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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
xÌ 82 89
Ï 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

xÌ ___ 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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