Question

Land's Bend 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 12 sales receipts for mail-order sales results in a mean sale amount of $94.50 with a standard deviation of $19.25. A random sample of 17 sales receipts for internet sales results in a mean sale amount of $81.80 with a standard deviation of $20.75. Using this data, find the 80% 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 standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places.

Answer 1

Using the t-distribution, it is found that the 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is (2.85, 22.55).

We have the standard deviations for the sample, thus, the t-distribution is used. The interval is:

$M \pm ts$

In which:

• M is the difference of the sample means.
• t is the critical value.
• s is the standard error of the sampling distribution.

The sample mean of mail-order purchases is of $94.50, while for internet sales is of $81.8, thus, the difference of the sample means is:

$M = 94.5 - 81.8 = 12.7$

The standard error for each sample is the standard deviation of the sample divided by the square root of the sample size, thus:

$s_M = \frac{19.25}{\sqrt{12}} = 5.557$

$s_I = \frac{20.75}{\sqrt{17}} = 5.0326$

The standard error of the sampling distribution is:

$s = \sqrt{s_M^2 + s_I^2} = \sqrt{5.557^2 + 5.0326^2} = 7.4972$.

Then, using a t-distribution calculator or the t-table, the critical value for a 80% confidence interval with 12 + 17 - 2 = 27 df is of t = 1.314.

Then, the interval is:

$M - ts = 12.7 - 1.314(7.4972) = 2.85$

$M + ts = 12.7 + 1.314(7.4972) = 22.55$

The 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is (2.85, 22.55).

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