A random sample of 49 lunch customers was taken at a restaurant. The average amo

Question

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 33 minutes. From past experience, the population standard deviation is 10 minutes. a. What is the point estimate of the mean. b. Determine the 95% confidence interval for the true mean time that customers spent in the restaurant? c. What is the margin of error? d. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant. e. How large of a sample would have to be taken to provide a margin of error of 2.5 minutes?

Explanation / Answer

a. What is the point estimate of the mean.
mean = average amount of time the customers in the sample stayed in the restaurant was 33 minutes.
b. Determine the 95% confidence interval for the true mean time that customers spent in the restaurant?
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33
Standard deviation( sd )=10
Sample Size(n)=49
Confidence Interval = [ 33 ± Z a/2 ( 10/ Sqrt ( 49) ) ]
= [ 33 - 1.96 * (1.43) , 33 + 1.96 * (1.43) ]
= [ 30.2,35.8 ]

c. What is the margin of error?
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=33
Standard deviation( sd )=10
Sample Size(n)=49
Margin of Error = Z a/2 * 10/ Sqrt ( 49)
= 1.96 * (1.43)
= 2.8

d. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33
Standard deviation( sd )=10
Sample Size(n)=49
Confidence Interval = [ 33 ± Z a/2 ( 10/ Sqrt ( 49) ) ]
= [ 33 - 2.58 * (1.43) , 33 + 2.58 * (1.43) ]
= [ 29.31,36.69 ]

e. How large of a sample would have to be taken to provide a margin of error of 2.5 minutes?
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 10
ME =2.5
n = ( 1.96*10/2.5) ^2
= (19.6/2.5 ) ^2
= 61.47 ~ 62      

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