In: Statistics and Probability
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time these customers stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes.
a. Compute the standard error of the mean.
b. Construct a 95% confidence interval for the true average amount of time customers spent in the restaurant.
c. With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less?
Solution :
Given that,
s = 14
n = 49
Degrees of freedom = df = n - 1 = 49- 1 = 48
a ) The standard error of the mean is
= 2
The standard error of the mean is = 2
b ) At 95% confidence level the t is ,
Margin of error = E = t/2,df * (s /n)
= 4.02
Margin of error = 4.02
The 95% confidence interval estimate of the population mean is,
c ) Given that,
margin of error = E = 2.5
At 95% confidence level the z is ,
Sample size = n = ((Z/2 * ) / E)2
= ((1.960 * 14) / 2.5)2
= 120.47
= 120
Sample size = 120