Question

A stationary store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 20 greeting cards indicates a sample mean value of $4.95 and a sample standard deviation of $0.82.

(a) Assuming the population follows a normal distribution, construct a 95% confidence interval estimate of the mean of all greeting cards in the store’s inventory. (Hint: use the t-tables.)

(b) Interpret the interval constructed in part (a). (Hint: an interpretation is to give a brief description of what the confidence interval tells the stationary store about the mean of all greeting cards in their inventory.)

Answer 1

Solution :

Given that,

= 4.95

s = 0.82

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

a ) At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,19 = 2.093

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

= 2.093 * ( 0.82 / 20 )

= 0.38

Margin of error = 0.38

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

- E < < + E

4.95 - 0.38 < < 4.95 + 0.38

4.57 < < 5.33

(4.57, 5.33 )