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 but ignore sampling error. Suppose that a simple random sample of 75 stores this month shows mean sales of 52 units of small appliance, with a standard deviation of 13 units. During the same month last year, a simple random sample of 53 stores gave mean sales of 49 units, with standard deviation 11 units. An increase from 49 to 52 is a rise of 6%. The marketing manager is happy because sales are up 6%.(a) Estimate a 95% confidence interval for the difference in the mean number of units sold in the population of retail stores. (b) Explain why you constructed the confidence interval in the way that you did. (c) Explain in language that the manager can understand why she cannot be confident that sales rose by 6% and that in fact sales may even have dropped. Please be descrpritve to all

Answer 1

Answer:

n1 = 75

n2 = 53

x1bar = 52

x2bar = 49

s1 = 13

s2 = 11

a)

To give 95% confidence interval

Here for 95% confidence interval , z value is 1.96

Interval = (x1bar - x2bar) +/- z*sqrt(s1^2/n1 + s2^2/n2)

substitute values

= (52 - 49) +/- 1.96*sqrt(13^2/75 + 11^2/53)

= 3 +/- 1.96*2.1299

= 3 +/- 4.1746

= (3 - 4.1746 , 3 + 4.1746)

= (-1.1746 , 7.1746)

b)

Now here we can say that,

n1 & n2 are greater than 30 i.e., large samples.

So that we use large sample test 'z'.

c)

Here the actual increase is 6%

i.e.,

from 49 to 52 is 6%

Out of 75, there are 52 mean sales

i.e.,

= 52/75 * 100

= 69.33%

Out of 53, there are 49 mean sales

49/53 * 100

= 92.45%

So we can say that, there is a sales drop of (92.45 - 69.33)% = 23%