Chapter 13, Problem 61CR

a.

To determine

Check whether the data contradict the prior belief.

State and test the appropriate hypotheses using a 0.10 level of significance.

Explanation of Solution

The data is on age (x) and percentage of cribriform area of the lamina scleralis occupied by pores (y). The researchers believed that the average decrease in percentage area associated with a 1 year age increase was 0.5%.

1.

Here, $\beta$ indicates the slope of the population regression line relating age and percentage of cribriform area of the lamina scleralis occupied by pores.

2.

Null hypothesis:

$H_0:\beta =-0.5$

That is, the average decrease in percentage area is 0.5.

3.

Alternative hypothesis:

$H_a:\beta \ne -0.5$

That is, the average decrease in percentage area is not 0.5.

4.

Here, the significance level is $\alpha =0.1$.

5.

Test Statistic:

The formula for test statistic is,

$t=\frac{b-\left(\text{hypothesized value}\right)}{s_b}$.

In the formula, b denotes the estimated slope, $s_b$ denotes the standard deviation of b.

6.

The data are plotted in the scatter plot below.

If the scatterplot of the data shows a linear pattern, and the vertical variability of points does not appear to be changing over the range of x values in the sample, then it can be said that the data is consistent with the use of the simple linear regression model.

Software procedure:

Step-by-step procedure to obtain the scatterplot using MINITAB software:

• Choose Graph>Scatter plot.
• Select Simple.
• Click OK.
• Under Y variables, enter the column of y.
• Under X variables, enter the column of x.
• Click OK.

The output using MINITAB software is given below:

The scatter plot shows no apparent curve and there are no extreme observations. There is no change in the y values as the value of x changes and there are no influential points. Therefore, the simple linear regression model seems appropriate for the data set.

Assumption:

Here, the assumption made is that, the simple linear regression model is appropriate.

7.

Calculation:

The values required for the calculation of t is obtained in the following steps:

Software procedure:

Step-by-step procedure to obtain regression line using MINITAB software:

• Choose Stat > Regression > Regression > Fit Regression Model.
• Under Responses, enter the column of values y.
• Under Continuous predictors, enter the column of values x.
• Click OK.

Output using MINITAB software is given below:

Since, the output provides test statistic value for a hypothesized mean of 0, the test statistic value for a hypothesized mean of –0.5 is obtained below.

For the given x values, ${\displaystyle \sum x^2=4,334.41\text{ and }{\left[{\displaystyle \sum x}\right]}^2}=3,588.01$. The value of $S_{xx}\text{ and }{s}_b$ is as follows.

$\begin{array}{c}{S}_{xx}={\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}\\ =43,447-\frac{{\left(821\right)}^2}{17}\\ =43.447-39,649.47\\ =3797.529\end{array}$

$\begin{array}{c}{s}_b=\frac{s_e}{\sqrt{S_{xx}}}\\ =\frac{6.75598}{\sqrt{3797.529}}\\ =0.1096\end{array}$

Substitute, $b=-0.447$, $s_e=6.75598\text{ and }{s}_{xx}=3797.529$ and hypothesized value as –0.5 in the test statistic formula.

$\begin{array}{c}t=\frac{-0.447-\left(-0.5\right)}{0.1096}\\ =0.488\end{array}$

Hence, the test statistic value is 0.488.

8.

Formula for Degrees of freedom:

The formula for degrees of freedom is as follows:

$df=n-2$

Degrees of freedom:

The number of data values given are 17, that is $n=17$. Substitute $n=17$ in the degrees of freedom formula.

$\begin{array}{c}df=17-2\\ =15\end{array}$

P-value:

Formula for p-value:

The P-value is, $2\times P\left(t>t_{cal}\right)$.

The test statistic value is 0.488. The value of $P\left(t>0.488\right)$ is to be obtained.

Software procedure:

Step-by-Step procedure to find probability using MINITAB software:

• Choose Graph > Probability Distribution Plot.
• Choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter Degrees of freedom as 15.
• Click the Shaded Area tab.
• Choose X Value and Right tail for the region of the curve to shade.
• Enter the X value as 0.488.
• Click OK.

Output using MINITAB software is as follows:

Thus, $P\left(t_{15}>0.488\right)=0.3163$

P-value:

$\begin{array}{c}2\times P\left(t>t_{cal}\right)=2\times 0.3163\\ =0.6326\\ \approx 0.633\end{array}$

Thus, the p-value is 0.633.

9.

Rejection rule:

If $P\text{-value}\le \alpha$ then reject the null hypothesis.

Conclusion:

The P-value is 0.633.

The level of significance is 0.1.

The P-value is greater than the level of significance.

That is, $P\text{-value}\left(=0.633\right)>\alpha \left(=0.1\right)$

Based on rejection rule, do not reject the null hypothesis.

Thus, there is no convincing evidence that the average decrease in percentage area associated with a 1-year age increase is not 0.5.

b.

To determine

Obtain an estimate of the average percentage area covered by pores for all 50-year olds in the population.

Answer to Problem 61CR

The estimate of the average percentage area covered by pores for all 50-year olds in the population is between 47.05 and 54.08.

Explanation of Solution

Calculation:

The confidence interval for $\alpha +\beta x^{*}$ is $(a+b{x}^{*})\pm (t\text{ critical value)}\cdot s_{a+b{x}^{*}}$, where $(a+b{x}^{*})$ is the point estimate value, t critical value is based on df = n – 2, and $s_{a+b{x}^{*}}$ is the estimated standard deviation.

From the MINITAB output in Part (a), the estimated linear regression line is  $\widehat{y}=72.92-0.447x$ and $s_e=6.75598$.

Point estimate:

The point estimate when the percentage area covered by pores for all 50-year olds in the population is calculated as follows.

$\begin{array}{c}\widehat{y}=72.92-0.447(50)\\ =72.92-22.35\\ =50.57\end{array}$

Estimated standard deviation:

Substitute, $s_{xx}=3797.529\text{ and }\overline{x}=48.294$. The estimated standard deviation of $a+b(50)$ is as follows.

$\begin{array}{c}{s}_{a+b{x}^{*}}=s_e\sqrt{\frac{1}{n}+\frac{{\left(20-\overline{x}\right)}^2}{s_{xx}}}\\ s_{a+b(50)}=6.75598\sqrt{\frac{1}{17}+\frac{{\left(50-48.294\right)}^2}{3797.529}}\\ =6.75598\sqrt{0.0588+\frac{2.9104}{3797.529}}\\ =6.75598\sqrt{0.0596+0.00008}\\ =6.75598\times 0.2441\\ =1.649\end{array}$

Critical value:

From the Appendix: Table 3 the t Critical Values:

• Locate the value 15 in the degrees of freedom (df) column.
• Locate the 0.95 in the row of central area captured.
• The intersecting value that corresponds to the df 15 with confidence level 0.95 is 2.13.

Thus, the critical value for $df=15$ with two-tailed test is $2.13$.

Substitute $a+b{x}^{*}=50.57$, $t\text{ critical value}=2.13$, and $s_{a+b(50)}=1.649$ in confidence interval as shown below.

$\begin{array}{c}50.57\pm (2.13\times 1.649)=\left(50.57-3.5124,50.57-3.5124\right)\\ =\left(47.05,54.08\right)\end{array}$

Therefore, one can be 95% confident that the estimate of the average percentage area covered by pores for all 50-year olds in the population is between 47.05 and 54.08.