Q 21. Bag lunch? Phoebe has a hunch th... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 10: Q 21. (page 641)

Bag lunch? Phoebe has a hunch that older students at her very large high
school are more likely to bring a bag lunch than younger students because they have grown tired of cafeteria food. She takes a simple random sample of 80 sophomores and finds that $52$ of them bring a bag lunch. A simple random sample of $104$ seniors reveals that $78$ of them bring a bag lunch.
a. Do these data give convincing evidence to support Phoebe鈥檚 hunch at the $伪 = 0.05$ significance level?
b. Interpret the $P$ -value from part (a) in the context of this study.

Short Answer

Expert verified

a. There is no convincing evidence.

b. $P$ value is $22.96 %$

Step by step solution

Given Information

It is given that $x_{1} = 52$

$x_{2} = 78$

$n_{1} = 80$

$n_{2} = 104$

$伪 = 0.05$

To explain do these data give convincing evidence to support Phoebe's hunch at theα=0.05 significance level or not.

Claim: Higher proportion for seniors.

Appropriate hypothesis is:

Null: $H_{0} : p_{1} = p_{2}$

Alternative: $H_{a} : p_{1} < p_{2}$

$p_{1}$ is the proportion of high school sophomores that brings a bag lunch and $p_{2}$ is proportion of high school seniors that brings a bag lunch.

Conditions are:

Random: Samples are independent random samples.

Independent: $80 sophomores < 10 %$ of all sophomores and $104$ seniors is less than $10 %$ of all seniors.

Normal: Success are $52, 78$ and failures are $80 - 52 = 28, 104 - 78 = 26$ which are less than ten. Hence all conditions are satisfied.

Sample proportion is ${\hat{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{52}{80} = 0.65$

${\hat{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{78}{104} = 0.75$

${\hat{p}}_{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}} = \frac{52 + 78}{80 + 104} = \frac{130}{184} = 0.7065$

Test Statistic:

$z = \frac{{\hat{p}}_{1} - {\hat{p}}_{2} - (p_{1} - p_{2})}{\sqrt{{\hat{p}}_{p} (1 - {\hat{p}}_{p})} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{0.65 - 0.70 - 0}{\sqrt{0.7065 (1 - 0.7065)} \sqrt{\frac{1}{80} + \frac{1}{104}}} 鈮� - 0.74$

Probability is $P = P (Z < - 0.74) = 0.2296$

Now, $P > 0.05 鈬� Fail to Reject H_{0}$

Hence, there is no convincing evidence to support Phoebe's hunch.

Determining P value

From above calculation, $P = 22.96 %$

There is $22.96 %$ to get similar or more extreme when there is no difference between the proportion of sophomores who bring a bag lunch and the proportion of seniors who bring a bag lunch.