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Bess [88]
3 years ago
8

The marketing manager of a large supermarket chain would like to use shelf space to predict the sales of pet food. For a random

sample of 12 similar stores, she gathered the following information regarding the shelf space, in feet, devoted to pet food and the weekly sales in hundreds of dollars. Use these data to answer questions 8 through 10.
Store 1 2 3 4 5 6
Shelf Space 5 5 5 10 10 10
Weekly Sales 1.6 2.2 1.4 1.9 2.4 2.6

Store 7 8 9 10 11 12
Shelf Space 15 15 15 20 20 20
Weekly Sales 2.3 2.7 2.8 2.6 2.9 3.1


What is the estimated regression equation?

A. = 2.63 + 0.724x
B. = 1.45 + 0.724x
C. = 1.45 + 0.074x
D. = 2.63 - 0.174x
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

m=\frac{27.75}{375}=0.074

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{150}{12}=12.5

\bar y= \frac{\sum y_i}{n}=\frac{28.5}{12}=2.375

And we can find the intercept using this:

b=\bar y -m \bar x=2.375-(0.074*12.5)=1.45

So the line would be given by:

y=0.074 x +1.45

C. = 1.45 + 0.074x

Step-by-step explanation:

The data given is:

x: 5,5,510,10,10, 15,15,15, 20,20,20

y: 1.6,2.2,1.4, 1.9, 2.4,2.6, 2.3,2.7, 2.8,2.6, 2.9, 3.1

For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i = 150

\sum_{i=1}^n y_i =28.5

\sum_{i=1}^n x^2_i =2250

\sum_{i=1}^n y^2_i =70.69

\sum_{i=1}^n x_i y_i =384

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=2250-\frac{150^2}{12}=375

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=384-\frac{150*28.5}{12}=27.75

And the slope would be:

m=\frac{27.75}{375}=0.074

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{150}{12}=12.5

\bar y= \frac{\sum y_i}{n}=\frac{28.5}{12}=2.375

And we can find the intercept using this:

b=\bar y -m \bar x=2.375-(0.074*12.5)=1.45

So the line would be given by:

y=0.074 x +1.45

C. = 1.45 + 0.074x

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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And for this case we know that the 95% confidence interval is given by:

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And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

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And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

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And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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