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## Karnataka 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference

### 2nd PUC Statistical Inference One and Two Marks Questions With Answers

Question 1.

What are parameter and statistic?/Define parameter and statistic.

Answer:

Statistical constants of the population such as Mean (µ), S.D (σ) are called parameter.

Ex:- The population has mean = µ, S.D = σ, population proportion P_{0
}Statistical measures computed from the samples such as Mean (x̄), S.D(σ) are called statistic.

Ex:- A random sample of size ‘n’ has mean=x̄ and S.D= s, sample proportion p

Question 2.

Define Statistical Hypothesis

Answer:

It is a statement or an assertion made regarding the parameters

Question 3.

What are null and alternative hypothesis

Answer:

Null Hypothesis is a hypothesis, which is being tested for possible rejection under the assumption that it is true. Denoted by H_{0}

The hypothesis, which is accepted when the null hypothesis is rejected, is called alternative hypothesis. Denoted by H_{1}.

Question 4.

Define Simple and Composite Hypothesis.

Answer:

A hypothesis, which completely specifies the parameter of the distribution, is a simple hypothesis.

Ex:- H: µ = µ_{0} (25) is a is a simple hypothesis,

H: The population is normally distributed with mean µ = 25 and. σ = 2

A hypothesis, which does not completely specify the parameter of the distribution, is a composite hypothesis.

Ex:- H: The population is normally distributed with mean µ = 25

Question 5.

Define Standard Error

Answer:

The standard deviation of the sampling distribution of sample statistic is called S.E of statistic

Question 6.

Give any two uses of Standard Error.

Answer:

- It is used in interval estimation, to write down the confidence intervals.
- It is used in testing of hypothesis, to test whether the difference between the sample statistic and the population parameter is significant or not.

Question 7.

Define Type I/Type II Error

Answer:

Type I Error is taking a wrong decision to reject the null hypothesis when it is actually true.

Type II Error is taking a wrong decision to accept the null hypothesis when it is actually not true.3

Question 8.

What is the Power of a test.

Answer:

It is the probability of rejecting the null hypothesis when it is not true. (1 – β ),

Here β = P (Type II Error) which is referred as ‘consumers risk

Question 9.

Define size of a test.

Answer:

Probability of occurrence of type I Error is called level of significance, denoted by ‘α’. It is also called as Size of the test.

Question 10.

Define critical Region and Critical value.

Answer:

The set of all those values of the test statistic, which lead to the rejection of the null hypothesis, is called critical region (ω).

The Critical Region is also called as Rejection Region.

The value of test statistic, which separates the critical region (ie. rejection region) and . acceptance regions is called the Critical value, denoted by + k or -k or – k_{1} / + k_{2}

Question 11.

Define level of significance, confidence level and confidence co-efficient.

Answer:

Probability of occurrence of type I Error is called level of significance, denoted by ‘α’

The confidence interval is the interval within which the unknown population parameter is expected to lie.

The prob. that the confidence interval contains the parameter is called confidence coefficient.

Question 12.

Define Point estimation and Interval estimation.

Answer:

‘While estimating the unknown parameter, if a specific’value is proposed as an estimate, which is called Point estimation’.

‘While estimating the unknown parameter instead of a specific value, an interval is proposed, which is likely to contain the parameter is called Interval estimation’

Question 13.

What is the best estimator of the population mean?

Answer:

x̄

Question 14.

Define sample space and parameter space.

Answer:

The set of all samples of size ‘n’ that can be drawn from population is called sample space(S)

The set of all the admissible values of the population parameters is called parameter space.

Question 15.

For a S.N.V. test statistic Z, in left tail test the calculated value is (- 2.63) and the table value is (-1.90).

Indicate suitable inference to be drawn based on this data.

Answer:

Here Z_{cal} < Z_{tab} H_{0} is rejected and H_{1} is accepted.

Question 16.

Mention any two conditions while applying %2 – test in Testing the Independent of attributes

Answer:

(a) Observations should be independent,

(b) Total frequency N should be large,

(c) Each cell frequencies ie ,a, b, c, d are all ≥ 5

Question 17.

Mention any two conditions of test for goodness of fit

Answer:

(a) Observations should be independent,0.6

(b) Total frequency N should be large,

(e) Each E_{i}‘s ≥ 5.

Question 18.

For the application of %2 – distribution, what is the condition for the expected cell frequency?when pooling is made done in testing of goodness of fit?

Answer:

Each cell frequency ie. E_{i}‘s ≥ 5

Question 19.

In a chi-square test for goodness of fit if there are 8 classes and if two parameters are’ estimated, that is the degrees of freedom of the test statistic.

Answer:

(n – 1 – 2) = (n – 3)d.f.

Question 20.

A random sample of size 25 is drawn from a population whose standard deviation is 3. Find the standard error of the sample mean.

Answer:

Question 21.

Given H, : < p2, Write H_{0}.

Answer:

µ_{1} = µ_{2 .}

Question 22.

The proportion of vegetarians in a city is 0.48 Find the standard Error of the proportion of vegetarians in a random sample of size 20.

Answer:

Question 23.

Given P = 2 / 3, n = 40, find SE (P).

Answer:

SE(p) = \(\sqrt{\frac{\mathrm{PQ}}{\mathrm{n}}}=\sqrt{\frac{2 / 3 \times 1 / 3}{20}}\) = 0.07

Question 24.

The proportion of vegetarians in village A is 0.42.The proportion of vegetarians in village B is 0.37. Among 70 randomly selected people from village A, if PI is the proportion of vegetarians and among 60 randomly selected people from village B, if P_{2} is the proportion of vegetarians, find Standard Error of (P_{1} = P_{2})

Answer:

25.

In paired t-test if n = 10, d = 4.3 and S_{d} = 1.3 What would be t_{obs}?

Answer:

Standard Errors:

Question 26.

Write down the standard of Proportion.

Answer:

Question 27.

Write down the standard of difference of sample means of two samples

Answer:

Question 28.

Write down the standard of difference of proportion of two samples.

Answer:

Question 29.

Write down the test statistics for test for mean for large sample:

Answer:

Large sample tests: ( n / n_{1} / n_{2} >30):

Question 30.

Write down the test statistics for test for equality of two means for large samples:

Answer:

Question 31.

Write down the test statistics for test for proportion:

Answer:

Test for proportion

Question 32.

Write down the test statistics for test for equality of two proportions.

Answer:

Test for equality of proportion:

Question 33.

Write down the chi-square test statistics for testing the variance with Hypothesis.

Answer:

Chi-square tests:

Test for population variance:

H_{0}: σ^{2} = σ^{2}_{0}; H_{1} : σ^{2} ≠ σ^{2}_{0}; OR H,: σ^{2} > σ^{2}_{0}; OR H_{1}: σ^{2} < σ^{2}_{0};

Question 34.

Write down the chi-square test statistics for testing the Goodness of fit with Hypothesis.

Answer:

Test for good ness of fit: H_{0} : O_{i} and E_{i} are good fit; H_{1}: O_{i} and E_{i }are not good fit

when any of the parameter (p / λ) is estimate d.f = (n – 1 – 1) = (n – 2).

Question 35.

Write down the chi-square test statistics for testing the Independence of attributes.

