(a) Calculate a 95 percent confidence interval for µ d = µ 1 – µ 2 . Can we be 9

Question

(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 ? 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.)

(c) The p-value for testing H0: µd < 3 versus Ha: µd > 3 equals .1373. Use the p-value to test these hypotheses with ? 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 = Reject H0 at ? equal to (Click to select)all test values0.1,and 0.001no test values0.10.05  (Click to select)somevery strongnoextremely strongstrong evidence that µ1 differs from µ2. a? = 4.2

Explanation / Answer

Solution:

We are given,

d-bar = 4.2

Sd = 7.6

n = 49

df = n – 1 = 49 – 1 = 48

Part a

Here, we have to find 95% confidence interval for difference between two population means.

Confidence interval formula is given as below:

Confidence internal = d-bar -/+ t*Sd/sqrt(n)

Where t is the two tailed critical value for 95% confidence level and n is sample size.

Degrees of freedom = n – 1 = 49 – 1 = 48

Critical value = t = 2.0106

Confidence interval = 4.2 -/+ 2.0106*7.6/sqrt(49)

Confidence interval = 4.2 -/+ 2.182937

Lower limit = 4.2 - 2.182937 = 2.017063

Upper limit = 4.2 + 2.182937 = 6.382937

Confidence interval = (2.02, 6.38)

Yes, we are 95% confident that the difference between µ1 and µ2 is greater than 0.

Part b

Here, we have to use t test for difference between two population means.

Null hypothesis: H0: µd = 0

Alternative hypothesis: Ha: µd 0

Test statistic = t = d-bar / [Sd/sqrt(n)]

Test statistic = t = 4.2/[7.6/sqrt(49)]

Test statistic = t = 3.868

Critical values = 2.0106 and -2.0106

P-value = 0.0003

For = 0.10, 0.05, 0.01, and 0.001, p-value is less.

So, we reject the null hypothesis H0

There is insufficient evidence to conclude that average paired difference is zero.

There is strong evidence that µ1 differs from µ2.

Part c

We are given

Test statistic = t = d-bar / [Sd/sqrt(n)]

Test statistic = t = 3/[7.6/sqrt(49)]

Test statistic = t = 1.105

P-value = 0.1373

For = 0.10, 0.05, 0.01, and 0.001, p-value is greater.

Therefore, we do not reject the null hypothesis H0.

There is no strong evidence to conclude that µ1 and µ2 differ by more than 3.

(Note: p-values and critical values are calculated by using t-distribution tables.)

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