, n₂ = 100, p₂ = 0.09, α = 5%
H₀ : There is no significant difference in the population proportions i.e., P₁ = P₂
H₁ : There is a significant difference in the population proportions i.e., P₁ ≠ P₂ (Two tailed test)
Under H₀, the test statistic is :

At α = 5% the two tail critical values are ± k = ± 1.96
Here Z_cal lies in Accepted Region (A. R)
i.e., – k < z_cal < k.
∴ H₀ is accepted

Conclusion: There is no significant difference in the population proportions.

Question 21.
A mechanist is making engine parts with axle diameters 0.700″ a random sample of 10 parts shows a means diameter of 0.742″ with a S.D of 0.04″. Test whether work is meeting the specification?
Answer:
Given: µ₀ =0.700, n = 10, x̄ = 0.742, s = 0.04
H ₀ : Mean diameter of axle is 0.7” i.e., µ = 0.700”
H₁ : Mean diameter of axle differs from 0.7” i.e., µ ≠ 0.700” (Two tail test)
Under H₀, the test statistic is : t = \(\frac{\bar{x}-\mu}{s \sqrt{n-1}}\) ~ t(n – l)d.f. ;
t_cal = \(\frac{0.742-0.700}{0.04 / \sqrt{10-1}}=\frac{0.042 \times 3}{0.04}=3.15\)
At α = 5% for (n – 1) d.f = 10 – 1 = 9 d.f.
the two tail critical values are ± k = ± 2.26.
Here t_cal lies in rejection region (R.R)
i.e., t_cal > K
∴ H₀ is rejected

Conclusion: Mean diameter of axle differs from 0.700″ i.e., µ ≠ 0.700.

Question 22.
A soap manufacturing company was distributing a particular brand of shop through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozens. After the campaign a sample of 26 shops was taken and mean sales was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective?
Answer:
Given: µ₀ = 140, n = 26, x̄ = 147, s = 16
H₀ : Advertisement is not effective i.e., µ = 140
H₁ : Advertisement is effective i.e., µ > 140, (upper tail test) (i.e., sales increased)
Under H₀, the t-test statistic is : t = \(\frac{\bar{x}-\mu}{s \sqrt{n-1}}\) ~ t(n – l)d.f.
t_cal = \(\frac{147-140}{16 / \sqrt{26-1}}=\frac{7 \times 5}{16}=2.1875\)
At α = 5% for (n – 1) = 26 – 1 = 25d.f
the upper tail critical value k = 1.71
Here t_cal > k,
i.e., t_cal lies in rejection region.
∴ H₀ is rejected

Conclusion: Advertisement is effective i.e., sales increased; µ >140.

Question 23.
A survey in a locality revealed that on an average 16% of persons bought a particular newspaper. But the newspaper agent felt that it is more. He conducted a survey in 10 locality and it was found that on an average 185 bought the newspaper with S.D 6. Conduct the t – test at α = 0.01.
Answer:
Given: µ = 180, n = 10, x̄ = 185, s = 6. α = 0.01
H₀ : Average is 180 i.e., µ = 180
H₁ : Average is more than 180 i.e., µ > 180 (upper tail test)
Under H₀, the t-test statistic is : t = \(\frac{\bar{x}-\mu}{s \sqrt{n-1}}\) ~ t(n – l)d.f.;
t_cal = \(\frac{185-180}{6 \sqrt{10-1}}=\frac{5 \times 3}{6}=2.5\)
At α = 0.01 the upper tail critical value for
(n – 1) = (10 – 1) = 9.d.f k = 2.82
Here t_cal lies in acceptance region (A.R)
∴ H₀ is accepted.

Conclusion: Average is 180 i.e., µ = 180

Question 24.
The weights of 5 students are 50, 48, 46, 49 and 50kgs. Test whether the average weight of the students is 50 kgs?
Answer:
Given: n = 5, µ₀ = 50
H₀: Average weight is 50 kg i.e., µ = 50
H₁ : Average weight differs from 50. i.e., µ ≠ 50 (Two tail test)
Under H₀, the t-test statistic is :

At α = 5% for (n – 1) d.f = 5 – 1 = 4 d.f
the two tall critical values are ±k = 2.78 Hear t_cal lies in AR
∴ H₀ is accepted.

