Question

A random sample of 29 lunch customers was taken at a restaurant The average amount of time the customers in the sample 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 68% confidence interval for the true average amount of time customers spent in the restaurant?

c) Construct a 90% confidence interval for the true average amount of time customers spent in the restaurant?

d) Discuss why the answers in parts (a) and (b) are different?

e) With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 3 minutes or less?

Answer 1

Given that,

Sample size (n) = 29

Sample mean (Xbar) = 45 minutes

Standard deviation (sd) = 14 minutes.

a) Compute the standard error of the mean?

Standard error of the mean = sd / sqrt(n) = 14 / sqrt(29) = 2.60

b) Construct a 68% confidence interval for the true average amount of time customers spent in the restaurant?

Here sample size is < 30 so we use one sample t-interval.

We can find one sample t-interval in ti-83 calculator.

steps :

STAT --> TESTS --> 8:TInterval --> ENTER --> STATS --> ENTER --> Input all the values --> C-level : 0.68 --> Calculate --> ENTER

68% confidence interval for population mean is (42.368, 47.632)

c) Construct a 90% confidence interval for the true average amount of time customers spent in the restaurant?

Here again we have to find confidence interval with C-level = 0.90

90% confidence interval for population mean is (40.578, 49.422)

d) Discuss why the answers in parts (a) and (b) are different?

Two confidence intervals are different because confidence level different.

e) With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 3 minutes or less?

The formula for sample size is,

n = [ (Zc*sd) / E]²

With 95% confidence level Zc = 1.96 - ---- By using statistical table

n = [ (1.96*14) / 3 ]² = 83.66 which is approximately equal to 84.