Answered

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A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.

Today Five years ago
x├М 82 89
├П 161 181.5
n 46 33

The point estimate for the difference between the means is -7 and the standard error is 3. Find the 98% confidence interval for the difference between the two population means.

a. -13.98 to -0.02
b. -12.88 to -1.12
c. 9.33 to -4.67
d. 0.67 to 5.33


Sagot :

fichoh

Answer:

(-13.98 ; - 0.022)

Step-by-step explanation:

Given :

____Today __ Five years ago

x├М ___ 82 _____ 89

s ___ 161 _____181.5

n ___ 46 _____ 33

Sunce the sample. Size are large ; n > 30 ;

(x1 - x2) ┬▒ Zcritical * standard error

Zcritical at 98% confidence interval = 2.326

(x1 - x2) = 82 - 89 = - 7

Standard Error = 3

Hence ; we use the Z distribution

-7 ┬▒ 2.326 * 3

-7 ┬▒ 6.978

Lower boundary = (-7 - 6.978) = - 13.978

Upper boundary = (-7 + 6.978 = - 0.022

(-13.98 ; - 0.022)

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