Conclusion: Average weight is 50 kg i.e., µ = 50

Question 25.
Ten students are selected at random from a college and their heights are found to be 138, 140, 144, 150, 160, 162, 164, 166 and 168 Cms. Test at 5% level of significance that the average height of the student of the college is 150 cms.
Answer:
Given: n = 5, µ₀ = 150 cms.
H₀ : The average height of the students is 150 cms i.e., µ = 150
H₁ : The average height of the students differs from 150 cms i.e., µ ≠ 150 (Two tail test)
Under H₀, the t-test statistic is : t = \(\frac{\bar{x}-\mu}{s \sqrt{n-1}}\) ~ t(n – l)d.f.


x̄ = 150 – \(\frac{50}{10}\) = 145 s = \(\sqrt{\frac{1364}{10}-\left(\frac{50}{10}\right)^{2}}\) = 10.55
∴ t_cal = \(\frac{145-150}{8.899 / \sqrt{10-1}}\) = -1.42
At α = 5% for (n – 1) d.f = 10 – 1 = 9 d.f
the two tail critical values are ±k = 2.26
Hear t_cal lies in acceptance region (A.R)
i.e., -k ≤ t_cal ≤ k
∴ H₀ is accepted.

Conclusion: Average height of the students is 150 cms. i.e., µ = 150.

Question 26.
A company states that it sells 5600 articles in a month, but an insepector feels that it is not 5600. He randomly selects 17 months and finds that on an average the sales is 5750 and S.D. = 175. conduct the test at 5% level of significance
Answer:
Given: n = 17, µ₀ = 5600, x̄ = 5750, s = 175, α = 5%
H₀ : The average sales is 5600 i.e., µ = 5600
H₁ : The average sales differs from 5600, i.e., µ ≠ 5600 (Two tailed test)
Under H₀, the t-test statistic is : t = \(\frac{\bar{x}-\mu}{s \sqrt{n-1}}\) ~ t(n – l)d.f.
∴ t_cal = \(\frac{5750-5600}{175 \sqrt{17-1}}=3.4286\)
At α = 5% for (n – 1) d.f = 17 – 1 = 16 d.f
the two tail critical values are ±k = ±2.12
Here t_cal lies in acceptance region (R.R)i.e., t_cal > k
∴ H₀ is accepted.

Conclusion: The average sales differs from 5600.
i.e., µ = 5600.

Question 27.
For the following data examine whether the means of two samples differ significantly

Answer:
Given: n₁ = 12, n₂ = 71, x̄₁ = 572,
x̄₂ =52.3, S₁ = 3.41, s₂ = 3.62
H₀ :Means does’t differ significantly i.e., µ₁ = µ₂
H₁ :Means differs significantly i.e., µ₁ ≠ µ₂ (Two tail test)
Under H₀, the t-test statistic is :

At α = 5% for (n₁ + n₂ – 2) d.f = 12 + 7 – 2 = 17 d.f.
the two tail critical values are ±k = ±2.11
Here t_cal lies in rejection region i.e., t_cal> k
∴ H₀ is rejected.

Conclusion: Means differs significantly i.e. µ₁ ≠ µ₂

Question 28.
The mean and variance of 4 observations are 2.075 and 1.022 respectively and that of 5 observations are 2.86 and 3.106 respectively. Test whether mean of the first set of observation is less than the second.
Answer:
Given: n₁ = 4, x̄₁ = 2.075, s₁²(variance) = 1.022
n₂ =5, x̄₂ = 2.86, s₂² = 3.106
H₀ : Mean of first and second set of observations same i.e., µ₁ = µ₂
H₁ : Mean of first set of observation is less than second are set of observation i.e., µ₁ < µ₂ (Lower tail test)
Under H₀, the t-test statistic is :


At α = 5% for (n₁ + n₂ – 2) d.f= 4 + 5 – 2 = 7 d.f.
The lower tail critical value – k = -1.90
Here t_cal lies in acceptance region (A.R)
∴ H₀ is accepted.

Conclusion: Means of first and second set of observations are same i.e., µ₁ = µ₂.

