In: Statistics and Probability
Data were collected on the amount spent for lunch by 64 customers at a major Houston restaurant. The sample provided a sample mean of $21.5. Based upon past studies the population standard deviation is known with σ = $6. Develop a 99% confidence interval estimate of the mean amount spent for lunch.
Solution :
Given that,
Sample size = n = 64
Z/2
= 2.576
Margin of error = E = Z/2*
(
/
n)
= 2.576 * (6 /
64)
= 1.93
At 99% confidence interval estimate of the population mean is,
- E <
<
+ E
21.5 - 1.93 <
< 21.5 + 1.93
19.57 <
< 23.43
( 19.57, 23.43)