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inna [77]
3 years ago
8

A copyeditor thinks the standard deviation for the number of pages in a romance novel is six. A sample of 25 novels has a standa

rd deviation of nine pages. At , is this higher than the editor hypothesized?
Mathematics
1 answer:
alina1380 [7]3 years ago
8 0

Answer:

No, the standard deviation for number of pages in a romance novel is six only.

Step-by-step explanation:

First we state our Null Hypothesis, H_o : \sigma = 6

             and Alternate Hypothesis, H_1 : \sigma > 6

We have taken these hypothesis because we have to check whether our population standard deviation is higher than what editor hypothesized of 6 pages in a romance novel.

Now given sample standard deviation, s = 9 and sample size, n = 25

To test this we use Test Statistics = \frac{(n-1)s^{2} }{\sigma^{2} } follows chi-square with (n-1) degree of freedom [\chi ^{2}_n__-1]

       Test Statistics = \frac{(25-1)9^{2} }{6^{2} } follows \chi ^{2}_2_4  = 54

and since the level of significance is not stated in question so we assume it to be 5%.

Now Using chi-square table we observe at 5% level of significance the \chi ^{2}_2_4 will give value of 36.42 which means if our test statistics will fall below 36.42 we will reject null hypothesis.

Since our Test statistics is more than the critical value i.e.(54>36.42) so we have sufficient evidence to accept null hypothesis and conclude that our population standard deviation is not more than 6 pages which the editor hypothesized.

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The first one for 56 and 26 would be 2

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The answer to your question is 55 :)
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The following data represent the number of flash drives sold per day at a localcomputer shop and their prices.Price Units Sold34
inysia [295]

Answer:

a)3145 x 0.01 = 31.45  3145- 31.45 = 3113.55

Compute the sample correlation 3113.55 -? we find the least square pressing at least 15x on the calculator then minus this from 3113.55 to find a better fit and minimum regression.

We add the differences of units then divide by distribution as seen below.

b) unsure.

c) = (see below) just test each number shown unit sold per day / price then x can show the differences in each number from day 1 to day 2.

d) = 16 sold.

Step-by-step explanation:

a) We count the units up and deduct from it from the equation p is recognized as units sold. R1 is cost R2 is total days.

b) The line of best fit is described by the equation ŷ = bX + a, where b is the slope of the line and a is the intercept (i.e., the value of Y when X = 0).

c) r 2= decimal ; the regression equation has accounted for percentage of the total sum of squares. You cna do this one.

d) = 16 sold at $28 each. - Why ? We using 7 day data and prove a how many units can be sold p/d if the price of flash drive is set to $28 each per unit.

Day 1 = 34  /  28  =  1      =    1.21428571429 = 1 no difference day prior.

Day 2 = 336   / 28   = 12  = 12 = difference day prior is 11

Day 3 = 432  / 28  = 15    =  15.4285714286   = 15 difference day prior is <em>3</em>

Day 4 = 635  / 28 =  23   =  22.6785714286  = 23 difference day prior is<em> 8</em>

Day 5  = 530  /  28 = 19    =  18.9285714286  = 19 difference day prior is minus <em>- 4</em>

Day 6 = 938  / 28  =  34   = 33.5 = 34 difference day prior is<em> 15</em>

Day 7 = 240  / 28  =   9    =   8.57142857143 = 9 difference day prior is minus <em>-25</em>

Total days 7 = Total revenue / price = average units sold

Average units sold total = 1+ 12+15 +23 +19+34+9 = 113 rounded.

Average units sold total = 1.21428571429 + 12 + 15.4285714286

+ 22.6785714286

+18.9285714286

+ 33.5

+ 8.57142857143 =  112.321428572 units sold weekly when priced at $28

To answer D we divide this by 7  to show;

112.321428572/ 7 = 16.0459183674

Daily units sold = 16

6 0
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Answer:

Step-by-step explanation:

Given that,

G(t) = 2+5

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Substitute 7 for(t)

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Answer:

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