A research firm supplies manufacturers with estimates of the sales of their products from samples of stores. Marketing managers often look at the sales estimates and ignore sampling error. An SRS of 50 stores this month shows mean sales of 41 units of a particular appliance with a standard deviation of 11 units. During the same month last year, an SRS of 52 stores gave mean sales of 38 units of the same appliances with a standard deviation of 13 units. An increase from 38 to 41 is a rise of 7.9 % . The marketing manager is happy because sales are up 7.9 % . (a) Give a 95 % confidence interval for the difference in mean number of units of the appliance sold at all retail stores. (Enter your answer rounded to three decimal places.)

The 95% confidence interval for the difference in mean number of units of the appliance sold at all retail stores for this case is [-1.667, 7.667]

How to find the confidence interval for difference in mean?

Supposing the samples are large, let we're given that:

For first sample:

[tex]n_1[/tex] =sample size
[tex]s_1[/tex] = sample standard devation
[tex]\overline{x}_1[/tex] = sample mean

For second sample:

[tex]n_2[/tex] =sample size
[tex]s_2[/tex] = sample standard devation
[tex]\overline{x}_2[/tex] = sample mean

Suppose the confidence level be p% = p/100 in decimal, then the level of significance would be [tex]\alpha = 1 - p/100[/tex]

Then, the margin of error would be:

[tex]MOE = Z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}[/tex]

where [tex]Z_{\alpha/2}[/tex] is the critical value of Z at level of significance [tex]\alpha[/tex]

The confidence interval would be:

[tex]CI = \overline{x}_1 - \overline{x}_2 \pm MOE[/tex]

Thus, for this case, we're specified that:

For first sample (SRS this month):

[tex]n_1[/tex] =sample size = 50
[tex]s_1[/tex] = sample standard devation = 11
[tex]\overline{x}_1[/tex] = sample mean = 41

For first sample(SRS last month):

[tex]n_2[/tex] =sample size = 52
[tex]s_2[/tex] = sample standard devation = 13
[tex]\overline{x}_2[/tex] = sample mean = 38

Level of significance = 1 - 95/100 = 0.05

The critical value of z at 0.05 level of significance (from the tables of critical value of Z) is: [tex]Z_{0.05/2} = 1.96[/tex]

Thus, the margin of error would be:

[tex]MOE = Z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}\\\\\\MOE = 1.96\sqrt{\dfrac{11^2}{50} + \dfrac{13^2}{52}}\\\\MOE \approx 4.667[/tex]

Thus, the confidence interval would be:

[tex]CI = \overline{x}_1 - \overline{x}_2 \pm MOE\\CI \approx 41 - 38 \pm 4.667\\CI \approx 3 \pm 4.667\\CI \approx [3 - 4.667, 3 + 4.667] = [-1.667, 7.667][/tex]

Thus, the 95% confidence interval for the difference in mean number of units of the appliance sold at all retail stores for this case is [-1.667, 7.667]

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