J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

Question

2 months ago

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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.

Mathematics

1 answer:

babunello [11.8K] · Answer 1 · 2026-03-13T13:09:25+03:00

The 99% confidence interval for the actual mean difference between average mail-order and internet purchase amounts falls within [$(-31.82), $12.02].

Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

P.Q. =

~

where,

= sample mean of mail-order sales = $74.50 $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ $t__n_1_+_n_2_-_2$

= sample mean of internet sales = $84.40

$\bar X_1$ = standard deviation for mail-order sales = $17.25

$\bar X_2$ = standard deviation for internet sales = $21.25

$s_1$ = number of mail-order sales receipts = 16

= number of internet sales receipts = 9 $s_2$

Furthermore,

= $n_1$ = 18.74

$n_2$

The actual mean difference between average mail-order and internet purchases is denoted by ( $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$

).

Thus, the 99% confidence interval for ( $\mu_1-\mu_2$ ) is expressed as;

= $\mu_1-\mu_2$ Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807. =

= [$-31.82, $12.02] $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

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

$(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$