Question

Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean   d⎯⎯ =4.3d¯ =4.3 of and a sample standard deviation of sd = 6.9.

(a) Calculate a 95 percent confidence interval for µd = µ1 – µ2. Can we be 95 percent confident that the difference between µ1 and µ2 is greater than 0? (Round your answers to 2 decimal places.)

Confidence interval = [  ,   ] ;   (Click to select)   Yes   No

(b) Test the null hypothesis H0: µd = 0 versus the alternative hypothesis Ha: µd ≠ 0 by setting α equal to .10, .05, .01, and .001. How much evidence is there that µd differs from 0? What does this say about how µ1 and µ2 compare? (Round your answer to 3 decimal places.)

t =

Reject H0 at ? equal to   (Click to select)   all test values   no test values   0.1   0.1,and 0.001   0.05      (Click to select)   no   some   strong   very strong   extremely strong   evidence that µ1 differs from µ2.

(c) The p-value for testing H0: µd < 3 versus Ha: µd > 3 equals .0967. Use the p-value to test these hypotheses with α equal to .10, .05, .01, and .001. How much evidence is there that µd exceeds 3? What does this say about the size of the difference between µ1 and µ2? (Round your answer to 3 decimal places.)

t =  ; p-value

Reject H0 at ? equal to   (Click to select)   no test values   0.05   0.10 and 0.05   .10 .05 .01 and .001   0.05 and 0.01  ,   (Click to select)   Very strong   extremely strong   some   Strong   No  evidence that µ1 and µ2 differ by more than 3.

Answer 1

The given information is,

The sample size (n) is 49

(a):

The degrees of freedom (df) = 49 – 1 = 48

At the significance level 0.05 and the degrees of freedom 48, the two tailed critical value obtained from the t-table is +/- 2.0106.

The 95percent confidence interval for µd = µ1 – µ2 can be calculated as,

Yes, we can be 95 percent confident that the difference between µ1 and µ2 is greater than 0 since, the value 0 does not lies between the confidence limits 2.3181 and 6.2819.

(b):

It is given that the null hypothesis H0: µd = 0 versus the alternative hypothesis Ha: µd ≠ 0.

The test statistic is,

Therefore, the test statistic is 4.3623.

If the test statistic (4.3623) is greater than the critical value, then the decision is to reject the null hypothesis which leads to statistically significant result.

The two tailed critical value obtained at the degrees of freedom 48 and the significance level (0.10, 0.05, 0.01, 0.001) from the t-table is shown below.

Significance level	Critical values	Decision	Evidence
0.10	+/- 1.2994	Reject Ho	Very strong evidence µ1 differs from µ2
0.05	+/- 2.0106	Reject Ho	Strong evidence µ1 differs from µ2
0.01	+/- 2.6822	Reject Ho	strong µ1 differs from µ2
0.001	+/- 3.5051	Reject Ho	some evidence µ1 differs from µ2

(c).

It is given that H0: µd < 3 versus Ha: µd > 3.

The p-value is 0.0967.

Decision: If the p-value is less than the significance level then the decision is to reject the null hypothesis which leads to statistically significant result.

Significance level	Decision	Evidence
0.10	Reject Ho	Some evidence that µ1 and µ2 differ by more than 3
0.05	Fail to reject Ho	No  evidence that µ1 and µ2 differ by more than 3
0.01	Fail to reject Ho	No  evidence that µ1 and µ2 differ by more than 3
0.001	Fail to reject Ho	No  evidence that µ1 and µ2 differ by more than 3