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olga_2 [115]
3 years ago
15

The following table shows the​ weights, in hundreds of​ pounds, for six selected cars. Also shown is the corresponding fuel​ eff

iciency, in miles per gallon​ (mpg), for the car in city driving. Use the data provided in the table to complete parts ​a) through​d).
Weight​ (hundreds of​ pounds)

28

33

35

32

30

29

Fuel efficiency​ (mpg)

20

22

19

22

23

21

​a) Determine the correlation coefficient between the weight of a car and the fuel efficiency.

The correlation coefficient between the weight of a car and the fuel efficiency is

r=

​(Type an integer or decimal rounded to three decimal places as​ needed.)
Mathematics
1 answer:
Reika [66]3 years ago
8 0

Answer:

Step-by-step explanation:

Given is a table showing  the​ weights, in hundreds of​ pounds, for six selected cars. Also shown is the corresponding fuel​ efficiency, in miles per gallon​ (mpg), for the car in city driving.

Weight Fuel eff. x^2 xy y^2

X Y    

28 20 784 560 400

3 22 9 66 484

35 19 1225 665 361

32 22 1024 704 484

30 23 900 690 529

29 21 841 609 441

     

Mean 26.16666667 21.16666667 797.1666667 549 449.8333333

     

Variance   112.4722222  1.805555556

Covariance -553.8611111    

     

     

r -0.341120235    

Correlaton coefficient =cov (xy)/S_x S_y

Covariance (x,y) = E(xy)-E(x)E(y)

The correlation coefficient between the weight of a car and the fuel efficiency is -0.341

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Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

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So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so z = 2.327

M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

M = z*\frac{\sigma}{\sqrt{n}}

0.4 = 1.96*\frac{0.78}{\sqrt{n}}

0.4\sqrt{n} = 1.96*0.78

\sqrt{n} = \frac{1.96*0.78}{0.4}

(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}

n = 14.6

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

M = z*\frac{\sigma}{\sqrt{n}}

0.2 = 2.575*\frac{0.78}{\sqrt{n}}

0.2\sqrt{n} = 2.575*0.78

\sqrt{n} = \frac{2.575*0.78}{0.2}

(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}

n = 100.85

Rounding up

A sample size of 101 is needed.

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