Question 29.
The average weight 6 randomly selected women is 68kgs and that of 10 randomly selected men is 67.8kgs. Their variance are 12 and 17.066 respectively. Test whether the average weight of women is more than men. Take α = 0.01.
Answer:
Given: n₁ = 6, x̄₁ = 68, n₂ = 10, x̄₂ = 67.8,
s₁² = 12, s₂² = 17.066, α = 0.01
H₀ The average weight of women and men are same i.e., µ₁ = µ₂
H₁ The average weight of women is more than men i.e., µ₁ > µ₂ (upper tail test)
Under H₀, the t-test statistic is :

At α = 0.01 the upper tail critical value for
(n₁ + n₂ – 2) d.f = (6 + (10 – 2) = 14. d.f. k = 2.62
Here t_cal is lies in acceptance region (A.R) i.e., t_cal > k.
∴ H₀ is accepted.

Conclusion: Means of first and second set of observations are same i.e., µ₁ = µ₂

Question 30.
A group of 5 students weight 41, 60, 39, 48, 42 kgs and another group of 7 students weight 42, 56, 64, 38, 68, 62, 69 kgs. Test whether the mean weight of first group is less than second groups.
Answer:
Given n₁ = 5, n₂ = 7, (Two independent groups)
H₀ : Mean weight of first group and second group is same i.e., µ₁ = µ₂
H₁ : Mean weight of first group is less than the second group, i.e., µ₁ < µ₂ (Lower tail test)
Under H₀, the t-test statistic is:

Let x₁ and x₂ be the weight of the groups.



At α = 5% for (n₁ + n₂ – 2) d.f = (5 + 7 – 2) = 10. d.f.
the Lower tail critical value – k = – 1.81
Here t_cal is lies in acceptance region (A.R)
i.e., t_cal > k.
∴ H₀ is accepted.

Conclusion: Mean weight of first and second group is same i.e., µ₁ = µ₂.

Question 31.
The heights of 6 randomly chosen sailors are: 63”, 65″, 58”, 69″, 71” and 72”. The heights of 8 randomly chosen soldiers are : 61″, 62”, 69″, 65”, 70″, 71”, 72″ and 73″ Do these figures show that soldiers are on an average shorter than sailors.
Answer:
Given n₁ = 6, n₂ = 8
H₀ : Soldiers and sailors have same height i.e., µ₁ = µ₂
H₁ : Soldiers are shorter than sailors, i.e., µ₁ < µ₂ (Lower tail test)
Under H₀, the t-test statistic is:

Let x₁ and x₂ be the heights.

At α = 5% for (n₁ + n₂ – 2) d.f = 6 + 8 – 2 = 12. d.f.
the Lower tail critical value – k = -1.78
Here t_cal is lies in acceptance region (A.R)
i.e., t_cal > – k.
∴ H₀ is accepted.

Conclusion: Soldiers and sailors have same height, i.e., µ₁ = µ₂.

Question 32.
An IQ Test was administered to 5 persons before and after training.

Test whether there is any significant difference between average IQ before and after training. Take α = 0.01.
Answer:
Given: n₁ = n₂ = 5
(Before and offer of same students so, use paired t – test)
H₀: Average IQ before and after training is same i.e., µ₁ = µ₂,
H₁ : Average IQ before and after training is not same i.e., µ₁ ≠ µ₂ (Two tailed test) under H₀, the paired t-test statistic is:

Let x and y be the I.Q. before and after training.

At α = 0.01 for (n – 1)d.f = 5 – 1 = 4 d.f.
the two tail critical values are ±k = ±4.60.
Here t_cal lies in acceptance region (A.R) i.e., – k < t_cal < k.
∴ H₀ is accepted.

Conclusion : Average I.Q before and after training is same i.e., µ₁ = µ₂.

Question 33.
Eleven school boys were given a test in geometry. They were given a month’s tuition and second test was held at the end of it. Test whether the tuition was benefited the students. (Use a = 0.05)

Answer:
n = 11, α = 0.05
H₀ : Mean marks before after getting tuition is same i.e., µ₁ = µ₂, (Tuitions not benefited)
H₁ : Mean marks has increased after getting tuition i.e., µ₁ < µ₂ (Tuitions benefited): Lower tail test under H₀, the paired t-test statistic is:

At α = 0.05 for (n – 1) d.f = (11 – 1) = 10d.f.
the two tail critical values -k = -1.81.
Here t_cal lies in rejection region (R.R)
i.e., t_cal < – k.
∴ H₀ is rejected.

