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Expert-verified Found in: Page 622 ### The Practice of Statistics for AP

Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339 # Who uses instant messaging? Do younger people use online instant messaging (IM) more often than older people? A random sample of IM users found that $73$ of the $158$ people in the sample aged $18$ to $27$ said they used IM more often than email. In the $28$ to $39$ age group, $26$ of $143$ people used IM more often than email.$9$ Construct and interpret a $90$% conﬁdence interval for the difference between the proportions of IM users in these age groups who use IM more often than email.

From the given information, it could be interpreted that there is $90%$ the probability that the difference in the proportion of IM users lies between $0.1959$ and $0.3641$

See the step by step solution

## Step 1: Given Information

It is given in the question that, ${x}_{1}=73,{n}_{1}=158,{x}_{2}=26,{n}_{2}=143$

## Step 2: Explanation

The formula to compute the confidence interval for the difference in population proportion is :

$CI=\left({\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}\right)±{z}_{\alpha /2}×\sqrt{\frac{{\stackrel{^}{p}}_{1}\left(1-{\stackrel{^}{p}}_{1}\right)}{{n}_{1}}+\frac{{\stackrel{^}{p}}_{2}\left(1-{\stackrel{^}{p}}_{2}\right)}{{n}_{2}}}$

The $95%$ confidence interval using Ti-$83$ plus calculator is computed as: Therefore, the confidence interval is ($0.1959,0.3641$)

Therefore, it could be interpreted that there is $90%$ probability that the difference in the proportion of IM users lies between $0.1959$ and $0.3641$ 