Probability and Statistics for Engineering and the Sciences
Probability and Statistics for Engineering and the Sciences
9th Edition
ISBN: 9781305251809
Author: Jay L. Devore
Publisher: Cengage Learning
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Question
Chapter 12.5, Problem 62E

a.

To determine

Test whether there is enough evidence to conclude that the variables cation exchange capacity and specific surface area are positively correlated.

a.

Expert Solution
Check Mark

Answer to Problem 62E

There is enough evidence to conclude that the variables cation exchange capacity and specific surface area are positively correlated.

Explanation of Solution

Given info:

The data represents the values of the variables cation exchange capacity (x) and specific surface area (y). The correlation between x and y is 0.853.

Calculation:

Linear regression model:

A linear regression model is given as y^=b0+b1x where y^ be the predicted values of response variable and x be the predictor variable. b1 be the slope and b0 be the intercept of the line.

A linear regression model is given as y^=β^0+β^1x where y^ be the predicted values of response variable and x be the predictor variable. The β^1 be the estimate of slope and β^0 be the estimate of intercept of the line.

The test hypotheses are given below:

Null hypothesis:

 H0:ρ=0

That is, there exists no linear relationship between the variables cation exchange capacity (x) and specific surface area (y).

Alternative hypothesis:

H1:ρ>0

That is, there exists a positive correlation between the variables cation exchange capacity (x) and specific surface area (y).

Test statistic:

The test statistic is,

t=r1r2n2t(n2)

Degrees of freedom:

The sample size is n=20.

The degrees of freedom is,

d.f=n2=202=18

Thus, the degree of freedom is 18.

From the accompanying MINITAB output, the correlation between the variables cation exchange capacity and specific surface area is 0.853.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

t=rρ1r2n2=0.853010.85322026.93

Thus, the test statistic is 6.93.

Level of significance:

Here, level of significance is not given.

So, the prior level of significance α=0.05 can be used.

Critical value:

Software procedure:

Step by step procedure to obtain the critical value using the MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution and enter 18 as degree of freedom.
  • Click the Shaded Area tab.
  • Choose Probability-Value and Right Tail for the region of the curve to shade.
  • Enter the X-value as 0.05.
  • Click OK.

Output using the MINITAB software is given below:

Probability and Statistics for Engineering and the Sciences, Chapter 12.5, Problem 62E , additional homework tip  1

From the MINITAB output, the critical value is 1.734.

Thus, the critical value is 1.734.

Decision rule based on classical approach:

  • If test statistic(t)>critical value(tα), then reject the null hypothesis H0.
  • If test statistic(t)<critical value(tα), then fail reject the null hypothesis H0.

Conclusion:

The test statistic is 6.93 and critical value is 1.734.

Here, test statistic value is greater than the critical value.

That is 6.93(=tα)>1.734(=t).

By the rejection rule, reject the null hypothesis.

Therefore, the population correlation coefficient is greater than zero.

That is, there exists positive linear relationship between the variables cation exchange capacity (x) and specific surface area (y).

Therefore, there is enough evidence to conclude that the variables cation exchange capacity and specific surface area are positively correlated.

b.

To determine

Find the value of r at which the null hypothesis will not be rejected at 0.01 level of significance in part (a).

b.

Expert Solution
Check Mark

Answer to Problem 62E

The correlation coefficient should be less than 0.5155 to not reject the null hypothesis in part (a).

Explanation of Solution

Calculation:

Here, the sample size is n=20 and the level of significance is α=0.01.

Critical value:

Software procedure:

Step by step procedure to obtain the critical value using the MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution and enter 18 as degree of freedom.
  • Click the Shaded Area tab.
  • Choose Probability-Value and Right Tail for the region of the curve to shade.
  • Enter the X-value as 0.01.
  • Click OK.

Output using the MINITAB software is given below:

Probability and Statistics for Engineering and the Sciences, Chapter 12.5, Problem 62E , additional homework tip  2

From the MINITAB output, the critical value is 2.552.

Thus, the critical value is 2.552.

Decision rule based on classical approach:

  • If test statistic(t)>critical value(tα), then reject the null hypothesis H0.
  • If test statistic(t)<critical value(tα), then fail reject the null hypothesis H0.

Here, the test statistic value should be less than the critical value.

That is, the t(=r1r2n2)<t0.01,18(=2.552)

Required correlation coefficient is obtained as follows:

rn21r2<2.55218r21r2<6.512718r2<6.51276.5127r224.5127r2<6.5127

r<0.5155

Thus, the correlation coefficient between the variables cation exchange capacity (x) and specific surface area (y) should be less than 0.5155.

Thus, the correlation coefficient should be less than 0.5155 to not reject the null hypothesis in part (a).

c.

To determine

Find the 95% confidence interval estimate for the population correlation coefficient.

c.

Expert Solution
Check Mark

Answer to Problem 62E

The 95% confidence interval estimate for the population correlation coefficient is 0.659ρ0.941_.

Explanation of Solution

Calculation:

Linear regression model:

A linear regression model is given as y^=b0+b1x where y^ be the predicted values of response variable and x be the predictor variable. b1 be the slope and b0 be the intercept of the line.

A linear regression model is given as y^=β^0+β^1x where y^ be the predicted values of response variable and x be the predictor variable. The β^1 be the estimate of slope and β^0 be the estimate of intercept of the line.

Confidence interval for ρ:

The general formula for 100(1α)% confidence interval for population correlation coefficient is,

CI=(e2c11e2c11,e2c21e2c21)

Where, .c1 and c2 are the lower and upper bounds of the confidence interval of μv.

Confidence interval for μv:

The random variable V is,

V=12ln(1+r1r)

Here, the random variable V follows normal distribution with mean μV=12ln(1+ρ1ρ) and variance σV2=1n3.

From the accompanying MINITAB output, the correlation between the variables cation exchange capacity and specific surface area is 0.853.

The random variable V is obtained as follows:

V=12ln(1+r1r)=12ln(1+0.85310.853)=1.267

Thus, the value of the random variable is v=1.267.

The confidence interval for μv is obtained as follows:

Critical value:

For 95% confidence level,

1α=10.95α=0.05α2=0.052=0.025

From Table A.3 of the standard normal distribution in Appendix A, the critical value corresponding to the right tail area 0.025 is 1.96.

Thus, the critical value is (z0.025)=1.96.

The 95% confidence interval for μv is,

C.I=(vzα2n3,v+zα2n3)=(1.2671.9618,1.267+1.9618)(0.792,1.742)

Thus, the lower and upper bounds of the 95% confidence interval of μv are c1=0.792 and c2=1.742.

Confidence interval for ρ:

The 95% confidence interval for population correlation coefficient ρ is,

CI=(e2c11e2c11,e2c21e2c21)=(e2×0.7921e2×0.7921,e2×1.7421e2×1.7421)=(0.659,0.941)

Thus, the 95% confidence interval for the population correlation coefficient is 0.659ρ0.941_

Interpretation:

There is 95% confident, that the population correlation coefficient of the variables cation exchange capacity and specific surface area lies between 0.659 and 0.941.

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