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Marta_Voda [28]
3 years ago
8

HELP FAST PLEASE!!!!!!

Mathematics
1 answer:
MakcuM [25]3 years ago
6 0
-5 just had to write more

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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.
melamori03 [73]

Answer:

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

Step-by-step explanation:

We are given that 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.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \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

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  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} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

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

          = [$-31.82 , $12.02]

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

5 0
3 years ago
Find the equation of the line. Use exact numbers. ​
postnew [5]

Answer:

y=1/2x-3

is the slope intercept form

Step-by-step explanation:

Hope this helps :) Have a great day!!!

3 0
3 years ago
Pre Calculus:
Gwar [14]
Let n be a number of friends at the perty. If <span>dinner costs $30 per person, then it cost $30n for all friends.
</span>
Robert has booked a banquet hall which costs <span>$200, then his total expenses are $200+30n.
</span>
If the final bill is p, then p=<span>200+30n.
</span>
Answer: p=<span>200+30n.</span>

3 0
3 years ago
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Evaluate the expresion 6² + x - x² =3​
lisov135 [29]

Answer:

46

Step-by-step explanation:

8 0
3 years ago
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A study involves a population of 400 tall women. This population has a mean height of 179.832 cm and a standard deviation of 12.
Fofino [41]

Answer:  0.0386

Step-by-step explanation:

Given: The population of 400 tall women has a mean height(\mu) of 179.832 cm and a standard deviation(\sigma) of 12.192 cm.

Let X be a random variable that represents the height of woman.

Sample size : n= 50

The probability that the mean for this sample group is above 182.88 will be :

P(\overline{X}>182.88)\\\\=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{182.88-179.832}{\dfrac{12.192}{\sqrt{50}}})\\\\ =P(Z>1.7678)\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(Z

Hence, Required probability =  0.0386

3 0
3 years ago
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