Answer:

To test for Independence of Attributes:

Question 36.

Write down the t- test statistics for testing the mean.

Answer:

Small sample tests

Question 37.

Write down the t- test statistics for testing the equality of meAnswer:

Answer:

Question 38.

Write down the t- test statistics for testing the equality of means when observation are paired.

Answer:

Paired t-test for equality of means:

Question 39.

What do you mean by an Estimator?

Answer:

The Statistic used for the purpose of estimation of unknown population parameter is called Estimator.

Question 40.

What is Estimate?

Answer:

Estimate is a numerical value of the estimator computed from a given set of sample values.

Question Question 41.

What are Confidence interval?

Answer:

A Confidence interval is a interval within which the unknown population parameter is expected to lie.

Question 42.

What is Test Statistic? ,

Answer:

Test Statistic is the statistic based on whose distribution testing is conducted.

Question 43.

Define sampling distribution.

Answer:

The distribution of values of the statistic for different samples of the same size is called sampling distribution of a statistic.

Question 44.

Given X̄ = 22 gm, μ_{0} = 20 gm, σ=10 gm and n = 64 .Calculate test statistic z

Answer:

Here statistic x̄ and parameter μ_{0} are given, the test statistic is

Question 45.

A hospital has 150 doctors out of whom 78 are ladies. When a sample of 35 doctors are taken, find mean and standard error of proportion of ladies.

Answer:

Given N=150, X = 78, n = 35

Population proportion = P_{0} = \(\frac { X }{ N }\) = \(\frac { 78 }{ 150 }\) = 0.52; Q_{0} = 1 -P_{0}= 0.48

Sample proportion = p = p_{0} = 0.52

Standard error of sample proportion is

SE = \(\sqrt{n}=\sqrt{\frac{0.52 \times 0.48}{35}}\) = 0.084

Question 46.

Size of two samples are 40 and SO. Population standard deviations are 10 and 20. Compute S.E. (x̄_{1} – x̄_{2})

Answer:

Given : n_{1} = 40, n_{2} = 50, σ_{1} = 10, σ_{2}= 20

Question 47.

What are meant by one tailed test and two tailed test in testing of hypothesis?

Answer:

- if the critical region is considered at one tail of the null distribution of the test statistic, the test is one tailed.
- if the critical region is considered at both the tails of the null distribution of the test statistic, the test is two tailed.

Question 48.

If Z_{cal} = 2.089 and Z_{tab} = 2.58, then write down the decision on null hypothesis.

Answer:

Here Z_{cal} < Z_{tab}. (H_{0}) Null hypothesis is accepted.

Question 49.

For a S.N.V. test statistics Z, the calculated value Z is 2.63 and the table value of Z is 2.58. Indicate suitable inference to be drawn based on this data.

Answer:

Given Z = 2.63 and Z_{tab} = 2.58.

Here Z_{cal} > Z_{tab}. Null hypothesis is rejected.

Question 50.

The standard deviation of weight of children is 2 kg. What is the standard error of mean weight of 16 children?

Answer:

Given σ =2 kg, n= 16.

Question 51.

From the following data find the S.E. of difference between means of samples A&B.

Answer:

Given n_{1} = 1000 n_{2} = 500

x̄_{1} = 150 x̄_{2} = 146

S_{1} = 40 S_{2} = 30

Question 52.

Write down any two conditons for X test for goodness of fit.

Answer:

(a) If any parameters are estimated from the data one degrees of freedom should be lessened.

(b) All E; ≥ 5 if any E; < 5, it should be pooled with the adjacent frequencies.

Question 53.

If are cell frequencies, write the χ^{2} test statistics with Yate’s correction.

Answer:

Question 54.

It (X̄_{1}– X̄_{2}) = 2.7 and S.E (X̄_{1} – X̄_{3}) = 1.3, find Z.

Answer:

If (X̄_{1} -X̄_{2}) = 2.7 and S.E (X̄_{1}-X̄_{2}) = 1.3, find Z.

Question 55.

Write down the mean and standard deviation of at variate with n degrees of freedom.

Answer:

Mean = 0, S.D. = \(\sqrt{\text { variance }}=\sqrt{\frac{n}{n-2}}\) ; n >2.

Question 56.

If p_{1} = 0.3 and p_{2} = 0.2, find mean of (p_{1} – p_{2}).

Answer:

Mean of (p_{1} – p_{2}) = E(p_{1} – p_{2}) = 0.3 – 0.2 = 0.1.

### 2nd PUC Statistical Inference Five Marks Question With Answers

**Large Sample Test**

Question 1.

A certain brand soap is known to have weight with variance25. A random sample of 256 soaps had mean weight 122 gms. Can we conclude that the mean weight of the soaps manufactured by a firm is 125 gms. (Use 5% level of significance)

Answer:

Given; n-256, σ_{2} = 25, ∴ σ = 5. x̄= 122, µ = 125 ; α = 5%.

H_{0}: Mean Weight of the soaps manufactured is 125 gms (i.e., H_{0}: µ = 125)

H_{1}: Mean Weight of the soaps manufactured is differ from 125 gms.

(i.e., H_{1} : µ ≠ 125) {two tailed test}

The test statistics is –

At α = 2.5% the two fail critical values are K_{1} = -1.96 and K_{2} = 1.96

Here Z_{cal}, = -9.6 lies outside -K_{1} and K_{2}

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion: Mean Weight of the soaps manufactured by a firm differs from 125 gms.

Question 2.

The standard deviation of marks scored by PUC students in an examination is 10.5. The mean marks scored by 64 students of a college is 38. Can we conclude at 1% level of significance that mean marks of PUC students is 40 ?

Answer:

Given: σ=10.5, n = 64 x̄=38, p = 40, µ=l%

H_{0} = Mean marks of PUC students is 40. ( i.e., H_{0} = : µ = 40)

H_{1} = Mean marks of PUC students differs from 40.( i.e,. H_{1} = : µ ≠ 40)

The test statistics is :

At α = 1% the two tail critical values are -K_{1} = 2.58 and + K_{2} = + 2.58

Here, Zeal = -1.52 is in between -K_{1} and K_{2}.

H_{0} is accepted.

Conclusion : Mean marks of PUC students is 40.

Question 3.

A random sample of 200 cans containing baby food has mean height 999.5 gms and standard deviation 15 gms. Test whether the contents of the cans can be considered as 1 kg. (Test at 1% level of significance.)

Answer:

Given: n = 200, x̄ = 999.5 gms, s = 15 gms, µ = 1 kg i.e., µ = 1000 gms.

H_{0}: The contents of the can is 1 kg. (i.e., H_{0}: µ = 1000 gms.)

H_{1}: The contents of the can is less than 1 kg (i.e., H_{1} : µ < 1000 gms)

The test statistic is:

At α = 1% the left/lower tail critical value is k = -2.33

Here, Z_{cal} = -0.47 is in acceptance region.

∴ H_{0} is accepted;

Conclusion: The contents of the can be considered as 1 kg.

Question 4.

