Question

25. Based on a random sample of 16 customer transactions, the average customer waiting time at a fast food restaurant is 8 minutes. Based on the same sample, the standard deviation is 4 minutes. What is the margin of error of the 90% CI for the average waiting time at the restaurant? Useful values: Zo1 1.282, Zo.05 1.645; for df-16: to.11.337, to.os-1.746; for df-15: to.1-1.341, to.os 1.753. A) 1.000 B) 1.282 C) 1.645 D) 1.746 E) 1.753

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Answer 1

Given that,

$\bar x$ = 8

s =4

n = 16

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

At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

t_{$\alpha$ /2,df} = t_0.05,15=1.753

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

= 1.753 * (4 / $\sqrt$ 16) = 1.753

answer = 1.753