91Ó°ÊÓ

It is given that $x_1=52$

$x_2=78$

$n_1=80$

$n_2=104$

$\mathrm{Î\pm }=0.05$

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: $80\mathrm{sophomores}<10\%$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:

$z=\frac{{\hat{p}}_1-{\hat{p}}_2-\left(p_1-p_2\right)}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.65-0.70-0}{\sqrt{0.7065(1-0.7065)}\sqrt{\frac{1}{80}+\frac{1}{104}}}≈-0.74$

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

Now, $P>0.05⇒\text{Fail to Reject}{H}_0$

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

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.