A sample of 100 students is taken from a large population. The mean height of these students is 64″ and standard deviation 4″. Can it be reasonably regarded that, the population mean height is more than 66″. (Take a = 5%)

Answer:

Given: n = 100, x̄ = 64″, S = 4″, µ = 66″, α = 5%

H_{0}: The mean height of the population is 66″ (i.e., H_{0}: µ = 66″)

H_{1} : The mean height of the population is more than (i.e., H_{1} : µ > 66″)

The test statistic is –

At α = 5% the right/upper tail critical value K = 1.645.

Here Zeal = -5 is in acceptance region

∴ H_{0} is accepted.

Conclusion : The mean height of the populations is 66″.

Question 5.

A random sample of 1000 apples has mean weight of 187 gms and S.D. 8 gms. Another random sample of 800 apples has mean weight of 188.4 gms and S.D. of 10 gms. Test the hypothesis that the mean weights differ significantly at 1% level of significance.

Answer:

Given:

n_{1} = 1000; x̄_{1} =187gms, S_{1} = 8 gms

n_{2} = 800; x̄_{2} = 188gms, S_{2} = 10 gms

H_{0} : means are same (i.e., H_{0} : µ<sub.1 = µ_{2})

H, : means are not same.( i.e., H, : µ_{1} ≠ µ_{2})

The test statistics is –

At α = 5% the critical values are – K_{1} = – 1.96, K_{2} = 1.96.

Here Zeal = – 3.22 is out side of -K_{1} and K_{2}

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion: Mean weight of apples are same.

Question 6.

From the following data, test at 1% level of significance whether the sample means differ significantly.

Answer:

Given:

n_{1}=200; x̄_{1} =219.3, S_{1}^{2} = 3,

n_{2} = 300; x̄_{2} =220.1, S_{2}^{2} = 2, a=l%

H_{0}: Means are equal/ means does not differ significantly (i.e., H_{1} : µ_{1} = µ_{2})

H_{1}: Means are not equal/means differ significantly (i.e. H_{1} > : µ<sub>1</SUB> ≠ µ<sub>2</sub>)

The test statistic is –

Z_{cal} = – 5.43

At α = 1 %, the two Tailed critical values are -K_{1} = -2.58 and K_{2} = 2.58

Here Zeal = -5.43 is in rejection region/ Z_{cal} is outside the interval -k_{1} and k_{2}

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : means are not equal.

Question 7.

Samples of electric lamps manufactured by two firms gave the following results:

Test whether the average life of bulbs manufactured by the firm A is more than firm B. Use 5% level of significance

Answer:

Given:

n_{1} = 60, x̄_{1} =1400, S_{1}= 106

n_{2} = 70, x̄_{2} = 1300, S_{2} =100, α = 5%.

H_{0} : Average life of bulbs of firm A and B are same.( i.e., H_{0}: µ_{1},, = µ,sub>2)

H_{1} : Average life of bulbs of firm A is more than firm B. (i.e., H_{1} : µ_{1} > µ_{2})

The test statistics is –

Zcal = 5.50

At α = 5% the upper tail critical value K = 1.645

Here Z_{cal} = 5.50 is in Rejection region / z_{cal} > K

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : The mean life of bulbs of firm A is more than firm B.

Question 8.

The following data gives pass percentage in two colleges :

Test at 5% level of significance. Whether the pass percentage of college A is more than college B.

Answer:

Given:

x̄_{1} =80, S_{1} = 5, n_{1}=100

x̄2 = 70, S_{2} = 4, n_{2} = 90, α = 5%.

H_{0} : The pass percentage of college A and B is same, i.e., H_{0} : µ_{1} = µ_{2}

H_{1} : The pass percentage of college A is more than college B. (i.e., H_{1} : µ_{1} > µ_{2})

The test statistic is –

At α = 5%, the upper/right Tail critical value K = 1.645.

Here Zeal = 15.29 is in rejection region/Z_{cal} > K.

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion: The pass percentage of college A is more than college B.

Question 9.

In an election to the parliament the leaders of a specific party contend that they would secure 40% of the votes. A pre-poll survey of 250 voters revealed that the percentages is 32. Does the survey support the leaders claim? (Test at 5% level of significance).

Answer:

Given: n = 250, P_{0} = 0.40 P = 0.32. ∴ Q_{0} = 1 – P_{0} = 0.60

H_{0}: The leader would secure 40% votes, (i.e, H_{0} : P_{1}= 0.40)

H_{1}: The leader would not secure 40% votes (i.e., H_{1} : P ≠ 0.40)

The test statistic is –

At α = 5% the two tail critical values are -K_{1} = 1.96 and K_{2} = 1.96

Here Zcal = 2.58 lies in rejection region/ Z_{cal} > Kr

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion: The Leader of a specific party does not secure 40% votes or The survey does not support the claim.

Question 10.

A manufacturer claims that less than 2% of his products are defective. A retailer buys a batch of 250 articles from the manufacturer and find that 13 are defectives. Test at 1% level of significance that whether the manufacturer’s claim is justifiable.

Answer:

Given P_{0} = 0.02, n = 250, x=13, α = l%

∴ P = \(\frac{x}{n}=\frac{13}{250}\) = 0.052; Q_{o} = 1 – p_{0} =1 – 0.02 = 05.98

H_{0} : proportion of defective products are 2%

(i.e. H_{0}: P_{0} = 2% = 0.02)

H_{1} : proportion of defective products are less than 2%……

(i.e H<sub.1 : p_{0} < 2% )

The test Statistic is

Z_{cal} =3.636

At α = 5% the lower tail critical value – k = – 2.33

Here Z_{cal} > – k / Z_{cal} lies in (AR) acceptance region.

∴ H_{0} is accepted

Conclusion : proportion of defective products are 0.02 or 2%.

Question 11.

From a large consignment of mangoes, a random sample of 50 mangoes were examined and among them four were found to be bad. Test whether it is reasonable to assume that more than 10% of mangoes were bad. [The critical value at 5% level of significance is 1.645].

Answer:

Given n = 4 x = 4 ∴p = \(\frac{x}{n}=\frac{4}{50}\) = 0.08

∴ Q_{0} = 1 – p_{0} = 0.9

H_{0}: The proportion of bad mangoes is 10% = 0.1 (i.e., H_{0}: P = 0.1)

H_{1} : The proportion of bad mangoes is more than 10% = 0.1 (i.e., H_{1} : P > 0.1)

The test statistic is

Zcal = -0.47

At 5% level of the upper tail critical value. K = 1.645.

Here Zeal = -0.47 is acceptance region/ Z_{cal} < K.

∴ H_{0} is accepted

Conclusion : The proportion of bad mangoes is 10%.

Question 12.

In a village, 200 out of 300 men are smokers. Does this data support the argument that majority of men in the village are smokers ?

Answer:

Given : n = 300 ; x = 200

∴ p = \(\frac{x}{n}=\frac{200}{300}\) = 0.67

H_{0}: Proportion of smokers is 0.5

(i.e., H_{0}: P_{0} = 0.5) .

H_{1} : Proportion of smokers is more than 0.5

(i.e., H_{1} : P_{0} > 0.5) {upper tail test}

Here it is assumed that proportion of smokers and non-smokers are equal. Therefore P_{0} = 0.5.

The test statistic is :

At α = 5% the upper tail critical value is K = 1.645.

