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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 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75. A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25. 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 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

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Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (
\mu_1-\mu_2) is [-9.132 , 23.332].

Explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. =
\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } ~
t__n__1-_n__2-2

where,
\bar X_1 = sample mean sales receipts for mail-order sales = $81.70


\bar X_2 = sample mean sales receipts for internet sales = $74.60


s_1 = sample standard deviation for mail-order sales = $18.75


s_2 = sample standard deviation for internet sales = $28.25


n_1 = size of sales receipts for mail-order sales = 7


n_2 = size of sales receipts for internet sales = 11

Also,
s_p=\sqrt{((n_1-1)s_1^(2) +(n_2-1)s_2^(2) )/(n_1+n_2-2) } =
\sqrt{((7-1)* 18.75^(2) +(11-1)* 28.25^(2) )/(7+11-2) } = 25.11

Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 80% confidence interval for the difference between population means, (
\mu_1-\mu_2) is ;

P(-1.337 <
t_1_6 < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 <
\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } < 1.337) = 0.80

P(
-1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } <
{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} <
1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } ) = 0.80

P(
(\bar X_1-\bar X_2)-1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } < (
\mu_1-\mu_2) <
(\bar X_1-\bar X_2)+1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } ) = 0.80

80% confidence interval for (
\mu_1-\mu_2) =

[
(\bar X_1-\bar X_2)-1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } ,
(\bar X_1-\bar X_2)+1.337 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } } ]

= [
(81.70-74.60)-1.337 * {25.11 * \sqrt{(1)/(7) +(1)/(11) } } ,
(81.70-74.60)+1.337 * {25.11 * \sqrt{(1)/(7) +(1)/(11) } } ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (
\mu_1-\mu_2) is [-9.132 , 23.332].

User Iordanis
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