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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25. A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25. Using this data, find the 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed. Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

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Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~ [tex]t__n_1_+_n_2_-_2[/tex]

where, [tex]\bar X_1[/tex] = sample mean for mail-order sales = $74.50

[tex]\bar X_2[/tex] = sample mean for internet sales = $84.40

[tex]s_1[/tex] = sample standard deviation for mail-order purchases = $17.25

[tex]s_2[/tex] = sample standard deviation for internet purchases = $21.25

[tex]n_1[/tex] = sample of sales receipts for mail-order purchases = 16

[tex]n_2[/tex] = sample of sales receipts for internet purchases = 9

Also,  [tex]s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }[/tex]  =  [tex]\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }[/tex] = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ([tex]\mu_1-\mu_2[/tex]).

Now, 99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) is given by;

             = [tex](\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}[/tex]

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = [tex](74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})[/tex]

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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