Here Z_{cal} > K.

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : Majority of men in the village are smokers i.e., proportion of smokers is more than 0.5.

Question 13.

In a random sample of 200 plastic bags manufactured by machine A, 16 were found to be defective. In a random sample of 300 plastic bags produced by a machines B, 30 were found to be defective. Test at 5% level of significance the hypothesis, that the proportion of the defective plastic bags produced by the machines are same.

Answer:

Given:

n_{1} =200, x_{1} = 16 ∴p_{1} = \(\frac{x_{1}}{n_{1}}=\frac{16}{200}\) = 0.08

n_{2} = 300, x_{2} = 30 ∴P_{2}= \(\frac{x_{2}}{n_{2}}=\frac{30}{300}\) = 0.1

H_{0}: The Proportion of defective plastic bags produced by machines are same.

(i.e„ H_{0}: P_{1} = P_{2}) .

H_{1} : The proportion of defective plastic bags produced by machines are not same.

(i.e., H_{1} = P_{1} ≠ P_{2}) {Two tailed test}

Under H_{0}, The test statistic is –

∴ Zcal = -0.758

At α = 5% the two tail critical values are -K_{1} = -1.96 and K_{2} = 1.96.

Here Zcal = -0.758 is in acceptance region/ Z_{cal} is in between -K_{1} and K_{2}

∴ H_{0} is accepted.

Conclusion : The proportion of defective plastic bags produced by machines are same.

Question 14.

Among 40 randomly selected people from urban area, 5 are interested in viewing football match. Among 50 randomly selected people from rural area, 4 are interested in viewing football match. Test st 5% level of significance, whether the proportion of viewers in urban and rural areas are the same.

Answer:

Given:

n_{1} = 40 ; x_{1} = 5, ∴p_{1} = \(\frac{x_{1}}{n_{1}}=\frac{5}{40}\) =0.125

n_{2} = 50: x_{2} = 4 ∴P_{2} = \(\frac{x_{2}}{n_{2}}=\frac{4}{40}\) = 0.08

H_{0}: The proportion of viewers in urban and rural areas are same.

(i e., H_{0}: P_{1}= P_{2})

H_{1} : The proportion of viewers in urban and rural areas are not same,

(i.e., H_{1} : P_{1} ≠ P_{2}) {two Tailed test}

The test Statistic is

Zcal = 0.7075

At α = 5%, the two tail critical values are -K_{1}= -1.96 and K_{2} = 1.96.

Here Zeal = 0.7075 is in acceptance region/ Z_{cal} is in between K_{1} and K_{2}

∴ H_{0} is accepted.

Conclusion : The proportion of viewers of football match in urban and rural areas are the same.

Question 15.

A machine produced 25 defective articles in a batch of 400. After over hauling it produced 15 defectives in a batch of 200. Test at 1% level of significance whether there is a reduction of defective articles after overhauling.

Answer:

Given:

n_{1} = 400, x_{1} = 25 ∴ P_{1} = \(\frac{x_{1}}{n_{1}}=\frac{25}{400}\) = 0.0625.

n_{2} = 200 x_{2} = 15 ∴P_{2} = \(\frac{x_{2}}{n_{2}}=\frac{15}{200}\) = 0.075

H_{0} : The proportion of defective articles before and after overhauling is the same

(i.e H_{0} : p_{1} = p_{2})

H_{1} : The proportion of defective articles are more before overhauling

(i.e H_{1} : p_{1} > p_{2}) { Right upprt Tailed test}

The test statistic is

∴ Zeal = -0.576

At α = 1 % the upper/right tail critical values K = 2.33

Here Zcal = -0.576 is in acceptance region/ Z_{cal}

∴ H_{0} is accepted.

Conclusion : The proportion of defective articles before and after .over hauling is same.

Question 16.

From the following data, test whether the difference between the proportions in the two samples is significant.

Answer:

Given:

n_{1} = 1000, P_{1} = 0.02

n_{2} = 1200, P_{2} = 0.01

p_{0} = 0.0145 ∴Q_{0} = 0.9855

H_{0}: The proportions are equal.

(i e., H_{0} = P_{1} = P_{2})

H, : The proportions are not equal,

(i.e., H_{1} : P_{1} ≠ P_{2})

Zcal = 1.953

At a= 5% the two Tail critical values are -K_{1} = -1.96 and K_{2} – 1.96

Here Zcal = 1.953 is in acceptance region.

∴ H_{0} is accepted.

Conclusion: The proportions are same.

**Chi Square Tests**

Question 1.

The weights of 10 babies at the time of birth are as follows :

Weight (Kgs) : 2.8, 3.4, 2.6, 3.0, 2.4, 2.7, 1.6, 3.1, 2.9, 2.4

Test the hypothesis that the standard deviation differs significantly from lkg. (use α = 1%)

Answer:

Given : n = 10, σ = population SD = lkg.

H_{0}: Sample SD does not differ significantly from population SD

i.e., H_{0}: σ = 1 kg

H_{1} : Sample SD differs significantly from population SD.

i.e., H_{1} : σ ≠ 1 kg. {Two Tailed .test}

The test statistics –

Where s^{2} = sample variance

Let ‘x’ be the weight

s_{2} = \(\frac { 14.03 }{ 10 }\) – \(\left(\frac{10.9}{10}\right)^{2}\) = 0.2149

∴ χ^{2} = \(\frac{10 \times 0.2149}{1}\) χ^{2}_{cal} = 2.149

For (n – 1) = (10 – 1) = 9 d.f at α = 1% the two tail critical values (from the χ^{2} – table) are + K_{1}= 1.73 and K_{2} = 23.59

Here χ^{2}_{cal} = 2.149 is in acceptance region.

∴ H_{0} is accepted.

Conclusion : The sample SD doesn’t differs from population S.D. = 1 kg.

Question 2.

Weight of 10 jack fruits are as follows :

Weight (kgs) : 6, 5, 8, 7, 4, 5, 8, 6, 5, 6

Test at 5% level of significance that SD differs significantly from 2 kgs.

Answer:

Given: n=10, σ = 2 kgs.

H_{0}: The sample SD doesnot differs significantly from population S.D. = 2 kgs.

i.e., H_{0}: σ = 2 kgs.

H_{1} : The sample SD differs significantly from population S.D. = 2 kgs.

i.e., H_{1} : σ ≠ 2 kgs

The test statistics is

Let x be the weight

∴ χ^{2} = \(\frac{10 \times 1.6}{2^{2}}\) = 4

χ^{2}_{cal} = 4

For (n – 1) = (10 – 1) = 9 d.f at α = 5 % the two Tail critical values (from the table)

K_{1} = 2.70 and K_{2} = 19.02. .

Here χ^{2}_{cal} = 4 is in acceptance region.

∴ H_{0} is accepted.

Conclusion : The sample SD doesn’t differs significantly from population SD=2 kgs.

Question 3.

25 randomly selected resistors have S.D. of resistance of 0.016 ohms. Test the hypothesis that the resistors in the lot have SD of resistance less than 0.01 ohm.

Answer:

Given: n = 25, S = 0.016ohms, σ = 0.01ohms.

H_{0} : The SD of resistance of resistors is 0.01 ohms,

i.e., H_{0} : σ = 0.01 ohm.