Conclusion : mean marks has increased after getting tuition i.e.,µ₁ < µ₂.(Tuition benefited)

2nd PUC Statistics chi – Square Test Pratical Assignments

Question 1.
Weights of 10 students are given below:
Weight (in kgs): 32,48, 50,47,49, 55,46, 51, 50. Can you conclude that standard deviation of the distribution of weights of students is 4kg? Use α = 0.01.
Answer:
Given: n = 10, = 4, α = 0.01
H₀ : S.D. of weight of students is 4kg i.e., σ = 4
H₁ : S.D. of weight of students differ from 4 kg : σ ≠ 4 (Two tail test)
Under H₀ the χ² – test statistic is χ² = \(\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \chi^{2}(\mathrm{n}-1)\)
Let x be the weight then s² – sample variance is calculated as below:


∴ χ²_cal = \(\frac{10 \times 32-84}{42}\) = 20.525
At α = 0.01 for (n – 1) d.f= 10 – 1 = 9 d.f.
the two tail critical values are k₁ = 1.73 and k₂ = 23.59
Here χ²_cal lies in acceptance region (A.R) i.e., k¹ ≤ χ² ≤ k²
∴ H₀ is accepted.

Conclusion: S.D. of weight of students is 45kg i.e., σ = 4

Question 2.
A manufacturer claims that the life time of a brand A batteries produced by the factory has variance less than 4000 hours2. To test this, a sample of 18 batteries of brand A was tested and found the variance of 600 hours2. Test the manufactures claim at α = 0.05.
Answer:
Given σ²(variance) = 4000, n = 18, s² = 600, α = 0.05
H₀ : variance of life of batteries is 4000 Hours², ie., σ² = 4000
H₁ : variance of life of batteries is less than 4000 Hours2 ie., σ² < 400 (Lower tail test) under H₀, the χ² – test statistic is:
χ² = \(\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}\)
χ²_cal = \(\frac{18 \times 600}{4000} = 2.7\)

At α = 0.05, for (n – 1) d.f = 18 – 1 = 17. d.f.
the: lower tail critical value k₁ = 8.67
Here χ²_cal lies in acceptance region (A.R)
ie., χ²_cal < k₁.
∴ H₀ is accepted
Conclusion : variance of life of batteries is σ² = 4000 hours² ie.,

Question 3.
Test the hypothesis that a = 5, given that sample standard deviation is 6 for a random sample of size 25 from a normal population. Use α = 0.05
Answer:
Given: σ = 5, S = 6, n = 25, α = 0.05
H₀ : σ = 5
H₁ : σ ≠ 5 (Two tailed test)

under H₀, the χ² – test ststistic is:
χ² = \(\frac{\mathrm{ns}^{2}}{\sigma} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}\) χ²_cal = \(\frac{25 \times \sigma^{2}}{5^{2}}=36\)
At α = 0.5 for (n-1) d.f = 25 – 1 = 24. d.f
the two tail critical values are k₁ = 12.40 and k₂ = 39.36
Here χ²_cal lies in acceptance region (A.R)
i.e., k₁ ≤ χ²_cal ≤ k₂
∴ H₀ is accepted
Conclusion: σ = 5.

Question 4.
The variance of the height of 20 SSLC students is 4Cms2. Test at 1% level of significant that the variance of height of SSLC students is more than 3cm².
Answer:
Given: n = 20, s² = 4cm², α = 1%, σ² = 3cm²
H₀ : variance of height is 3cm² ie., σ² = 3
H₁ : variance of height is more than 3cm2 ie., σ² > 3 (upper tail test)
Under H₀, the χ² – test statistic is:
χ² = \(\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}\)
χ²_cal = \(\frac{20 \times 4}{3}=26.67\)

At a = 1% the upper tail critical value k₂ for (n – 1)
d.f = (20 – 1) = 19.d.f is k₂ = 36.19
Here χ²_cal lies in a acceptance region (A.R)
ie., χ²_cal < K₂
∴ H₀ is accepted
Conclusion: variance of height is 3cm². ie., σ² = 3.

Question 5.
Binomial distribution is fitted to an observed frequency distribution after estimating ‘p’ from the observed data. The observed and the expected frequencies are given below.

