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n200080 [17]
3 years ago
8

Sum about cylinder PLZ HELP ME ASAP PLZ

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
5 0
Volume 48
6x 8y

Multiple 6 times 8
nirvana33 [79]3 years ago
4 0
Lol ok here for moral support
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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 intern
Naya [18.7K]

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].

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. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{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{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} } = 25.11

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

<em />

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{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.337) = 0.80

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

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

<u>80% confidence interval for</u> (\mu_1-\mu_2) =

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

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

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