2. A principal reported that the same percent of juniors and seniors felt that the homecoming should have been postponed due tò rain. A student checks this claim by rándomly surveying 85 juniors and 57 seniors. Respectively, 66 and 40 of those students said homecoming should have been postponed. a. Find the 95% confidence interval for difference between the proportions. b. What do you think of the principal's claim?

2. A principal reported that the same percent of juniors and seniors felt that the homecoming should have been postponed due tò rain. A student checks this claim by rándomly surveying 85 juniors and 57 seniors. Respectively, 66 and 40 of those students said homecoming should have been postponed. a. Find the 95% confidence interval for difference between the proportions. b. What do you think of the principal's claim?

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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2. A principal reported that the same percent of juniors and seniors felt that the
homecoming should have been postponed due tò rain. A student checks this claim by
rándomly surveying 85 juniors and 57 seniors. Respectively, 66 and 40 of those
students said homecoming should have been
a. Find the 95% confidence interval for difference between the proportions.
postponed.
b. What do you think of the principal's claim?
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Expert Solution

Step 1

Given Data:

${\hat{p}}_{1} = \frac{66}{85} {\hat{p}}_{2} = \frac{40}{57}$

Calculating the pooled standard deviation

$S d = \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{1}}} = \sqrt{\frac{66 / 85 (1 - 66 / 85)}{85} + \frac{40 / 57 (1 - 40 / 57)}{57}} = 0.0756$

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