Question

In: Statistics and Probability

Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed...


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⎯⎯ =4.0d¯ =4.0 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)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 .1578. 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.

Solutions

Expert Solution

(a) 95% confidence interval for d

= 4.0 1.6772 *6.9/ 7

= 4.0    1.6532

=(2.35, 5.65)

NOTE: for 95% CI and df = 48 , critical value of t is 1.6772

since the confidence interval is in the right side of 0, at 95% confidence we can say that difference is greater than zero

(b)

  

= 4.058

p value = 0.0001

reject H0 at all alpha values

there is extremely strong evidence that differs from

(c)

= 1.015

p value = 0.1578

fail to reject H0

or

reject H0 at no test values

no evidence that   differs from by more than 3

   


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