Question

A research firm supplies manufacturers with estimates of the sales of their products from samples of stores. Marketing managers often look at the sales estimates and ignore sampling error. An SRS of 50 stores this month shows mean sales of 41 units of a particular appliance with a standard deviation of 11 units. During the same month last year, an SRS of 52 stores gave mean sales of 38 units of the same appliances with a standard deviation of 13 units. An increase from 38 to 41 is a rise of 7.9% . The marketing manager is happy because sales are up 7.9% . (a) Give a 95% confidence interval for the difference in mean number of units of the appliance sold at all retail stores. (Enter your answer rounded to three decimal places.) lower interval: upper interval:

Answer 1

Mean difference = 41 - 38 = 3

The  standard error (SE) of the sampling distribution.

SE = sqrt[ (s1²/n1) + (s2²/n2) ]

where s1 is the  standard deviation of current year, s2 is the standard deviation of last year, n1 is the size of sample of current year, and n2 is the size of sample of last year.

SE = sqrt[ (11²/50) + (13²/52) ] = 2.381176

Assuming population standard deviations of sales in current ans last year are equal, degree of freedom = n1 + n2 - 2

= 50 + 52 - 2 = 100

t statistic at 95% confidence interval and df = 100 is 1.98

95% confidence interval is,

(3 - 1.98 * 2.381176, 3 + 1.98 * 2.381176)

= (-1.715, 7.715)

lower interval = -1.715

upper interval = 7.715

As, the confidence interval contains 0, there is no significant difference in mean number of units of the appliance sold at all retail stores.