A market research firm supplies manufacturers with estimates of the retail sales

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 23 stores this year shows mean sales of 68 units of a small appliance, with a standard deviation of 9.4 units. During the same point in time last year, an SRS of 22 stores had mean sales of 78.246 units, with standard deviation 9.5 units. A decrease from 78.246 to 68 is a drop of about 15%.

1. Construct a 95% confidence interval estimate of the difference ?1??2, where ?1 is the mean of this year's sales and ?2 is the mean of last year's sales.

(a) <(?1??2)<

(b) The margin of error is what?

Explanation / Answer

Ans:

Assume equal population variances.

df=23+22-2=43

critical t value=tinv(0.05,43)=2.017

pooled standard deviation=sqrt(((23-1)*9.4^2+(22-1)*9.5^2)/(23+22-2))=9.449

standard error for difference=9.449*sqrt((1/23)+(1/22))=2.8178

a)

95% confidence interval for difference of means

=(68-78.246)+/-2.017*2.8178

=-10.246+/-5.683

=(-15.929,-4.563)

b)

Margin of error=2.017*2.8178=5.683

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