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2 years ago

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A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

EastWind [94]

Answer:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

Step-by-step explanation:

We have the following info given from the problem

$\bar X_1 = 52$ sample mean for this year

$s_1= 13$ sample deviation for this year

$n_1 = 75$ random sample selected for this year

$\bar X_2 = 49$ sample mean for last year

$s_2= 11$ sample deviation for last year

$n_1 = 53$ random sample selected for last year

And we want to 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

For this case the formula that we need to use is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$