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 25 stores this year shows mean sales of 83 units of a small appliance, with a standard deviation of 13.6 units. During the same point in time last year, an SRS of 14 stores had mean sales of 68.176 units, with standard deviation 3.4 units. An increase from 68.176 to 83 is a rise of about 18%.

1. Construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales.

(a)____ <(μ1−μ2)<____

(b) The margin of error is ___

2. At a 0.05 significance level, is there sufficient evidence to show that sales this year are different from last year? Yes or No?

Answer 1

Solution:

1) a) The 95% confidence interval is :

CI= (x1bar-x2bar) +_ t0.05 * S. E(x1bar-x2bar)

= ( 83-68.176) +_ 2.026* (13.6/√25+ 3.4/√14)

=(7.4723,22.1757)

b) The margin of error is t0. 05 * S. E( x1bar- x2bar)

= 7.3517

2)Sample Mean= x1bar= 83

Sample Mean = x2bar= 68.176

Sample Std deviation: s1= 13.6, s2= 3.4,

Sample size : n1= 25, n2= 14

The following null and alternative hypotheses need to be tested:

H0: mu1= mu2 vs

H1: mu1≠mu2

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Test statistic :

t= (x1bar-x2bar) /(s1/√n1 +s2/√n2)

=(83-68.176)/(13.6/√25 + 3.4/√14)

=4.085

Since it is observed that calculated |t| = 4.085>tabulated t= 2.026 for 37 d. f ( n1+n2-2) , hence null hypothesis is rejected.

Therefore, Yes there is enough evidence to claim that sales this year are different from last year.