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Chapter: 12th Statistics : Chapter 2 : Tests Based on Sampling Distributions I

Student’s t Distribution and Its Applications

General Procedure for Student’s t Distribution and Its Applications, Properties, Procedure Steps, Example Solved Problems

STUDENT’S t DISTRIBUTION AND ITS APPLICATIONS

 

1. Student’s t-distribution

If X~N(0,1) and Y~χn2 are independent random variables, then


is said to have t-distribution with n degrees of freedom. This can be denoted by tn.

Note 1: The degrees of freedom of t is the same as the degrees of freedom of the corresponding chi-square random variable.

Note 2: The t-distribution is used as the sampling distribution(s) of the statistics(s) defined based on random sample(s) drawn from normal population(s).

(i) If X1, X2 , …, Xn is a random sample drawn from N (μ, σ2) population then


(ii) If (X1 , X2 , …, Xm) and (Y1 , Y2 , …, Yn) are independent random samples drawn from N (μ, σ2) and N (μY , σ2) populations respectively, then


(iii) If (X1 , Y1), (X2 , Y2), …, (Xn , Yn) is a random sample of n paired observations drawn from a bivariate normal population, then Di = Xi – Yi , i = 1, 2, …, n is a random sample drawn from N (μD , σD2). Here μD = μX μY.

Hence,


 

2. Properties of the Student’s t-distribution

1. t–distribution is symmetrical distribution with mean zero.

2. The graph of t-distribution is similar to normal distribution except for the following two reasons:

(i) The normal distribution curve is higher in the middle than t-distribution curve.

(ii) t–distribution has a greater spread sideways than the normal distribution curve. It means that there is more area in the tails of t-distribution.


3. The t-distribution curve is asymptotic to X-axis, that is, it extends to infinity on either side.

4. The shape of t-distribution curve varies with the degrees of freedom. The larger is the number of degrees of freedom, closeness of its shape to standard normal distribution (fig. 2.1).

5. Sampling distribution of t does not depend on population parameter. It depends on degrees of freedom (n–1).

 

3. Applications of t-distribution

The t-distribution has the following important applications in testing the hypotheses for small samples.

1. To test significance of a single population mean, when population variance is unknown, using T1.

2. To test the equality of two population means when population variances are equal and unknown, using T2.

3. To test the equality of two means – paired t-test, based on dependent samples, T3.

 

4. Test of Hypotheses for Normal Population Mean (Population Variance is Unknown)

Procedure:

Step 1 : Let µ and σ2 be respectively the mean and variance of the population under study, where σ2 is unknown. If µ0 is an admissible value of µ, then frame the null hypothesis as

H0: µ = µ0 and choose the suitable alternative hypothesis from

(i) H1: µ ≠ µ0 (ii) H1: µ > µ0 (iii) H1: µ < µ0

Step 2 : Describe the sample/data and its descriptive measures. Let ( X1, X2, …, Xn) be a random sample of n observations drawn from the population, where n is small (n < 30).

Step 3 : Specify the level of significance, α.

Step 4 : Consider the test statistic , under H0, where  and S are the sample mean and sample standard deviation respectively. The approximate sampling distribution of the test statistic under H0 is the t-distribution with (n–1) degrees of freedom.

Step 5 : Calculate the value of t for the given sample ( x1 , x2 ,... xn ) as . here  is the sample mean and s =  is the sample standard deviation.

Step 6 : Choose the critical value, te, corresponding to α and H1 from the following table


Step 7 : Decide on H0 choosing the suitable rejection rule from the following table corresponding to H1.


 

Example 2.1

The average monthly sales, based on past experience of a particular brand of tooth paste in departmental stores is ₹ 200. An advertisement campaign was made by the company and then a sample of 26 departmental stores was taken at random and found that the average sales of the particular brand of tooth paste is 216 with a standard deviation of 8. Does the campaign have helped in promoting the sales of a particular brand of tooth paste?

Solution:

Step 1 : Hypotheses

Null Hypothesis H0: µ = 200

i.e., the average monthly sales of a particular brand of tooth paste is not significantly different from ₹ 200.

Alternative Hypothesis H1: µ > 200

i.e., the average monthly sales of a particular brand of tooth paste are significantly different from ₹ 200. It is one-sided (right) alternative hypothesis.

Step 2 : Data

The given sample information are:

Size of the sample (n) = 26. Hence, it is a small sample.

Sample mean ( ) = 216, Standard deviation of the sample = 8.

Step 3 : Level of significance

α = 5%

Step 4 : Test statistic

The test statistic under H0 is T = 

Since n is small, the sampling distribution of T is the t-distribution with (n–1) degrees of freedom.

