Question

1. 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 sample standard deviation of 14 minutes.
   
   1. Compute the standard error of the mean.
   2. At 95% confidence, what is the confidence interval estimate?
   3. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant.
   4. Compare the margin of errors and width of your confidence intervals from parts (b) and (c), what do you notice?

Answer 1

Solution :

Given that,

= 45

s = 14

n = 49

The standard error = (s /n)

= (14 / 49)

= 2

The standard error = 2

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

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

(40.98, 49.02 )

c ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,48 = 2.682

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

= 2.682 * (14 / 49)

= 5.36

Margin of error = 5.36

The 98% confidence interval estimate of the population mean is,

- E < < + E

45 - 5.36 < < 45 + 5.36

39.64< < 50.36

(39.64, 50.36 )

c ) The confidence interval level increase of the margin of errors is wider.