Question

In studying of high school students, Mrs. Miller wishes to estimate the difference between two groups of highschool students regarding who helps students the most with financial issues. She asks two groups of random independent samples to find the 98% confidence interval for the difference between the proportions of group one and group two who get help from their mothers instead of fathers. A sample of 100 Students was taken from sullivan north highschool, with 43 students saying there mother helped most. Another sample of 100 students was taken from rogersville highschool, with 47 students saying their mother helped the most.

A) type of interval

B) Find the confidence interval (round to 3 decimal places)

C Using the confidence interval from part A) is there a difference between the proportion of sullivan north students and rogersville high students who say their mother help the most (Yes/No)

Answer 1

(A) The 98% Confidence interval for difference in 2 proportions. We will use the Z distribution

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(B) For the 98% Confidence interval

= 43/100 = 0.43 and 1 - = 0.57

= 0.47/100 = 0.47 and 1 - = 0.53

The Zcritical (2 tail) for = 0.02, is 2.326

The Confidence Interval is given by (- ) ME, where

(- ) = 0.43 – 0.47 = -0.04

The Lower Limit = -0.04 - 0.1635 = - 0.2035      -0.204 (Rounding to 3 decimal places)

The Upper Limit = 0.04 + 0.1635 = 0.1235    0.124 (Rounding to 3 decimal places)

The 98% Confidence Interval is -0.204 < p1 - p2 < 0.124

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(c) No, we cannot say that there is a difference in proportion as the confidence interval contains 0.

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