In: Statistics and Probability
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 ≤ ________
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 = t0.025,9 = 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