H_{1} : The SD of resistance of resistors is less than 0.01 ohm.

i.e., H_{1} : σ < 0.01 ohm [left/lower tail test]

χ^{2} = \(\frac{25(0.016)^{2}}{(0.01)^{2}}\)

χ^{2}_{cal} = 64

For (n – 1) = (25 – 1) = 24, d.f at a = 5% the left/lower tail critical value is K = 13.85.

Here χ^{2}_{cal} = 64 is in acceptance region χ^{2}_{cal} > K_{2}

∴ H_{0} is accepted.

Conclusion : The SD of resistance of resistors is 0.01 ohm.

Question 4.

A normal variate has standard deviation 2.6. Twenty sample observations of the variate have standard deviation 2. Test at 1% level of significance that population standard deviation is less than 2.6.

Answer:

Given: σ_{0} = 2.6, n = 20, S = 2, a=l%

H_{0}: σ_{0} =2.6(Standard deviation is 2.6)

H_{1} : σ_{0} < 2.6 (Standard deviation less than 2.6) {Lower tailed test}

Under H_{0}, the test statistic is :

\(=\frac{20 \times 2^{2}}{(2.6)^{2}}\)

χ^{2}_{cal} = 11.8343

For (n- 1) = (20 – 1) = 19 d.f the lower tail critical value at 5% level of significance is

K_{1} – 10.12;

Here χ^{2}_{cal} > K_{1}

∴ H_{0} is accepted.

Conclusion: Standard deviation is 2.6.

Question 5.

From the following data, test whether there is any significant difference between observed and expected frequencies, using χ^{2} test at 5% level of significance.

Answer:

H_{0} : There is no significant difference between observed and expected frequencies.

H_{1} : There is a significant difference between observed and expected frequencies.

The test statistics is –

χ^{2}_{cal} = 4.69

For (n – 1) = (6 – 1) = 5 d.f at α = 5% the upper Tail critical value K2/K = 11.07.

Here χ^{2}_{cal} = 4.69 is in acceptance region/ χ^{2}_{cal} < K_{2}.

H_{0} is accepted.

Conclusion: There is no significant difference between observed and expected frequencies. [Note: In the above table the first one and last two expected frequencies are less than 5, they are pooled with the adjacent frequencies such that finally all the frequencies are 5 or more.]

Question 6.

Four coins are tossed 80 times. The distribution of number of heads is given:

Apply χ^{2} – test and test at 1% level of significant if the coin is unbiased.

Answer:

H_{0}: The coin is unbiased.

∴ p = \(\frac { 1 }{ 2 }\)

H_{1} : The coin is biased.

i.e., p ≠ \(\frac { 1 }{ 2 }\) [upper Tail test only]

The test statistics is –

Let O and E be the observed and expected frequencies, and let x be the number of heads.

∴ χ^{2}_{cal}=0.73

For (n-1) = (5-1) = 4 d.f, at 1% level of significance the upper Tail critical value {From table} K/K_{2} = 13.28

Here χ^{2}_{cal} = 0.73 is in acceptance region/ χ^{2}_{cal} < K_{2}

∴ H_{0} is accepted.

Conclusion : The coin unbiased.

[Note : In the above table none of the expected frequencies are less than 5. So, pooling is not necessary.]

Question 7.

The following is the data regarding day-wise accidents occured in Bangalore during 2000:

Test whether there is day-wise equidistribution of accidents

Answer:

H_{0} : There is day-wise equidistribution of accidents.

H_{1} : There is no day-wise equidistribution of accidents. [Upper tail test only]

The test statistics is –

Since there is a day-wise equidistribution of accidents.

Therefore the expected frequencies, each is \(\frac { 105 }{ 7 }\) =15.

Let O and E be the observed and expected number of accidents.

∴ χ^{2} = 6.42

For (n – 1) = 7 – 1 = 6 d.f at 5% level of significance the upper tail critical value K/K_{2}= 12.59

Here χ^{2}_{cal} = 6.46 is in acceptance region/ χ^{2}_{cal} < K_{2}

∴ H_{0} is accepted.

Conclusion: There is day-wise equidistribution of accidents.

Question 8.

70 accidents that have occurred in a state in a week are tabulated as follows:

Test whether accidents occur uniformly throughout the week-

[χ^{2}_{tab} is 16.81]

Answer:

H_{0} : Accidents occur uniformly throughout the week

H_{1} : Accidents does not occur uniformly throughout the week…(upper tail test)

The test statistic is :

Let ‘O’ and ‘E’ be the observed and expected number of accidents.

Since accidents occur uniformly throughout the week.

Expected frequencies each is \(\frac { 70 }{ 7 }\) = 10

∴ χ^{2}_{cal}= 6.8

For (n – 1) = (7 – 1) = 6 degrees of freedom at 1% level of significance, the upper tail critical value given k_{2}= 16.81 (χ^{2}_{tab})

Here χ^{2}_{cal} < k_{2}/χ^{2}_{cal} lies on (AR-) acceptance region.

∴ H_{0} is accepted.

Conclusion: Accidents occur uniformly throughout the week.

Question 9.

A die is rolled 90 times and the fol lowing distribution is obtained:

Test whether the die is unbiased use 5% level of significance.

Answer:

H_{0} : The die is unbiased

H_{1} : The die is biased.

The test statistic is –

Since, the die is unbiased, the numbers of throws should be equally distributed on all 6 faces.

∴ The expected frequencies each is \(\frac { 90 }{ 6 }\) = 15.

Let ‘O’ and ‘E’ be the observed number of throws and expected frequencies and let x be the face value.

∴ χ^{2}_{cal} = 5.868

For (n – 1) = (6 – 1) = 5 d.f. At 5% level of sigificance the upper Tail critical value K/K_{2} = 11.07

Here = 5.868 is in acceptance region./ χ^{2}_{cal} < K

∴ H_{0} is accepted.

Conclusion : The die is unbiased.

Question 10.

According to Mendal’s law, the first generation of cross breeding of red and white 4 o’clock (flowers) will have 1:2:1 red, pink and white 4’oclocks. In an experiment, the frequencies happen to be 13, 21 and 10 respectively. Docs the data support Mendal’s law?

Answer:

H_{0}: The data support Mendal’s law.

H_{1} : The data does not support Mendal’s law.

The test statistics is –

Let ‘O’ be the experiment/observed frequencies (i.e., 13, 21 and 10)

Here total frequencies = 44

∴ The Expected frequencies ‘E’ should be in ratio of 1:2:1

i.e., divide 44 in to 3 out of 1 + 2 + 1 = 4 parts as :

E_{1} = \(\frac { 44 }{ 4 }\) × 1 = 11 E_{2} = \(\frac { 44 }{ 4 }\) × 2 = 22 E_{3} = \(\frac { 44 }{ 4 }\) × 3 = 11

∴ χ^{2}_{cal} = 0.496

For (n – 1) = (3 – 1) = 2 d.f at 5% level of significance upper tail critical value K_{2} = 5.99

Hereχ^{2}_{cal} = 0.496 is in Acceptance region./ χ^{2}_{cal} < K_{2}

∴ H_{0} is accepted

Conclusion : The data support Mendal’s Law.

