Question

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?

Answer 1

Solution :

Given that,

= 45

s = 14

n = 49

Degrees of freedom = df = n - 1 = 49- 1 = 48

a ) The standard error of the mean is

  ₌   / n .

= 14 49

= 2

The standard error of the mean is = 2

b ) At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,48 =2.011

Margin of error = E = t_/2,df * (s /n)

= 2.011 * (14 / 49)

= 4.02

Margin of error = 4.02

The 95% confidence interval estimate of the population mean is,

- E < < + E

45 - 4.02 < < 45 + 4.02

40.98 < < 49.02

c ) Given that,

standard deviation = = 14

margin of error = E = 2.5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Sample size = n = ((Z_/2 * ) / E)²

= ((1.960 * 14) / 2.5)²

= 120.47

= 120

Sample size = 120