Question

Annie is a researcher who wants to find a confidence interval estimate of the mean time it takes to complete an order over the phone at a call center for a large retail company. She has selected a random sample of 18 orders and recorded the time to complete the order. The sample mean time was 4.27 minutes and the sample standard deviation was 0.78 minutes. Assuming the conditions needed for inference are met,which of the following is the appropriate 98% confidence interval? OA. 4.27+2.3261078 08.4.27 +2.05410,79 OC.4.27+2.552976 OD.4.27+2.5676076

Answer 1

Solution :

Given that,

$\bar x$ = 4.27

s =0.78

n = Degrees of freedom = df = n - 1 = 18 - 1 = 17

a ) At 98% confidence level the t is

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02/ 2 = 0.01

t$\alpha$ /2,df = t0.01,17 =2.567 ( using student t table)

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

= 2.567 * ( 0.78/ $\sqrt$ 18)

The 98% confidence interval estimate of the population mean is,

$\bar x$ - E

4.27 +-  2.567 * ( 0.78/ $\sqrt$ 18)