Question

4. A survey asked a random sample of 363 first-year students how many hours they studied during a particular week. The mean was 15.3 hours. Suppose we know that the population standard deviation is 8.5 hours. Construct a 90%, 95% and 99% confidence interval for the mean study time of all first year students at this university. Interpret the 90% confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 15.3

Population standard deviation =   $\sigma$ = 8.5

Sample size = n = 363

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z_{0.05 = 1.645}

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.645 * ( 8.5 /  $\sqrt$363 )

= 0.73

At 90% confidence interval estimate of the population mean is,

T
- E < $\mu$ <
T
+ E

15.3 - 0.73 <  $\mu$ < 15.3 + 0.73

14.57 <  $\mu$ < 16.03

( 14.57 , 16.03 )

The 90% confidence interval for the population mean : ( 14.57 , 16.03 )

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 = 0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.96}

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 8.5 /  $\sqrt$363 )

= 0.87

At 95% confidence interval estimate of the population mean is,

T
- E < $\mu$ <
T
+ E

15.3 - 0.87 <  $\mu$ < 15.3 + 0.87

14.43 <  $\mu$ < 16.17

( 14.43 , 16.17 )

The 95% confidence interval for the population mean : ( 14.43 , 16.17 )

At 99% confidence level

$\alpha$ = 1 - 99%  

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z$\alpha$/2 = Z_{0.005 = 2.576}

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 2.576 * ( 8.5 /  $\sqrt$363 )

= 1.15

At 99% confidence interval estimate of the population mean is,

T
- E < $\mu$ <
T
+ E

15.3 - 1.15 <  $\mu$ < 15.3 + 1.15

14.15 <  $\mu$ < 16.45

( 14.15 ,16.45 )

The 99% confidence interval for the population mean : ( 14.15 ,16.45 )