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andreyandreev [35.5K]
2 years ago
10

The data below are the final exam scores of 10 randomly selected statistics students and the number of hours they studied for th

e exam. What is the best predicted value for y given x = 7? Assume that the variables x and y have a significant correlation.
Hours, x 3 5 2 8 2 4 4 5 6 3
Scores, y 65 80 60 88 66 78 85 90 90 71
Choose one answer.

A. 89
B. 90
C. 91
D. 92
Mathematics
1 answer:
laiz [17]2 years ago
4 0

Answer:

Correct option: (C) 91.

Step-by-step explanation:

To predict the score received by a student if the number of hours studied is 7 can be done using a regression line.

The general form of a regression line is:

y=\alpha +\beta x

Here,

<em>Y</em> = dependent variable

<em>X</em> = independent variable

<em>α</em> = intercept

<em>β</em> = slope

The formula to compute the slope and intercept are:

\alpha =\frac{\sum Y.\sum X^{2}-\sum X\sum XY}{n.\sum X^{2}-(\sum X)^{2}}

\beta=\frac{n.\sum XY-\sum X\sum Y}{n.\sum X^{2}-(\sum X)^{2}}

Consider the table below.

Compute the value of <em>α</em> as follows:

\alpha =\frac{\sum Y.\sum X^{2}-\sum X\sum XY}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(773\times 208)-(42\times 3406)}{(10\times 208)-(42)^{2}}=56.114

Compute the value of <em>β</em> as follows:

\beta=\frac{n.\sum XY-\sum X\sum Y}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(10\times 3406)-(42\times 773)}{(10\times 208)-(42)^{2}}=5.044

The regression equation is:

y=56.114+5.044\ x

For <em>x</em> = 7 compute the value of <em>y</em> as follows:

y=56.114+5.004\ x\\=56.114+(5.044\times 7)\\=91.422\\\approx 91

Thus, the score received by a student who studied 7 hours is 91.

The correct option is (C).

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6 0
2 years ago
After a number of complaints about its tech assistance, a computer manufacturer examined samples of calls to determine the frequ
icang [17]

Answer:

a) Upper Control Limit = 0.10

   Lower Control Limit = 0.01

b) The tech assistance process is stable and in control.

Step-by-step explanation:

Given - After a number of complaints about its tech assistance, a

            computer manufacturer examined samples of calls to determine

            the frequency of wrong advice given to callers. Each sample

            consisted of 100 calls.

SAMPLE             1   2   3   4   5   6   7   8   9  10 11 12 13 14 15 16

Number of errors 5  3   5   7   4   6   8   4   5    9   3   4   5   6   6   7

To find - a. Determine 95 percent limits.

              b. Is the tech assistance process stable (i.e., in control)

Proof -

z = 95% confidence interval

  = 1.96

⇒z = 1.96

Proportion of defects,P = total defectives/total observations

                                     = 0.0544

⇒P = 0.0544

Now,

Q = 1-P

   = 0.9456

⇒Q = 0.9456

Now,

Average sample size, N = 100

Standard deviation = \sqrt{\frac{P.Q}{N} }

                             = 0.0227

Now,

Upper Control Limit = P + z(Standard deviation)

        = 0.0988

⇒Upper Control Limit = 0.0988 ≈ 0.10

And

Lower Control Limit = P - z(Standard deviation )

       = 0.0099

⇒Lower Control Limit = 0.0099 ≈ 0.01

∴ we get

Upper Control Limit = 0.10

Lower Control Limit = 0.01

b.)

Now,

it is clear that the fraction defective values are wit in upper an lower control limits.

So, The tech assistance process is stable and in control.

5 0
2 years ago
over the course of numbering every page in a book, a mechanical stamp printed 2929 individual digits. How many pages does the bo
Kazeer [188]

1. Page 1-9  (1digit/page x 9page) = 9 digits = 9 page

2. Page 10-99 (2digits/page x 90page)= 180 digits = 90 page

3. Page 100-999 (3digits/page x 900page)= 2700 digits = 900 page

2,929 total digits – (9 digits + 180 digits + 2700 digits)= 40 digits

Since there are only 4 digits after page 999 we divide 40/4 = 10, thus there are 10 more pages after 999

So the book has 1009 pages.

Hope this answer will be a good help for you.

<span> </span>

4 0
2 years ago
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