Test whether B.D is a good fit.
Answer:
H₀ : Binomial distribution is good fit
H₁ : Binomial distribution is not good fit

Given that p is estimated from the data.


χ²_cal = 7.16
At α = 5% for (n – 2)d.f = 6 – 2 = 4.d.f
the upper tail critical value k₂ = 9.49
Here χ²_cal lies in acceptance region
ie., χ²_cal < k.
∴ H₀ is accepted
Conclusion: Binominal distribution is good fit.

Question 6.
From the following data, test whether the Poisson distribution is a good fit. The values are tabulated after estimating the parameter use α = 0.01

Answer:
H₀ : poisson distribution is good fit.
H₁ : poisson distribution is not good fit
under H₀, the χ² – test statistic is :

Given that parameter is estimated from the data.


∴ χ²_cal = 1.243
At α = 0.01 for(n – 2) = 4 – 2 = 2 d.f.
the upper tail critical value k₂ = 9.21
Here χ²_cal lies in acceptance region (A.R)
ie., χ²_cal < k₂
∴ H₀ is rejected
Conclusion: poisson distribution is good fit.

Question 7.
Five coins are tossed 250 times and the following distribution is obtained.

fit a binomial distribution to the data and test the goodness of fit at 1% level of significance.
Answer:
Let x be the number of tails and f be the no. of tosses, then from the observed frequency
distribtion : mean = x̄ = np = \(\frac{\sum f x}{N}\) : Here n = 5

5p = \(\frac{648}{250}\); p = \(\frac{2.592}{5}\)
∴ p = 0.52 and q = 1 – p = 1 – 0.52 = 0.48
The p.m.f is: p (x) = ⁿC_xP^x q^{n – x}; x = 0, 1, 2….n
P(x) = ⁵C_x(0.52)^x (0.48)^{5 – x}; x = 0,1, 2….5
Theoretical frequency:
T_x = p(x)N
T₀ = p(x)250
T₀ = 50(0.52)⁰(0.43)^{5 – 0} × 250 = 6.35
Using recurrence relation : T_x = \(\frac{n+1-x}{x} \frac{P}{q} T_{x}-1\)

The observed and theoretical frequency distribution is

Chi – square test:
H₀ : Binomial distribution is good fit.
H₁ : Binomial distribution is not good fit.


χ²_cal = 134.606
At a = 1% for ( n – 2) = 6 – 2= 4 d.f
the upper tail critical value k₂ = 13.28
Here χ²_cal lies in rejection region (R.R)
i.e., χ²_cal > k₂,
∴ H₀ is rejected and H₁ is accepted
Conclusion: B.D is not good fit.

Question 8.
A die is thrown 120 times and each time the number of the upper most face is noted
The resulted are as follows:

At 5% level of significance test whether the die is fair.
Answer:
Let 0/0₁ be the observed no. of frequency. If the die is assumed as fair, then outcomes / frequencies are equally distributed through out the faces of a die and so, each
E₁ = \(\frac{120}{6}\) = 20 each.
H₀ : The die is fair
H₁ : The die is not fair
Under H₀, the χ² – test statistic is:
χ² = \(\Sigma \frac{(0-E)^{2}}{E}-\chi^{2}(n-1) d . f\)


∴ χ²_cal = 10.8
At a = 5% for (n – 1) = 6 – 1 = 5d.f
the upper tail critical value k₂ = 11.07
Here χ²_cal lies in acceptance region (A.R)
ie., χ²_cal > k₂
∴ H₀ is accepted
Conclusion: The die is fair.

Question 9.
Consider the following data

Fit a poisson distribution to the data and test at 5% whether the distribution is a good fit.
Answer:
From the given frequency distribution λ is obtained as:



The distribution of observed and theoretical frequencies is:

Chi – square test:
H₀ : Poisson distribution is good fit
H₁ : Poisson distribution is
under H₀ χ² – test statistic is

λ is estimated from the data so (n – 2) d.f.

