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. A random sample of 36 stores this year shows mean sales of 83 units of a small appliance with a standard deviation of 5 units. During the same point in time last year, a random sample of 49 stores had mean sales of 78 units with standard deviation 3 units.Required:Construct a 95 percent confidence interval for the difference in population means.

Answer 1

The 95% confidence interval for the difference in population means is (−26.325175 , 36.325175)

Explanation:

Given that :

sample size n₁ = 36

sample mean
`\over\ x`₁ = 83

standard deviation
`\sigma`₁ = 5

sample size n₂ = 49

sample mean
`\over\ x`₂= 78

standard deviation
`\sigma`₂ = 3

The objective is to construct a 95% confidence interval for the difference in the population means

Let the population means be
`\mu_1` and
`\mu_2`

The 95% confidence interval or the difference in population means can be calculated by using the formula;

`(\overline{x_1} - \overline{x_2}) \pm t_(\alpha /2) \ * s_(p)`

where;

the pooled standard deviation
`s_(p) = ((n_1-1)s_1^2+(n_2-1)s^2_2)/(n_1+n_2-2)`

`s_(p) = ((36-1)5^2+(49-1)3^2)/(36+49-2)`

`s_(p) = ((35)25+(48)9)/(83)`

`s_(p) = (875+432)/(83)`

`s_(p) = (1307)/(83)`

`s_p` = 15.75

degree of freedom =
`n_1 +n_2 -2`

degree of freedom = 36+49 -2

degree of freedom = 85 - 2

degree of freedom = 83

The Critical t- value 95% CI at df = 83 is

t critical = T.INV.2T(0.05, 83) = 1.9889

Therefore, for the population mean , we have:

= (83 - 78) ± (1.9889 × 15.75)

= 5 ± 31.325175

= 5 - 31.325175 , 5 + 31.325175

= (−26.325175 , 36.325175)