Answer to Question #268553 in Statistics and Probability for muqaddas zulfiqar

Question

question 1:Work Interruptions A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted three or more times an hour

question 2:High School Graduates Who Take the SAT The national average for the percentage of high school graduates taking the SAT is 49%, but the state averages vary from a low of 4% to a high of 92%. A random sample of 300 graduating high school seniors was polled across a particular tristate area, and it was found that 195 of them had taken the SAT. Estimate the true proportion of high school graduates in this region who take the SAT with 95% confidence.

Answer 1

Answer 1:

Find the 90% confidence interval.

From the given information the sample size is 200 . The formula for confidence interval is as follows:

$\hat{p}-\left(z_{\frac{a}{2}} \times \sqrt{\frac{\hat{p} \hat{q}}{n}}\right)$

$<p<\hat{p}+\left(z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p} \hat{q}}{n}}\right)$

where $z_{\frac{a}{2}}$ is denoted as tabulated value from the standard normal distribution.

Obtain the value of $\hat{p}$ and $\hat{q}$ for the given value of X is 168 and n is 200 .

The value for $\hat{p}$ is calculated as follows:

$\begin{aligned} \hat{p} &=\frac{X}{n} \\ &=\frac{168}{200} \\ &=0.84 \end{aligned}$

The value for $\hat{q}$ is calculated as follows:

$\begin{aligned} \hat{q} &=1-\hat{p} \\ &=1-0.84 \\ &=0.16 \end{aligned}$

Hence, the value for $\hat{p}$ , that is, the true proportion of workers who are interrupted three or more times an hour is 0.84 and $\hat{q}$ is 0.16.

Therefore. the confidence interval is obtained as follows:

$\begin{aligned} &0.84-\left(z_{0.10} \times \sqrt{\frac{(0.84 \times 0.16)}{200}}\right)<p<0.84+\left(z_{0.10}^{\circ} \times \sqrt{\frac{(0.84 \times 0.16)}{2}}\right) \\ &0.84-(1.65 \times 0.026)<p<0.84+(1.65 \times 0.026) \\ &(0.84-0.043)<p<(0.84+0.043) \\ &0.797<p<0.883 \end{aligned}$

Thus, the 90% confidence interval for the true proportion of workers who are interrupted three or

more times an hour is (0.797, 0.883). 

Answer 2:

$\begin{aligned} &{\left[\hat{p}-Z_{0.025} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+Z_{0.025} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]} \\ &n=300, \hat{p}=195 / 300=0.65, Z_{0.025}=1.96 \end{aligned}$

The 95% confidence interval is given by

$\begin{aligned} &{[0.65-1.96 * \sqrt{(0.65 *(1-0.65) / 300)}, 0.65-1.96 * \sqrt{(0.65 *(1-0.65) / 300)}]} \\ &=[0.596,0.704] \end{aligned}$

The 95% confidence interval for true proportion of SAT takers in this region is $[ 0.596,0.704 ]$