Step 5 : Calculation of test statistic

The value of T for the given sample information is calculated from


Step 6 : Critical value

Since H1 is one-sided (right) alternative hypothesis, the critical value at α =0.05 is

te = tn-1, α =t25,0.05 = 1.708

Step 7 : Decision

Since it is right-tailed test, elements of critical region are defined by the rejection rule t0 >te = tn-1, α = t25,0.05 = 1.708. For the given sample information t0 = 10.20 > te =1.708. It indicates that given sample contains sufficient evidence to reject H0. Hence, the campaign has helped in promoting the increase in sales of a particular brand of tooth paste.

 

Example 2.2

A sample of 10 students from a school was selected. Their scores in a particular subject are 72, 82, 96, 85, 84, 75, 76, 93, 94 and 93. Can we support the claim that the class average scores is 90?

Solution:

Step 1 : Hypotheses

Null Hypothesis H0: µ = 90

i.e., the class average scores is not significantly different from 90.

Alternative Hypothesis H1 : µ ≠ 90

i.e., the class means scores is significantly different from 90.

It is a two-sided alternative hypothesis.

Step 2 : Data

The given sample information are

Size of the sample (n) = 10. Hence, it is a small sample.

Step 3 : Level of significance

 α= 5%

Step 4 : Test statistic

The test statistic under H0 is T =  

Since n is small, the sampling distribution of T is the t - distribution with (n–1) degrees of freedom.

Step 5 : Calculation of test statistic

The value of T for the given sample information is calculated from t0 as under:


Sample mean


 

Step 6 : Critical value

Since H1 is two-sided alternative hypothesis, the critical value at α  = 0.05 is te = tn-1, α/2 = t9,0.025 = 2.262

Step 7 : Decision

Since it is two-tailed test, elements of critical region are defined by the rejection rule |t0| > te = t n-1, α/2 = t =2.262. For the given sample information |t0| = 1.806 < te = 2.262.

It indicates that given sample does not provide sufficient evidence to reject H0. Hence, we conclude that the class average scores is 90.

 

5. Test of Hypotheses for Equality of Means of Two Normal Populations (Independent Random Samples)

Procedure:

Step 1 : Let μX and μY be respectively the means of population-1 and population-2 under study. The variances of the population-1 and population-2 are assumed to be equal and unknown given by σ2.

Frame the null hypothesis as H0 : µX = µY and choose the suitable alternative hypothesis from (i) H1 : μX ≠ μY (ii) H1 : μX > μY (iii) H1 : μX < μY

Step 2 : Describe the sample/data. Let (X1, X2 , …, Xm) be a random sample of m observations drawn from Population-1 and (Y1, Y2 , …, Yn) be a random sample of n observations drawn from Population-2, where m and n are small (i.e., m < 30 and n < 30). Here, these two samples are assumed to be independent.

Step 3 : Set up level of significance (α)

Step 4 : Consider the test statistic


where Sp is the “pooled” standard deviation (combined standard deviation) given by


The approximate sampling distribution of the test statistic


 is the t-distribution with m+n–2 degrees of freedom i.e., t ~ tm+n–2.

Step 5 : Calculate the value of T for the given sample ( x1 , x2 ,... xm ) and ( y1 , y 2 ,... yn ) as


Step 6 : Choose the critical value, te, corresponding to α and H1 from the following table


Step 7 : Decide on H0 choosing the suitable rejection rule from the following table corresponding to H1.


 

Example 2.3

The following table gives the scores (out of 15) of two batches of students in an examination.


Test at 1% level of significance the average performance of the students in Batch I and Batch II are equal.

Solution:

Step 1 : Hypotheses: Let µX and µY denote respectively the average performance of students in Batch I and Batch II. Then the null and alternative hypotheses are :

Null Hypothesis H0 : µ X = µY

i.e., the average performance of the students in Batch I and Batch II are equal.

Alternative Hypothesis H1 : µ X µY

i.e., the average performance of the students in Batch I and Batch II are not equal.

Step 2 : Data

The given sample information are:

Sample size for Batch I : m =10

Sample size for Batch II : n = 8

Step 3 : Level of significance

α= 1%

Step 4 : Test statistic

The test statistic under H0 is


The sampling distribution of T under H0 is the t-distribution with m+n–2 degrees of freedom i.e., t ~ tm+n–2

Step 5 : Calculation of test statistic

To find sample mean and sample standard deviation:


To find sample means:

Let (x1 , x2 ,..., x10) and (y1, y2 ,..., y8) denote the scores of students in Batch I and Batch II respectively.


