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 51, 32, 41, 42, 56, 80, 48, 29, 32, and 69. The analyst assumes the number of arrivals is normally distributed. Using these data, the analyst computed a 95% confidence interval to estimate the mean value for all five-minute intervals. What interval values does she get?

Answer 1

Answer:

Given,

51,32,41,42,56,80,48,29,32,69

Sample size n = 10

Mean = x / n

= (51 + 32 + 41 + 42 + 56 + 80 + 48 + 29 + 32 + 69)/10

= 480/10

= 48

xbar = 48

Standard deviation = sqrt((1/(n-1))*(xi - xbar)^2)

= sqrt(((51-48)^2 + .......+(69 - 48)^2)/(10-1))

s = 16.6533

Degree of freedom = n - 1

= 10 - 1

= 9

Critical value corresponding to df 9 & alpha 0.05 is 2.2621

Consider,

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 48 +/- 2.2621*16.6533/sqrt(10)

= 48 +/- 11.9128

Interval = (36.0872 , 59.9128)