Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
Question
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Chapter 11.3, Problem 18E

a.

To determine

Find the value of b1.

a.

Expert Solution
Check Mark

Answer to Problem 18E

The slope b1 is –0.063.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 7 values for x and y.

Slope or b1:

b1=rsysx

where,

r represents the correlation coefficient between x and y.

sy represents the standard deviation of y.

sx represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

  • • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • • In Variables enter the columns of x and y.
  • • Choose Options Statistics, and select Mean and Standard deviation.
  • • Click OK.

Output obtained from MINITAB is given below:

Essential Statistics, Chapter 11.3, Problem 18E , additional homework tip  1

Correlation:

r=1n1(xx¯sx)(yy¯sy)

The table shows the calculation of correlation:

xyxx¯  yy¯ xx¯sx  yy¯sy(xx¯sx)(yy¯sy)
1218–1.140.57–1.06640.3144–0.3353
1319–0.141.57–0.13100.86597–0.1134
15181.860.571.73990.31440.547
1316–0.14–1.43–0.1310–0.78870.1033
1216–1.14–1.43–1.0664–0.78870.8411
14150.86–2.430.8045–1.3403–1.0783
1320–0.142.57–0.13101.41754–0.1856
 –0.2212

Thus, the correlation is

r=0.221271=0.22126=0.0369

b1=rsysx

Substitute r as –0.0369, sy as 1.813 and sx as 1.069.

b1=0.0369(1.8131.069)=0.0369(1.696)=0.063

Thus, the slope b1 is –0.063.

b.

To determine

Find the residual standard deviation se.

b.

Expert Solution
Check Mark

Answer to Problem 18E

The residual standard deviation se is 1.98.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or b0:

b0=y¯b1x¯

y¯ represents the mean of y values.

x¯ represents the mean of x values.

b1 represents the slope coefficient.

Substitute y¯ as 17.43, x¯ as 13.14 and b1 as 0.063.

b0=17.43(0.063)(13.14)=17.43+0.83=18.26

Thus, the intercept b0 is 18.26.

The residual standard deviation se is calculated using the formula,

se=(yy^)2n2

Where,

(yy^)2 represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is y^=18.2600.063x

Use the estimated regression equation to find the predicted value of y for each value of x.

xyy^yy^(yy^)2
121817.5040.4960.24602
131917.4411.5592.43048
151817.3150.6850.46922
131617.441-1.4412.07648
121617.504-1.5042.26202
141517.378-2.3785.65488
132017.4412.5596.54848
Total   19.6876

Substitute (yy^)2 as 19.6876 and n as 7.

se=19.687672=19.68765=3.938=1.98

Thus, the residual standard deviation se is 1.98.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution
Check Mark

Answer to Problem 18E

The sum of squares for x is 6.8572.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

xxx¯(xx¯)2
12–1.141.2996
13–0.140.0196
151.863.4596
13–0.140.0196
12–1.141.2996
140.860.7396
13–0.140.0196
Total 6.8572

Thus, the sum of squares for x is 6.8572.

d.

To determine

Find the standard error of b1,sb.

d.

Expert Solution
Check Mark

Answer to Problem 18E

The standard error of b1sb is 0.756.

Explanation of Solution

Calculation:

The standard error of b1 is calculated by using the formula:

sb=se(xx¯)2

Where,

se represents the residual standard deviation.

(xx¯)2 represents the sum of squares due to x.

Substitute (xx¯)2 as 6.8572 and se as 1.98.

sb=1.986.8572=1.982.619=0.756

Thus, the standard error of b1, sb is 0.756.

e.

To determine

Find the critical value for a 95% confidence interval of β1.

e.

Expert Solution
Check Mark

Answer to Problem 18E

The critical value for a 95% confidence interval of β1 is 3.182.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

  • • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • • From Distribution, choose ‘t’ distribution.
  • • In Degrees of freedom, enter 5.
  • • Click the Shaded Area tab.
  • • Choose Probability and Two tail for the region of the curve to shade.
  • • Enter the Probability value as 0.05.
  • • Click OK.

Output obtained from MINITAB is given below:

Essential Statistics, Chapter 11.3, Problem 18E , additional homework tip  2

Thus, the critical value for a 95% confidence interval of β1 is 2.572.

f.

To determine

Find the margin of error for a 95% confidence interval of β1.

f.

Expert Solution
Check Mark

Answer to Problem 18E

The margin of error of b1 is 1.944.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of β1 is calculated using the formula:

Margin of error=tα2sb1

Where,

tα2 represents the two tailed critical value for a given level of significance.

sb1 represents the standard error of b1.

Substitute tα2 as 2.572 and sb1 as 0.756.

Margin of error=(2.571)(0.756)=1.944

Thus, the margin of error of b1 is 1.944.

g.

To determine

Construct the 95% confidence interval for β1.

g.

Expert Solution
Check Mark

Answer to Problem 18E

The 95% confidence interval for β1 is (2.00,1.88).

Explanation of Solution

Calculation:

The confidence interval for β1 is calculated by using the formula:

Confidence interval=b1±tα2sb1

Where,

b1 represents the estimated slope coefficient.

tα2sb1 represents the margin of error.

Confidence interval=0.063±(2.571)(0.756)=0.063±1.944=2.00,1.88

Thus, the 95% confidence interval for β1 is (2.00,1.88).

h.

To determine

Test the significance of β1 using 5% level of significance.

h.

Expert Solution
Check Mark

Answer to Problem 18E

There is no support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

H0:β1=0

That is, there is no linear relationship between x and y.

Alternate hypothesis:

H1:β10

That is, there is a linear relationship between x and y.

Test statistic:

t=β^1β1sb1

Where,

β^1 represents the estimated slope coefficient.

β1 represents the hypothesized value of slope coefficient.

sb1 represents the standard error of slope coefficient.

Substitute β^1 as –0.063, β1 as 0 and sb1 as 0.756.

t=0.06300.756=0.0630.756=0.083

From part e, the critical value is identified as 2.571.

Decision Rule:

Reject the null hypothesis when the test statistic value is greater than the critical value for a given level of significance. Otherwise, do not reject the null hypothesis.

Conclusion:

The test statistic value is 0.083 and the critical value is 2.571.

The test statistic value is lesser than the critical value.

That is, 0.083(=test statistic)<2.571(=critical value)

Thus, the null hypothesis is not rejected.

Hence, there is no support of evidence to conclude that there is a linear relationship between x and y.

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Chapter 11 Solutions

Essential Statistics

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