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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 = $6.

Click on the datafile logo to reference the data.



Round your answers to two 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.
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Respuesta :

Answer:

a) ME= 1.93

b) confidence interval= (19.59,23.45)

Explanation:

a) Sample of customers is 64, population standard deviation is 6 and confidence level is 99%

Sample mean= 21.52

Sample size= 64

Confidence level= 99%

Population standard deviation= 6

Standard error of the mean= 0.75

Z-value= -2.5758 (From Z table)

Interval half width= 1.9319

Margin of error at 99% confidence interval is 1.93 from the output.

b) Confidence interval

Interval upper limit= 19.59

Interval lower limit= 23.45

99% confidence interval is (19.59, 23.45) from the output.

ME= [tex]\frac{23.45-19.59}{2}[/tex]= 1.93

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