Question

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores. Marketing managers are prone to look at the estimate and ignore sampling error. An SRS of 28

stores this year shows mean sales of 64 units of a small appliance, with a standard deviation of 13.8 units. During the same point in time last year, an SRS of 28 stores had mean sales of 48.958 units, with standard deviation 15.4 units. An increase from 48.958 to 64 is a rise of about 24%.


1. Construct a 99% confidence interval estimate of the difference μ1−μ2

μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales.


(b) The margin of error is



2. At a 0.01 significance level, is there sufficient evidence to show that sales this year are different from last year?

Answer 1

Confidence interval = (4.61972, 25.46428)

Margin of Error = 3.9078675

Explanation:

Given that :

This year:

Mean (μ1) = 64

Standard deviation (s1) = 13.8

Sample size (n1) = 28

Last year:

Mean (μ2) = 48.958

Standard deviation (s2) = 15.4

Sample size (n2) = 28

99% confidence interval estimate of the difference μ1−μ2

α = 1 - 99% = 0.01

(μ1−μ2) ± t0.01,27 * (√s1²/n1 + s2²/n2)

t0.01, 27 = 2.770683 (t value calculator)

√s1²/n1 + s2²/n2 = √13.8^2/28 + 15.4^2/28 = 3.9078675

(64 - 48.958) ± 2.667(3.9078675)

15.042 ± 10.42228

(15.042 - 10.42228), (15.042 + 10.42228)

(4.61972, 25.46428)

The margin of error :

√(s1²/n1) + (s2²/n2)

√(13.8^2/28) + (15.4^2/28)

√(6.8014285 + 8.47)

√15.2714285

= 3.9078675