Question

7. A researcher surveys 40 college students to determine how much they spend on textbooks per semester. The responses have a sample mean of $426. The population standard deviation is known to be $83. Find the 95% confidence interval of the true mean. Assume the variable is normally distributed.

Answer 1

Solution :

Given that,

Point estimate = sample mean =  $\bar x$   = 426

Population standard deviation =   $\sigma$ =83

Sample size n =40

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z$\alpha$/2    * ( $\sigma$ /$\sqrt$n)
= 1.96 * (83 / $\sqrt$40)

= 25.7
At 95% confidence interval estimate of the population mean
is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

426-  25.7 <  $\mu$ < 426 + 25.7

400.3 <  $\mu$ < 451.7