Chapter 10.1, Problem 21E

(a)

To determine

To explain do these data give convincing evidence to support Phoebe’s hunch at the $\alpha =0.05$ significance level or not.

Answer to Problem 21E

No, there is no convincing evidence.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=52\\ x_2=78\\ n_1=80\\ n_2=104\end{array}$

  $\alpha =0.05$

Define Hypotheses:

So, the given claim that: higher proportions for seniors.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Where we have,

  $p_1=$ the proportion of high school sophomores that brings a bag lunch.

  $p_2=$ the proportion of high school seniors that brings a bag lunch.

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the samples are independent random samples.

Independent: It is satisfied because the $80$ sophomores are less than $10\%$ of all sophomores and $104$ seniors are less than $10\%$ of all seniors.

Normal: It is satisfied because there are $52$ successes and $80-52=28$ failures in the first sample and there are $78$ successes and $104-78=26$ failures in the second sample, which are all at least ten.

Thus, all the conditions are satisfied.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{52}{80}\\ =0.65\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{78}{104}\\ =0.75\\ {\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{52+78}{80+104}\\ =\frac{130}{184}\\ =0.7065\end{array}$

Now, we will calculate the value of test statistics as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.65-0.70-0}{\sqrt{0.7065(1-0.7065)}\sqrt{\frac{1}{80}+\frac{1}{104}}}\\ \approx -0.74\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-0.74)\\ =0.2296\end{array}$

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is no convincing evidence to support the Phoebe’s hunch.

(b)

To determine

To interpret the P value from part (a) in the context of this study.

Answer to Problem 21E

There is a $22.96\%$ chance to obtain similar test results 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.

Explanation of Solution

From part (a), we have that,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Where we have,

  $p_1=$ the proportion of high school sophomores that brings a bag lunch.

  $p_2=$ the proportion of high school seniors that brings a bag lunch.

And, the value of test statistics is as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.65-0.70-0}{\sqrt{0.7065(1-0.7065)}\sqrt{\frac{1}{80}+\frac{1}{104}}}\\ \approx -0.74\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-0.74)\\ =0.2296\\ =22.96\%\end{array}$

From this we can conclude that there is a $22.96\%$ chance to obtain similar test results 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.