χ²_cal = 11.05
At α = 5% for (n – 2) = 4 – 2 = 2 d.f
the upper tail critical value k₂ = 5.99
Here χ²_cal lies in rejection region (R.R)
i.e., χ²_cal > k₂
∴ H₀ is accepted
Conclusion : poission distribution is not good fit i.e., O_i ≠ E_i

Question 10.
A sample analysis of examination results of 500 students was made. It was found that 220 student had failed, 170 has secured 3rd class, 90 were placed in 2nd class, and 20 got 1st class. Do these figures commensurate with the general examination result which is in the ratio of 4:3:2:1 for various categories respectively?
Answer:
Let 0/0 be observed no, of students: 220,170,90,20.
The expected frequencies are in the ratio of 4:3:2:1
∴ E₁ = \(\frac{500}{4+3+2+1} \times 4=\frac{500}{10} \times 4=200\)
E₂ = \(\frac{500}{10} \times 3=150\)
E₃ = \(\frac{500}{100} \times 2=100\) and
E₄ = \(\frac{500}{10} \times 1=50\)
H₀ : Figures commensurate with the general result.
H₁ : Figures doesn’t commensurate with the general result under H₀, the

∴ χ²_cal = 23.67
At a = 5% for (n – 1) = 4 – 1 = 3d.f
the upper tail critical value k₂ = 7.81
Here χ²_cal lies in rejected region (R.R)
ie., χ²_cal > k.
∴ H₀ is rejected
Conclusion: Figures does’t commensurate with the general result.

Question 11.
Consider the following 2 × 2 contingences table. Test whether X and Y are independent at 1% level of

Given :
H₀ : X and Y are independent
H₁ : X and Y are dependent.(upper tail test)
The 2 × 2 contingency table is:



At α = 1% forl.d.f the upper tail critical value
k₂ = 6.63
Here χ²_cal lies in acceptance region (A.R)
ie., χ²_cal < k₂
∴ H₀ is accepted.
Conclusion : X and Y are independent.

Question 12.
For the following data, test the effect of vaccine in controlling the independence of a certain disease.

Test at 5% level of significance.
Answer:
H₀: Inoculation and effect of disease are independent
H₁ : Inoculation and effect of disease are dependent
The 2 × 2 contingency table is:


At α = 5% for l.d.f the upper tail critical value k₂ = 3.84
Here χ²_cal lies in acceptance region (A.R)
ie., χ²_cal < k₂.
∴ H₀ is accepted.
Conclusion: Inoculation and affect of disease are independent.

Question 13.
40 students of college A and 120 students of college B are appeared for an examination. The results are as follows.

Answer:
H₀ : Result of the examination and colleges are independent
H₁ : Result of the examination and colleges are dependent.
Under H₀, the χ² – test statistic is:


χ²_cal = \(\frac{160(25 \times 20-100 \times 15)^{2}}{125 \times 35 \times 400+120}\)
= \(\frac{160 \times 1000 \times 1000}{21000000}\) = 7.619
At α = 5% for l.d.f the upper tail critical value
k₂ = 3.84
Here χ²_cal lies in rejection region (A.R)
∴ H₀ is rejected.
Conclusion: Result of examination is dependent on college.

Question 14.
In an experiment an immunization of cattle from tuberculosis, the following results were obtained.

Test whether the vaccine is effective in controlling tuberculosis.
Answer:
H₀ : vaccline is not effective in controlling T.B.
H₁ : vaccline is effective in controlling T.B
The 2 × 2 contingency table as:

At α = 5% for 1. D.f the upper tail critical values
k₂ = 3.84
Here χ²_cal lies in rejection region (R.R)
ie., χ²_cal > K₂
∴ H₀ is rejected.

Conclusion: vaccine is effective in controlling the disease.

Question 15.
In a survey of 200 boys of which 75 were intelligent, 40 had skilled fathers, while 85 of the unintelligent boys had unskilled fathers. Do these figures support the hypothesis that skilled fathers have intelligents boys.
Answer:
H₀ : skill offathers and intelligence of boys are independent
H₀ : skill offathers and intelligence of boys are dependent
The 2 × 2 contingency tables as below:

given data; 200 – 75 = 125; 125 – 85 = 40 etc.,
The test statistic is:

At α = 5% for 1 d.f the upper tail critical value
k₂ = 3.84.
Here χ²_cal lies in rejection region (R.R)
i.e., χ²_cal > k₂
∴ H₀ is rejected.

Conclusion : Skill of fathers and intelligence of boys are dependent.