Question

An industrial psychologist consulting with a chain of music stores knows that the average number of complaints management receives each month throughout the industry is 4, but the variance is unknown. Nine of the chain's stores were randomly selected to record complaints for one month; they received 15, 4, 2, 5, 11, 12, 7, 6, and 10 complaints. Using the .05 significance level, is the number of complaints received by the chain different from the number of complaints received by music stores in

general? What is the confidence interval?

Answer 1

From Given sample

= X / n = 72 / 9 = 8

Sample standard deviation S = sqrt [ ( X2 - n 2 ) / n - 1 ]

= sqrt [ (720 - 9 * 82 ) / 8 ]

= 4.2426

95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 9- 1 ) = 2.306 (From T table)
8 ± t(0.05/2, 9 -1) * 4.2426/√(9)
Lower Limit = 8 - 2.306 * 4.2426/√(9)
Lower Limit = 4.74
Upper Limit = 8 + 2.306 * 4.2426/√(9)
Upper Limit = 11.26
95% Confidence interval is ( 4.74 , 11.26 )

Since claimed mean does not contained in confidence interval , we have sufficient evidence

to support the claim that the number of complaints received by the chain is different from the number of

complaints received by music stores in general.