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 57, 32, 41, 48, 56, 80, 42, 29, 32, and 74. 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?

Answer 1

Sample mean = X / n = 49.1

Standard deviation = sqrt [ ( X² - n ² ) / n-1 ] = 17.6097

df = n - 1 = 10 - 1 = 9

t critical value at 0.05 significance level with 9 df = 2.262

95% confidence interval for is

- t * S / sqrt (n) < < + t * S / sqrt (n)

49.1 - 2.262 * 17.6097 / sqrt(10) < < 49.1 + 2.262 * 17.6097 / sqrt(10)

36.504 < < 61.696

95% CI is ( 36.504 , 61.696 )