Suppose a sample of 49 paired differences that have been randomly selected from
ID: 3045244 • Letter: S
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 of a-5.7 and a sample standard deviation of sd 7.8 (a) Calculate a 95 percent confidence interval for ,-M1-H2. Can we be 95 percent confident that the difference between 1 and 2 is greater than 07(Round your answers to 2 decimal places.) Corldence interval . 1: Cick to select: b) Test the null hypothesis H e* 0 ve sus the alte native hypothesis H ud70 by setting ? equal to 10, 05, 01, and .001. How much evidende is the e that d differs fr m ? What does this say about how and 2 compare? (Round your answer to 3 decimal places.) Reject Ho at ? equal to (Click to select) (Click to select) evidence that 1 differs from 2- 10 05, 01, and .001. How much evidence is there that d exceeds 3? what does this c The value for testing H dS 3 versus H : -3 equals 0096. Use the p value to test these hypotheses with ? equal to say about the size of the difference between 1 and 27(Round your answer to 3 decimal places.) p-value Reject Ho at ?equal to (Click to select) (Click to select) , evidence that mand 2 differ by more than 3Explanation / Answer
a)
Paired T-Test and CI
N Mean StDev SE Mean
Difference 49 5.70 7.80 1.11
95% CI for mean difference: (3.46, 7.94)
T-Test of mean difference = 0 (vs 0): T-Value = 5.12 P-Value = 0.000
yes .
since every value in confidence interval is more than 0 .
b)
p-value = 0.000
we reject the null hypothesis at alpha = 0.10,0.05,0.01,0.001
when alpha = 0.05
df = n-1 = 48
critical value = 2.0106
c)
Paired T-Test and CI
N Mean StDev SE Mean
Difference 49 5.70 7.80 1.11
95% lower bound for mean difference: 3.83
T-Test of mean difference = 3 (vs > 3): T-Value = 2.42 P-Value = 0.010
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