To find combined sample standard deviation:


Pooled standard deviation is:


The value of T is calculated for the given information as


Step 6 : Critical value

Since H1 is two-sided alternative hypothesis, the critical value at α = 0.01 is te = tm+n-2, α/2 = t16,0.005 = 2.921

Step 7 : Decision

Since it is two-tailed test, elements of critical region are defined by the rejection rule |t0| < te = tm+n-2, α = t16,0.005 = 2.921. For the given sample information |t0| = 1.3957 < te = 2.921.

It indicates that2 given sample contains insufficient evidence to reject H0. Hence, the mean performance of the students in these batches are equal.

 

Example 2.4

Two types of batteries are tested for their length of life (in hours). The following data is the summary descriptive statistics.


Is there any significant difference between the average life of the two batteries at 5% level of significance?

Solution:

Step 1 : Hypotheses

Null Hypothesis H0 : μX = μY

i.e., there is no significant difference in average life of two types of batteries A and B.

Alternative Hypothesis H0 : μX μY

i.e., there is significant difference in average life of two types of batteries A and B. It is a two-sided alternative hypothesis

Step 2 : Data

The given sample information are :

m = number of batteries under type A = 14

n = number of batteries under type B = 13

 = Average life (in hours) of type A battery = 94

 = Average life (in hours) of type B battery = 86

sX = standard deviation of type A battery =16

sY = standard deviation of type B battery = 20

Step 3 : Level of significance

 α= 5%

Step 4 : Test statistic

The test statistic under H0 is


The sampling distribution of T under H0 is the t-distribution with m+n–2 degrees of freedom i.e., t ~ tm+n–2

Step 5 : Calculation of test statistic

Under null hypotheses H0:


where s is the pooled standard deviation given by,


The value of T is calculated for the given information as


Step 6 : Critical value

Since H1 is two-sided alternative hypothesis, the critical value at α = 0.05 is te = tm+n-2, α/2 = t25, 0.025 = 2.060.

Step 7 : Decision

Since it is a two-tailed test, elements of critical region are defined by the rejection rule |t0| < te = tm+n-2,α/2 = t25, 0.025 = 2.060. For the given sample information |t0| = 1.15 < te = 2.060. It indicates that2 given sample contains insufficient evidence to reject H0. Hence, there is no significant difference between the average life of the two types of batteries.

 

6. To test the equality of two means – paired t-test

Procedure:

Step 1 : Let X and Y be two correlated random variables having the distributions respectively N(μX , σX2) (Population-1) and N(μY , σY2) (Population-2). Let D = XY, then it has normal distribution N(μD = μXμY, σD2).

Frame null hypothesis as

H0 : μD= 0

And choose alternative hypothesis from

(i) H1 : μD 0 (ii) H1 : μD> 0 (iii) H1 : μD< 0

Step 2 : Describe the sample/data. Let (X1, X2, …, Xm) be a random sample of m observations drawn from Population-1 and (Y1, Y2, …, Yn) be a random sample of n observations drawn from Population-2. Here, these two samples are correlated in pairs.

Step 3 : Set up level of significance (α)

Step 4 : Consider the test statistic


The approximate sampling distribution of the test statistic T under H0 is t - distribution with (n–1) degrees of freedom.

Step 5 : Calculate the value of T for the given data as


Step 6 : Choose the critical value, te, corresponding to α and H1 from the following table


Step 7 : Decideon H0 choosing the suitable rejection rule from the following table corresponding to H1.


 

Example 2.5

A company gave an intensive training to its salesmen to increase the sales. A random sample of 10 salesmen was selected and the value (in lakhs of Rupees) of their sales per month, made before and after the training is recorded in the following table. Test whether there is any increase in mean sales at 5% level of significance.


Solution:

Step 1 : Hypotheses

Null Hypothesis H0 : μD= 0

i.e., there is no significant increase in the mean sales after the training.

Alternative Hypothesis H1 : μD> 0

i.e., there is significant increase in the mean sales after the training. It is a one-sided alternative hypothesis.

Step 2 : Data

Sample size n = 10

Step 3 : Level of significance

 α= 5%

Step 4 : Test statistic

Test statistic under the null hypothesis is


The sampling distribution of T under H0 is t - distribution with (10–1) = 9 degrees of freedom.

Step 5 : Calculation of test statistic

To find  and s:

Let x denote sales before training and y denote sales after training


Here instead of di = xi – yi it is assumed di = yi – xi for calculations to be simpler.


Step 6 : Critical value

Since H0 is a one-sided alternative hypothesis, the critical value at 5% level of significance is te = tn-1, α =t9,0.05 = 1.833

Step 7 : Decision

It is a one-tailed test. Since |t0| = 1.6052 < te = tn-1, α = t9,0.05 = 1.833, H0 is not rejected.

Hence, there is no evidence that the mean sales has increased after the training.

 

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