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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 ([tex]\mu_1-\mu_2[/tex]) is [-9.132 , 23.332].

Step-by-step 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. =  [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\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 sales receipts for mail-order sales = $81.70

[tex]\bar X_2[/tex] = sample mean sales receipts for internet sales = $74.60

[tex]s_1[/tex] = sample standard deviation for mail-order sales = $18.75

[tex]s_2[/tex] = sample standard deviation for internet sales = $28.25

[tex]n_1[/tex] = size of sales receipts for mail-order sales = 7

[tex]n_2[/tex] = size of sales receipts for internet sales = 11

Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }[/tex] = 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, ([tex]\mu_1-\mu_2[/tex]) is ;

P(-1.337 < [tex]t_1_6[/tex] < 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 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < 1.337) = 0.80

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

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

80% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =

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

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

= [-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 ([tex]\mu_1-\mu_2[/tex]) is [-9.132 , 23.332].

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