Question

A sample of 28 items provides a sample standard deviation of 10. Test the following hypotheses using a = 0.5. What is your conclusion? Use both the p-value approach and th critical value approach.

H0 : 2 <_ 54

Ha : 2 > 54

What is the p-value and conclusion? What is the critical value (x^2 .05) and conclusion?

Answer 1

Let's write the given information.

Sample standard deviaion = s = 10

So sample variance = 10² = 100

Level of significance = = 005

The given null hypothesis ( H₀ ) and the alternative hypothesis ( H_a) from the above claim is as folloes:

Let's use minitab:

Step 1) Click on Stat>>>Basic Statistics >>1 variance...

Data : Select "sample variance"

Sample size: n = 28

Sample variance: 100

Then select "Perform hypothesis test"

Look the following image:

Step 2) Click on Option

Confidence level = 95

Alternative: greater than

Then click on OK

Again Click on OK

So we get the following output:

From the above output, we get

p- value = 0.005

- test statistic value = 50.00

Let's find critical value in minitab:

Click on Graph >>> Probability distribution plot >>> Select fourth image >>> OK

Distribution:

Chi-square

degrees of freedom = 28 - 1 = 27

Look the following image:

Then click on Shaded area:

Look the following image:

Then click on OK, so we get the following output:

From the above output, the criticl value = 40.11

Decision rule based on critical value.

1) If - test statistic value > -critical value then we reject the null hypothesis.

2) If - test statistic value < -critical value then we fail to reject the null hypothesis.

Here - test statistic value = 50.00 > -critical value = 40.11, so we use first rule.

That is we reject the null hypothesis.

Decision rule based on p-value:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value =0.005 which is less than 0.05 so we used first rule.

That is we reject null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to conclude that the population variance is greater than 54.