Question 11.

5 coins are tossed 128 times. the result are tabulated as below: (10 Marks)

Answer:

Let x be the number heads obtained is a Binomial variable with the parameters n = 5 and

p = probability of getting head = \(\frac { 1 }{ 2 }\) ∴ q = \(\frac { 1 }{ 2 }\)

The p.m.f is –

p (x) = nC_{n} p^{x}q^{n-x}; x = 0, 1, 2 ………….. n.

= 5C_{x} (0.5)^{x} (0.5)^{5-x}; x = 0, 1,2 ………….. 5

Theoretical/expected frequency T_{x} = p(x) × N

∴ T(x = 0) = p(x = 0) ∴ 128

= 5C_{0}(0.5)°(0.5)^{5-0} × 128

T (0) = 4

Using recurrence relation for theoretical frequency T_{x} = \(\frac{n+1-x}{x} \frac{p}{q} T_{x-1}\)

T(x – 1) = \(\frac{5+1-1}{1} \times \frac{0.5}{0.5} \times \mathrm{T}_{(1-1)}\)

= 5 × 1 × T_{0}

= 5 × 4

T(1) = 20

similarly,

∴ The fitted observed theoretical frequency distribution is :

Chi-square Test is :

H_{0}: Binomial distribution is good fit

H_{1} : Binomial distribution is not a good fit

The Test statistic is –

Let ‘O’ and ‘E’ be the observed and expected frequencies.

Here first two and last two expected frequencies are less than 4 and are pooled with adjacent frequencies :

∴ χ^{2}_{cal} = 100.932

For (n – 1) = (4 – 1) = 3 d.f at 5% level of significance the upper Tail critical value K/K_{2} = 7.81

Here χ^{2}_{cal} = 100.932 is in rejection region/ χ^{2}_{cal} > K_{2}

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : Binomial distribution is not a good fit.

Question 12.

For the following data fit a Poisson distribution and test whether it is a good fit (test at 5% level of significance) (10 marks)

Answer:

Let x be the no. of T. V. Sold is a Poisson variate. The parameter λ can be obtained as below:

Let f be the no. of days.

i.e., λ

λ = 1.9

Theoretical/expected frequency T_{x} = P(x) × N.

∴ T(x = o) = p(x= o) × 150

\(=\frac{\mathrm{e}^{1.9}(1.9)^{0}}{0 !} \times 150\)

= 0.1496 × 150

T_{0} = 22.44

using recurring relation for theoretical frequency

∴ T(x – 1) = \(\frac { 1.9 }{ 1 }\) T_{1-1} = \(\frac { 1.9 }{ 1 }\) × 22.44

similarly

T_{1} = 42.64

∴ The fitted poisson distribution is :

Chi- square test:

H_{0}: Poisson distribution is good fit

H_{1} : Poisson distribution is not a good fit.

The test statistic is –

Since the parameter X is estimated so, one more d.f is lessened.

Let ‘O’ and ‘E’ be the observed and expected frequencies.

∴ χ^{2}_{cal} = 1.938

For (n – 2) = (6 – 2) = 4 d.f at 5% level of significance the upper tail critical value K_{2} = 9.49

Here χ^{2}_{cal} = 1.938 is in acceptance region. (χ^{2}_{cal} < K)

∴ H_{0} is accepted.

Conclusion : Poisson distributions is good fit.

Question 13.

From the following data regarding the intelligence of fathers and sons, test at 1% level of significance, whether father’s intelligence and son’s intelligence are independent.

Answer:

H_{0} : Father’s intelligence and sons intelligence are independent

H_{1} : Father’s intelligence and sons intelligence are dependent.

The test statistic is – {Right/upper tail test}

The given data can be represented in 2 × 2 table as

For 1 d.f. at 5% level of significance the appear

Tail critical value K = 3.84

Here χ^{2}_{cal} is in rejection region χ^{2}_{cal} > χ^{2}_{tab}

H_{0} is rejected and H_{1} is accepted.

Conclusion: Father’s intelligence and sons intelligence are dependent.

Question 14.

To test the effectiveness of inoculation against Cholera the following table was obtained:

Test at 5% level of significance whether inoculation and attack of Cholera are independent.

Answer:

H_{0} : Inoculation and attack of cholera are independent

H_{1} : Inoculation and attack of cholera are dependent

The given data can be presented in the form of 2 × 2 contingency table as below :

The test statistic is

χ^{2}_{cal} = 0.6349

For 1 d.f at 5% level of significance the upper tail critical value k= 3.84

Here χ^{2}_{cal} < k / χ^{2}_{cal} lies in (AR) acceptance region.

∴ H_{0} is accepted.

Conclusion: Inoculation and attack of cholera are independent.

Question 15.

The members of a driving school examined the result of 200 candidates who where taking their driving test for the first time. They found that, out of 90 men, 62 passed and out of 110 women, 49 passed. Do these results indicate at 5% level of significance a relationship between the sex of a candidate and the ability to pass the driving test for the first time?(Use χ^{2} test).

Answer:

H_{0}: Sex and ability to pass the driving test are independent.

H_{1} : Sex and ability to pass the driving test are dependent. {upper tail test}

The test statistic is –

The given data can be presented in 2 × 2 contigency table as below:

For ld.f. at 5% level of significance the upper tail critical value K = 3.84

Here χ^{2}_{cal} = 11.88 is in rejection region/ χ^{2}_{cal} > χ^{2}_{tab}

∴ H_{0} is rejected and H_{1} is accepted

Conclusion : sex and ability to pass the driving test are dependent.

Question 16.

Out of the 75 patients admitted to a hospital with Hepatities symtoms, 40 were administered a certain new drug, of these 34 recovered. Of the remaining who were give the traditional drug, 19 recovered. Test at 5% and 1% level of significance that the new drug is an improvement over the traditional drug.

Answer:

H_{0}: Type of drug and recovery from the disease are independent.

H_{1} : Type of drug and recovery from the disease are dependent.

The test statistic is –

The given data can be presented in 2 × 2 contigency table as below:

χ^{2} = \(\frac{75(34 \times 16-6 \times 19)^{2}}{40 \times 35 \times 53 \times 22}\)

χ^{2}_{cal} = 8.495

At 5% level: for 1 d.f at 5% level of significance the upper tail critical value is K = 3.84.

Here χ^{2}_{cal} = 8.495 is in rejection region/χ^{2}_{cal} > χ^{2}_{tab}

∴ H_{0} rejected and H_{1} is accepted.

At 1% level : Here K = 6.63

Here χ^{2}_{cal} = 8.495 is in rejection region/ χ^{2}_{cal} > χ^{2}_{tab}

∴ H_{0} rejected and H_{1} is accepted.

Conclusion: type of drug and recovery from the disease are dependent, t – Tests.

Question 17.

A random sample of 8 envelopes are taken from a letter box of a post office and their weights in grams are found to be 12.1, 11.9, 12.4, 12.3, 11.9, 12.1, 12.4, 12.1. Does this sample indicate an average weight of the envelopes received at that post office is less than 12.35 gms at 1% level of significance.

