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¯ =4.6d¯ =4.6 of and a sample standard deviation of sd = 7.6.

(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)YesNo

(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 valuesno test values0.10.1,and 0.0010.05  (Click to select)nosomestrongvery strongextremely strong evidence that µ1 differs from µ2.

(c) The p-value for testing H0: µd < 3 versus Ha: µd > 3 equals .0735. 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 values0.050.10 and 0.05.10 .05 .01 and .0010.05 and 0.01, (Click to select)Very strongextremely strongsomeStrongNo evidence that µ1 and µ2 differ by more than 3.

rev: 07_14_2017_QC_CS-93578, 12_08_2018_QC_CS-150993

Answer 1

(a)

n = 49

= 4.6

s_d = 7.6

SE = s_d/

= 7.6/

= 1.0857

= 0.05

ndf = n - 1 = 49 - 1 = 48

From Table, critical values of t = 2.0106

Confidence Interval:

4.6 (2.0106 X 1.0857)

= 4.6 2.1829

= ( 2.42 ,6.78)

Confidence Interval:

2.42 <   <    6.78

Since 0 is not included in the Confidence Interval, we can be 95 percent confident that the difference between µ1 and µ2 is greater than 0.

So,

Answers are:

[ 2.42 ,6.78]

Yes

(b)

(i)

= 0.10

ndf = n - 1 = 49 - 1 = 48

From Table, critical values of t = 1.6772

Test Statistic is given by:

t = 4.6/1.0857

= 4.237

Since calculated value of t = 4.237 is greater than critical value of t = 1.6772, the difference is significant. Reject null hypothesis. There is sufficient evidence that differs from 0.

(ii)

= 0.05

ndf = n - 1 = 49 - 1 = 48

From Table, critical values of t = 2.0106

Test Statistic is given by:

t = 4.6/1.0857

= 4.237

Since calculated value of t = 4.237 is greater than critical value of t = 2.0106, the difference is significant. Reject null hypothesis. There is sufficient evidence that differs from 0.

(iii)

= 0.01

ndf = n - 1 = 49 - 1 = 48

From Table, critical values of t = 2.6822

Test Statistic is given by:

t = 4.6/1.0857

= 4.24

Since calculated value of t = 4.24 is greater than critical value of t = 1.6822, the difference is significant. Reject null hypothesis. There is sufficient evidence that differs from 0.

(iv)

= 0.001

ndf = n - 1 = 49 - 1 = 48

From Table, critical values of t = 3.5051

Test Statistic is given by:

t = 4.6/1.0857

= 4.24

Since calculated value of t = 4.24 is greater than critical value of t = 3.5051, the difference is significant. Reject null hypothesis. There is sufficient evidence that differs from 0.

Cohen's d is given by:

d = 4.6/7.6

= 0.6053

Since d value is near 0.5, we conclude that there is "medium" effect size".

So, Answers are:

t = 4.237

Reject H0 at all test values. Some strong evidence evidence that 1 differs from 2.

(c)

(i)

= 0.10

Since p - value = 0.0735 is less than = 0.10, the difference is significant. Reject null hypothesis. There is sufficient evidence that exceeds 3.

(ii)

= 0.05

Since p - value = 0.0735 is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis. There is no sufficient evidence that exceeds 3.

(iii)

= 0.01

Since p - value = 0.0735 is greater than = 0.01, the difference is not significant. Fail to reject null hypothesis. There is no sufficient evidence that exceeds 3.

(iv)

= 0.001

Since p - value = 0.0735 is greater than = 0.001, the difference is not significant. Fail to reject null hypothesis. There is no sufficient evidence that exceeds 3.

So,

Answers are:

t = 4.237;   p - value = 0.0735

Reject H0 at = 0.10. Some strong evidence evidence that exceeds 3.