Question

A consumer affairs investigator records the repair cost for 16 randomly selected VCRs. A sample mean of $77.34 and standard deviation of $24.56 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Answer(How to Enter) 2 Points T Tables Keypad

Answer 1

Solution :

Given that,

$\bar x$ = $77.34

s =$24.56

n = 16

Degrees of freedom = df = n - 1 = 16- 1 =15

At 80% confidence level the t is ,

$\alpha$ = 1 - 80% = 1 - 0.80 = 0.20

  $\alpha$/ 2= 0.20 / 2 = 0.10

t$\alpha$ /2,df = t0.10, 15 = 1.341 ( using student t table)

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 1.341* ( 24.56/ $\sqrt$ 16)

= 8.2337

The 80% confidence interval estimate of the population mean is,

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

77.34 - 8.2337 < $\mu$ < 77.34+ 8.2337

69.1063 < $\mu$ < 85.8737

( 69.1063 , 85.8737)