Answer:

Given: n=8, µ = 12.35, a= 1%

Here n is small use t-test (i.e., n < 30)

H_{0} : Average weight of the envelops is less than 12.35 gms.

i.e., H_{0} : µ = 12.35 gms.

H_{1} : Average weight of the envelope is less than 12.53 gms. i.e.,

H_{1} : (µ< 12.35 gms {left Tailed test}

The test statistic is –

Here, S = Sample S.D.

X̄ = Sample mean.

Let X be the weight in gms

\(=\sqrt{\frac{0.78}{8}-\left(\frac{2}{8}\right)^{2}}\)

∴ s = 0.187

\(t=\frac{12.15-12.35}{0.187 / \sqrt{8-1}}\)

∴ t_{cal} = -2.83

For (n – 1) = (8 – 1) = 7. d.f the lower left tail crtical value, (from t-table) at 5% level of significance is -k_{1} =-1.90

Here t_{cal} = -2.83 is in rejection region (t_{cal} < t_{tab})

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : Average weight of envelops is less than 12.35 gms.

Question 18.

The following table gives the lengths of 12 samples of Egyptian cotton taken from a consginment. 48, 46, 49, 46, 52, 45, 43, 47, 47, 46, 45, 50. Test if the mean length of the consignment can be taken as more than 46.

Answer:

Given: n=12, µ = 46, Here n is small use t-test (i.e., n < 30)

H_{0}: Mean length of consignment of cotton is 46.

i.e., H_{0}: µ = 46.

H_{1} : Mean length of consignment of cotton is more than 46.

i.e., H_{1} : µ > 46. {Upper/right tailed test} .

The test statistic is –

Where x̄ = sample mean and

s = sample standard deviation.

‘

\(=\sqrt{\frac{0.78}{8}-\left(\frac{2}{8}\right)^{2}}\)

∴ S =0.187.

\(t=\frac{12.15-12.35}{0.187 / \sqrt{8-1}}\)

∴ t_{cal} = -283.

For (n – 1) = (8 – 1) = 7. d.f the lower left tail crtical value, (from t-table) at 5% level of significance is -k_{1} = -1.90

Here t_{cal} = -2.83 is in rejection region (t_{cal} < t_{tab>})

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : Average weight of envelops is less than 12.35 gms.

Question 19.

The following table gives the lengths of 12 samples of Egyptian cotton taken from a consginment. 48, 46, 49, 46, 52, 45, 43, 47, 47, 46, 45, 50. Test if the mean length of the consignment can be taken as more than 46.

Answer:

Given: n = 12, µ = 46, Here n is small use t-test (i.e., n < 30)

H_{0}: Mean length of consignment of cotton is 46.

i.e., H_{0} : µ = 46.

H_{1} : Mean length of consignment of cotton is more than 46.

i.e., H, : µ > 46. {Upper/right tailed test} .

The test statistic is –

Where x̄ = sample mean and

s = sample standard deviation

= 43 + \(\frac { 48 }{ 12 }\) = 47

∴ s = 2.345

\(t=\frac{47-46}{2.345 / \sqrt{12-1}}\)

∴ t_{cal} = 1.41

For (n – 1) = (12 – 1) = 11 d.f at 5% level of significance the upper/right tail critical value

k_{2}=1.80

Here t_{cal} = 1.41 is in acceptance region (i.e., t_{cal} < t_{tab})

∴ H_{0} is accepted.

Conclusion : Mean length of the consignment of cotton is 46.

Question 20.

It is required to test whether those who practice yoga have average blood sugar less than 120. A sample of consisting of 18 persons who practice yoga is observed. If their mean blood sugar is 108 and S.D is 8, what would you conclude ?

Answer:

Given: n=18, x̄ =108, S = 8, µ =120.

H_{0} : The average blood sugar is 120.

i.e., H_{0}: µ = 120.

H_{1} : The average blood sugar is less than 120.

i.e H_{1}: µ < 120 {left tailed test}

The test statistic is –

\(t=\frac{108-120}{s / \sqrt{18-1}}\)

t _{cal} = -6.18

For (n- 1) = (18 – 1)= 17 d.f at 5% level of significance the left/upper lower Tail critical value -k = -1.74

Here t_{cal} = -6.18 is in rejection region. (t_{cal} < t_{tab})

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : The average blood sugar is less than 120.

Question 20.

It was found that a machine has produced pipes having a thickness 0.50 mm. To determine whether the machine is in proper working order a sample of 10 pipes is chosen for which the mean thickness is 0.53 mm and S.D. is 0.03 mm. Test the hypothesis that the machine is in proper working order using a level of significance of 0.01.

Answer:

Given: n= 10, x̄ =0.53, μ = 0.50, S = 0.03, α = 0.01 = l%

H_{0}: The machine is in proper working order,

i.e., H_{0}: µ = 0.50

H_{1} : The machine is not in proper working order.

i.e., H_{1} : µ ≠ 0.50 {two tailed test}

The test statistic is –

\(t=\frac{0.53-0.50}{0.03 / \sqrt{10-1}}\)

For (n-1) = (10 -1) = 9 d.f at 1% level of significance the two tail critical values are

k_{1} =-3.25 and k_{2} = 3.25

Here t_{cal} = 3 is in acceptance region.

∴ H_{0} is accepted.

Conclusion : The machine is in proper working order.

Question 21.

To compare the prices of a certain product in two cities ten shops were selected at

Test whether the average prices can be said to be same in the towns.

Answer:

Given: n_{1} = 10 n_{2} = 10.;

{Here n_{1} n_{2} are less than 30 use small sample test for equality of meAnswer:}

H_{0} : The average prices in two cities are same,

i.e., H_{0} µ_{1} – µ_{2}.

H_{1} : The average prices in two cities are not same.

i.e., H_{1} : µ_{1} ≠ µ_{2} {two tailed test}

The test statistic is –

where

x̄_{1} , x̄_{2} – sample means of city A & B

S_{1}^{2}, S_{2}^{2} – sample variances of city A & B

Let x and y be the prices of city A and B. For City A

\(=45+\frac{124}{10} \quad=45+\frac{107}{10}\)

x̄_{1} = 57.4 x̄_{2} = 55.7

\(=\frac{1786}{10}-\left(\frac{124}{10}\right)^{2}\)

S_{1}^{2} = 24.84

S_{2}^{2} = 25.01

t_{cal} = 0.723

For (n_{1} + n_{2} – 2) = (10 + 10 -2) = 18 d.f at 5% level of sigficance the two Tail critical values are -k_{1} = -2.10 and k_{2} = 2.10.

Here t_{cal} = 0.723 is in acceptance region.

∴ H_{0} is accepted

Conclusion.: The average prices in two cities are same.

H_{1} : The increase in weight due to diet A is more than diet B.

i.e., : H_{1} : µ_{1} > µ_{2}r {Right tailed test}

The test statistic is –

Where x̄_{1}, x̄_{2} – Sample means of diet A and diet B and S_{1}^{2}, S_{1}^{2} – Sample variances of Diet A and B.

Let x and y be the weights of diet A and B.

