Question

The marketing director of a large department store wants to estimate the average number of customers who enter the store every five minutes. She randomly selects five-minute intervals and counts the number of arrivals at the store. She obtains the figures 56, 32, 41, 49, 56, 80, 42, 29, 32, and 70. The analyst assumes the number of arrivals is normally distributed. Using these data, the analyst computes a 95% confidence interval to estimate the mean value for all five-minute intervals. What interval values does she get?

_____ ≤ μ ≤ ______

Answer 1

Solution:

From the given data,

sample size = n = 10

Point estimate = sample mean = = (Xi )/n = 487/10 = 48.7

sample standard deviation = s = (Xi - ) / n- 1

= 152.64 / 9

= 16.96

Degrees of freedom = df = n - 1 = 9

At 95% confidence level the t is ,

t _/2,df = t_0.025,9 = 2.262

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

= 2.262 * ( 16.96/ 10)

= 12.1

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

- E < < + E

48.7 - 12.1 < < 48.7 + 12.1

36.6 60.1