Data were collected on the amount spent by 64 customers for lunch at a major Hou
ID: 3357641 • Letter: D
Question
Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the file named Houston. Based upon past studies the population standard deviation is known with = $12. Click on the webfile logo to reference the data. Round your answers to two decimal places. Use the critical value with three decimal places. a. At 99% confidence, what is the margin of error? $ b. Develop a 99% confidence interval estimate of the mean amount spent for lunch. $ to $
Explanation / Answer
Solution:
From the given information
Standard deviation = 12
Sample size n = 64
a) Z/2 = Z0.005 = 2.576
For 99% confidence ineterval,
Margin of Error = Z/2 (SE)
= 2.576 (/n)
= 2.576 (12/64)
= 3.8640
b) The mean spent for lunch is, x = 21.52 with the margin of error 3.8640.
The 99% confidence interval estimate of the mean amount spent for lunch is,
CI = x ± Margin of error
= 21.52 ± 3.8640
= ( 17.656, 25.384)
Hence, the required confidence interval is, 17.656 < < 25.384
Therefore, it is say that we are 99% confident that on customers will spend between $17.656 and $25.384 for lunch.
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