= 2 + \(\frac { 54 }{ 10 }\) = 2 + \(\frac { 32 }{ 8 }\)

x̄_{1} = 7.4 x̄_{2} = 6

\(=\frac{394}{10}-\left(\frac{54}{10}\right)^{2} \quad=\frac{208}{8}-\left(\frac{32}{8}\right)^{2}\)

S_{1}^{2} = 10.24 s_{2}^{2} = 10.

Question 22.

From the following data regarding weights of randomly selected boys and girls of P.U.C. classes, test whether P.U.C. boys weigh more than P.U.C. girls. [Test at 5% level of significance]

Answer:

Given n_{1} = 10, n_{2} = 12 (less than 30, use t test)

x̄_{1} = 71, x̄_{2} = 69

s_{1}^{2} = 3 s_{2}^{2} = 2

H_{0} :mean weight of boys and girls are same,

(i.e. H_{0}: µ_{1}= µ_{2})

H_{1} : mean weight of boys is more than girls

(i.e. H_{1}: µ_{1} > µ_{2}) …(upper tail test)

The test statistic is

For (n_{1} + n_{2} – 2) = (10 + 12 -2) = 20 degrees of freedom the upper tail critical value for 5% level of significance k_{2} = 1.73

Herz t_{cal} > k_{2}/t_{cal} lies in (RR) rejection region.

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : Mean weight of boys is more than girls.

Question 23.

A group of 10 rabbits fed on diet A and another group of 8 rats fed on a different diet B recorded the following increase in weig it in gms.

Test whether diet A is superior to diet B.

Answer:

Given : n_{1} =10, n_{2} = 8. {Here n_{1},n_{2} < 30 use t-test.}

H_{0} : There is no significant difference between mean weight due to diet A and B.

i.e., : H_{0} : µ_{1} = µ_{2}.

H_{1} : The increase in weight due to diet A is more than diet B.

i.e., : H_{1}: µ_{1} > µ_{2} {Right tailed test}

The test statistic is –

Where x̄_{1}, x̄_{2} – Sample means of diet A and diet B and S_{1}^{2}, S_{2}^{2} – Sample variances of Diet A and B.

Let x and y be the weights of diet A and B.

= 2 + \(\frac { 54 }{ 10 }\) = 2 \(\frac { 32 }{ 8 }\)

x̄̄_{1} = 7.4 x̄_{2} = 6

\(=\frac{394}{10}-\left(\frac{54}{10}\right)^{2} \quad=\frac{208}{8}-\left(\frac{32}{8}\right)^{2}\)

s_{1}^{2} = 10.24 s_{2}^{2} = 10

t_{cal} =0.874

For (n_{1} + n_{2} – 2) = (10 + 8 – 2) = 16 d.f at 5% level of significance the upper tail critical

value k= 1.75

Here t_{cal} = 0.874 is in acceptance region (t_{cal} < t_{tab})

∴ H_{0} is accepted.

Conclusion : There is no siginificant difference betwee mean weight due to diet A and B.

Question 24.

Two types of batteries are tested for their length of life and the following data is obtained. Is there a significant difference in the two means?

Answer:

Given:

n_{1} = 9, x̄_{1} =500, S_{1}^{2} = 100

n_{2} = 8, x̄_{2} =540, S_{2}^{2} = 121 :

H_{0} : There is no significant difference in the two means,

i.e., : H_{0}: µ_{1}, = µ_{2}

H_{1} : There is a significant difference in the two meAnswer:

i.e., : H_{1} : µ_{1} ≠ µ_{2}.

The test statistic is –

For (n_{1} + n_{2} – 2) = (9 + 8 – 2) = 15 d.f. at 5%

level of signficance the two tail critical values are -k_{1} = -2.13 and k_{2} = 2.13.

Here t_{cal} = -7.38 is in rejection region.

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion: There is no significant difference in the two meAnswer:

Question 25.

The following are the B.P. (Blood Pressure) of 8 persons before and after performing ‘yoga’.

Can we conclude that yoga reduces BP?

Answer:

Here B.P. before and after yoga of the same persons are given, so use paired t-test for equality of means :

H_{0} : B.P. before and after yoga is same.

i.e., H_{0}: µ_{1} = µ_{2}

H_{1} : BP reduced after yoga.

i.e., H_{1} : µ_{1} > µ_{2} {Right Tailed test}

Let x and y be the BP before and after yoga.

∴ d̅ = \(\frac { 38 }{ 8 }\)

\(S=\sqrt{\frac{252}{8}-\left(\frac{38}{8}\right)^{2}}=2.99\)

\(t=\frac{4.75}{2.99 / \sqrt{8-1}}\)

∴ t_{cal} = 4.203

For (n – 1) = 8 – 1 = 7 d.f. at 5% level of significance the upper tail critical value k = 1.90.

Here t_{cal} = 4.203 is in rejection region (t_{cal} > t_{tab})

∴ H_{0} is rejected and H_{1} is accepted.

Conclusion : B.P reduced after yoga.

Question 26.

The following are the marks obtained by 5 students before and after attending coaching classes.

Test whether marks of students has improved after attending coaching classes.

Answer:

Here marks of same students before and after attending coaching class has given so use paired t-test for equality of meAnswer:

H_{0} : There is no change in marks of the students before and after attending coaching class,

i.e., H_{0}: µ_{1} = µ_{2}

H_{1} : The marks of the students have improved after attending coaching class i.e.,

H_{1} : µ_{1}, < µ_{2} {Left Tail test}

The test statistic is –

Let x and y be the marks of students before and after attending coaching classes

∴ d̅ = \(\frac { -10 }{ 5 }\) = 2

s = \(\sqrt{\frac{140}{5}-\left(\frac{-10}{5}\right)^{2}}\)

s = 4.89

\(t=\frac{-2}{4.89 / \sqrt{5-1}}\)

∴ t_{cal} = -0.818

For (n – 1) = (5 – 1) = 4 d.f the lower critical value at 5% level of significance

-k = -2.13

Here t_{cal} = -0.818 is in acceptance region.

∴ H_{0} is accepted.

Conclusion : There is no change in the marks of the students before and after attending coaching classes.

Question 27.

The manufacturer of a cattle feed claims that, cows fed on his product yield more milk. To substaintiate his claim, 8 cows were fed on the manufacturer’s feed and their milk yield in litres. When they were under usual feed were given below:

Answer:

Here milk yield of the same set of cows are given, use paired t-test for equality of meAnswer:

H_{0} : Milk yield of cows before and after manufacturers feed is same.

i.e., H_{0}: µ_{1} = µ_{2}.

H_{1} : Milk yield of cows after manufacturers feed is more.

i.e., H_{1} : µ_{1} < µ_{2} {left tailed test}

The test statistic is –

∴ d̅ = \(\frac { -10 }{ 5 }\) = -1.25

s = \(\sqrt{\frac{31.7}{8}-\left(\frac{-10}{8}\right)^{2}}\)

s = 1.55

\(t=\frac{1.25}{1.55 / \sqrt{8-1}}\)

t_{cal} = 2.13

For (n – 1) = (8 – 1) = 7 d.f. The upper tail critical value at 5% level of significance is k = 1.90

Here t_{cal} = -2.13 is in rejection region./ t_{cal} < K

∴ H_{0} is rejected and H_{1} accepted.

Conclusion : Milk yield of cows after manufacturers feed has increased.