**Question 1: **The heights of the hypothetical population of Police personnel is
normally distributed with a standard deviation of 2 ft. If a sample of
size 16 is randomly selected and the mean is computed to be 6.2 ft,
calculate the 90% confidence interval for the mean height of the cops.

** Solution: **

Given that,

X = 6.2 ft , σ = 2ft and n =16 and α = 1 - 0.90 = 0.10

We know from the critical value tables $Z_{\frac{\alpha }{2}}$ = Z_{0.05} = 1.645

The Maximum error is calculated using the formula,

E = $Z_{\frac{\alpha }{2}}(\frac{\sigma }{\sqrt{n}})$

= 1.645 ($\frac{2}{\sqrt{16}}$) = 1.645 x $\frac{1}{2}$ = 0.8225 ≈ 0.823

The Lower limit for the confidence interval = 6.2 - 0.823 = 5.377

The upper limit for the confidence interval = 6.2 + 0.823 = 7.023

Hence the confidence interval for the true mean is

5.377 < μ < 7.023.

**Question 2: **A hypothetical survey of 200 women University students found 62 of them
are married. Based on this, calculate the confidence interval for
population proportion of married woman students at 95% confidence
interval.

** Solution: **

From the info given $\hat{p}$ = $\frac{62}{200}$ = 0.31, $\hat{q}$ = 1 - 0.31 = 0.69, n =200, α = 1 - 095 = 0.05

We also note that both np and nq > 5.

Hence we can use the formula given to calculate the interval estimate at the confidence level = 95%

The critical value to be used from the table is

$Z_{\frac{\alpha }{2}}$ = Z_{0.025} = 1.96

The maximum error for the interval estimate is

E = $Z_{\frac{\alpha }{2}}$($\sqrt{\frac{\hat{p}\hat{q}}{n}}$)

= 1.96($\sqrt{\frac{0.31\times 0.69}{200}}$)

= 1.96 x 0.0327 ≈ 0.064

The Lower limit for the confidence interval = 0.31 - 0.064 = 0.246

The upper limit for the confidence interval = 0.31+ 0.064 = 0.374

Hence the confidence interval for the proportion of married among woman studying in Colleges at 95% confidence level

0.246 < p < 0.374