Chapter 27, Problem 41E

a)

To determine

To check the assumptions and conditions for inference.

Answer to Problem 41E

All conditions satisfied.

Explanation of Solution

Given:

  

  

The conditions are: straight enough, Independence, Randomization, Random residual, Does plot thicken? and Nearly normal condition.

We will check it one by one.

Straight enough condition: Satisfied, because no curvature is present in the scatterplot.

Independence assumption: Satisfied, because the 58 are less than 10% of the population of all cities.

Randomization condition: Satisfied, as cities were randomly selected.

Random residuals condition: Satisfied, because there is no obvious pattern in the residual plot.

Does the plot thicken? condition: Satisfied, because the vertical spread in the residual plot is not increasing or decreasing.

Nearly normal condition: Satisfied, because the histogram of the residual is roughly symmetric and unimodal.

b)

To determine

To perform the hypothesis test.

Answer to Problem 41E

There is sufficient evidence that the slope is non zero and there is association between mortality rate and education level.

Explanation of Solution

Given:

  

Formula:

Test statistic:

  $t=\frac{b-{\beta }_0}{S{E}_b}$

The null and alternative hypotheses:

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_a:{\beta }_1\ne 0\end{array}$

Test statistic:

  $t=\frac{-49.9202-0}{8}=-6.24$

The degrees of freedom = df = 56

Therefore, p-value would be,

P-value = 0.0000 …Using excel formula, =TDIST(6.24,56,2)

Decision: P-value < 0.05, reject H₀.

Conclusion: There is sufficient evidence that the slope is non zero and there is association between mortality rate and education level.

c)

To determine

To explain whether to conclude that getting more education is likely to prolong your life.

Answer to Problem 41E

No.

Explanation of Solution

Given:

  

The data are on cities, not individuals. Also, these are observational data. Therefore, we can not predict causal consequences from them.

d)

To determine

To find the 95% confidence interval for the slope of the regression line.

Answer to Problem 41E

We are 95% confident that the slope of the population of regression line is between -65.9442 and -33.8962.

Explanation of Solution

Given:

  

Formula:

Confidence interval for slope is,

  $(Coefficient(b)\mp t_c\times S{E}_b)$

The confidence level = 0.95

So, level of significance = $\alpha$ = 0.05

First need to find critical t value.

tc = 2.003 …Using excel formula, =TINV(0.05,56)

Using formula, 95% confidence interval for the slope of the regression line.

  $\begin{array}{l}(-49.9202-2.003\times 8,-49.9202+2.003\times 8)\\ (-65.9442,-33.8962)\end{array}$

Therefore, we are 95% confident that the slope of the population of regression line is between -65.9442 and -33.8962.

e)

To determine

To explain the 95% confidence interval for the slope of the regression line.

Answer to Problem 41E

The mortality decreases on average between 33.8946 and 65.9442 deaths per 100,000 for each extra year of average Education.

Explanation of Solution

Given:

  

Therefore, we are 95% confident that the slope of the population of regression line is between -65.9442 and -33.8962.

The mortality decreases on average between 33.8946 and 65.9442 deaths per 100,000 for each extra year of average Education.

f)

To determine

To find the 95% confidence interval for the average Mortality rate in cities where the adult population completed an average of 12 years of school.

Answer to Problem 41E

We are 95% confidence that the average Mortality for cities with an average of 12 years of Education will be between 874.23 and 914.19 deaths per 100,000 people.

Explanation of Solution

Given:

  

Formula:

Confidence interval for average of response variable is,

  $(\widehat{y}\mp t_c\times S{E}_{\widehat{\mu }})$

Standard error is,

  $S{E}_{\widehat{\mu }}=\sqrt{S{E}_b{}^2{(x-\overline{x})}^2+\frac{s{e}^2}{n}}$

Therefore, predicted value becomes,

  $\widehat{y}=1493.26-49.9202\times 12=894.2176$

So, standard error is,

  $\begin{array}{l}S{E}_{\widehat{\mu }}=\sqrt{8^2\times {(12-11.0328)}^2+\frac{{47.92}^2}{58}}\\ S{E}_{\widehat{\mu }}=9.97\end{array}$

The confidence level = 0.95

So, level of significance = $\alpha$ = 0.05

First need to find critical t value.

tc = 2.003 …Using excel formula, =TINV(0.05,56)

Using formula, 95% confidence interval is,

  $\begin{array}{l}(894.2176-2.003\times 9.97,894.2176+2.003\times 9.97)\\ (874.23,914.19)\end{array}$

Hence, we are 95% confidence that the average Mortality for cities with an average of 12 years of Education will be between 874.23 and 914.19 deaths per 100,000 people.