Question

Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and find the proportion of presidents who were taller than their opponents. Use that result to construct a 95% confidence interval estimate of the population percentage. Based on the result, does it appear that greater height is an advantage for presidential candidates? Why or why not? Click the icon to view the table of heights. Construct a 95% confidence interval estimate of the percentage of presidents who were taller than their opponents (Round to one decimal place as needed.) 50% of the elections. In this case, greater height to be an advantage for presidential candidates because the confidence If greater height was an advantage, then taller candidates should have won interval include 50%

Answer 1

We first find the presidents who are taller than their opponent

Height	Height Opp	Taller?
173	168	Yes
178	196	No
173	185	No
171	191	No
170	178	No
170	189	No
168	180	No
189	170	Yes
188	188	No
180	182	No
183	182	Yes
183	178	Yes
180	180	No
188	182	Yes
173	178	No
178	175	Yes
183	187	No
178	180	No
179	178	Yes
188	173	Yes

We can see that 8 out of 20 presidents are taller than their opponent

n=20 is the sample size

is the sample proportion of presidents are taller than their opponent

p= = 0.4 20

The estimated standard error of proportions is

$\hat{\sigma}_{\hat{p}}=\sqrt\frac{\hat{p}(1-\hat{p})}{n}=\sqrt\frac{0.4(1-0.4)}{20}=0.1095$

The values

$n\hat{p}=20\times 0.4=8\text{ and}\\ n(1-\hat{p})=20\times (1-0.4)=12$

are both greater than 5 and hence we can use normal distribution as the sampling distribution of proportions

The 95% confidence level is $\alpha=1-95/100=0.05$ level of significance

The right tail critical value is

P(Z > 2a/2) = a/2 = 0.05/2 = 0.025 P(Z < Za/2) = 1 -0.025 = 0.975

Using the standard normal tables, we get for z=1.96, P(Z<1.96)=0.975

Hence, $z_{\alpha/2}=1.96$

The 95% confidence interval of proportions is

$\begin{align*} &\hat{p}\pm z_{\alpha/2}\hat{\sigma}_{\hat{p}}\\ \implies& 0.4\pm 1.96\times 0.1095\\ \implies &(0.185, 0.615) \end{align*}$

ans: The 95% confidence interval estimate is

$\begin{align*} 18.5\%<p< 61.5\% \end{align*}$

ans:

If greater height was an advantage, the taller candidates should have won more than 50%of the elections. In this case, greater height does not seem to be an advantage for the presidential candidates because the confidence interval does include 50%

The sample details are

n=291+55=346 is the sample size of drive through orders

$\begin{align*} \hat{p}=\frac{55}{346}=0.1590 \end{align*}$ is the sample proportion of orders that are not accurate

The estimated standard error of proportions is

$\hat{\sigma}_{\hat{p}}=\sqrt\frac{\hat{p}(1-\hat{p})}{n}=\sqrt\frac{0.1590(1-0.1590)}{346}=0.0197$

The values

$n\hat{p}=346\times 0.1590=55\text{ and}\\ n(1-\hat{p})=346\times (1-0.1590)=291$

are both greater than 5 and hence we can use normal distribution as the sampling distribution of proportions

The 95% confidence level is $\alpha=1-95/100=0.05$ level of significance

The right tail critical value is

P(Z > 2a/2) = a/2 = 0.05/2 = 0.025 P(Z < Za/2) = 1 -0.025 = 0.975

Using the standard normal tables, we get for z=1.96, P(Z<1.96)=0.975

Hence, $z_{\alpha/2}=1.96$

The 95% confidence interval of proportions is

$\begin{align*} &\hat{p}\pm z_{\alpha/2}\hat{\sigma}_{\hat{p}}\\ \implies& 0.1590\pm 1.96\times 0.0197\\ \implies &(0.120, 0.197) \end{align*}$

a) the 95% confidence interval is

ans:

$\begin{align*} 0.120<p< 0.197 \end{align*}$

b) The 95% confidence interval for Restaurant B is 0.136<p<0.216 and they overlap

ans:

O C. Since the two confidence intervals overlap, neither restaurant appears to have a significantly different percentage of orders that are not accurate