# 13.11 Review Questions

Researchers examined the controversial issue of the human vomeronasal organ regarding its structure, function, and identity. The following table shows the age of fetuses ([latex]x[/latex]) in weeks and the length of crown-rump ([latex]y[/latex]) in millimeters.

Age ([latex]x[/latex]) | 10 | 10 | 13 | 13 | 18 | 19 | 19 | 23 | 25 | 28 |
---|---|---|---|---|---|---|---|---|---|---|

Length ([latex]y[/latex]) | 66 | 66 | 108 | 106 | 161 | 166 | 177 | 228 | 235 | 280 |

The summaries of the data are given by [latex]n = 10, \sum x_i = 178, \sum x_i^2 = 3522, \sum y_i = 1593, \sum y_i^2 = 302027, \sum x_iy_i = 32476.[/latex]

- Given the summaries of the data, find the least-squares regression equation.
- Graph the regression equation and the data points.
- Interpret the slope of the regression equation obtained in part (a) in the context of the study.
- Calculate [latex]r[/latex], the correlation coefficient between [latex]y[/latex] and [latex]x[/latex]. Interpret the number.
- Calculate the coefficient of determination [latex]r^2[/latex]. Interpret the number.
- Test at the 1% significant level whether the age of fetuses is a useful predictor for the length of the crown-rump. You could use [latex]s_e = 5.518.[/latex]
- Predict the crown-rump length of a 19-week-old fetus.
- What is the residual for the last observation with response [latex]y = 280[/latex] and [latex]x = 28[/latex]?

## Show/Hide Answer

- [latex]\begin{align*} S_{xx}&=\sum x_i^2-\frac{(\sum x_i)^2}{n} = 3522-\frac{(178)^2}{10} =353.6; \\ S_{xy}&=\sum x_iy_i-\frac{(\sum x_i)(\sum y_i)}{n} =32476- \frac{(178)(1593)}{10}=4120.6; \\ S_{yy}&=\sum y_i^2-\frac{(\sum y_i)^2}{n} = 302027- \frac{(1593)^2}{10} = 48262.1; \\ b_1&=\frac{S_{xy}}{S_{xx}}=\frac{4120.6}{353.6}=11.65328;\\ b_0&=\frac{\sum y_i}{n}-b_1\times \frac{\sum x_i}{n}=\frac{1593}{10}-11.65328\times \frac{178}{10}=-48.12838. \end{align*}[/latex]

Therefore, the least-squares regression equation [latex]\hat y=b_0+b_1 x=-48.12838+11.65328x[/latex] or [latex]\widehat {length}=-48.12838+11.65328age[/latex]. - Left as an exercise for the reader.
- The slope is [latex]b_1=11.65328[/latex].

Interpretation: The average length of the crown rump increases by 11.65328 millimeters when the age of the fetus increases by 1 week. In other words, for each week the fetus ages, the expected increase in crown-rump length is 11.65328 mm. - The correlation coefficient [latex]r[/latex] is given by [latex]r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{4120.6}{\sqrt{(353.6)(48262.1)}}=0.9974732.[/latex]

Interpretation: there is a very strong, positive, linear association between the length of crown-rump [latex](y)[/latex] and the age [latex](x)[/latex] of the fetus. - The coefficient of determination is [latex]r^2=0.9974732^2=0.9949528.[/latex]

Interpretation: 99.50% of the variation in the length of the crown rump is due to the age of the fetus. Or 99.50% of the variation in the length of crown-rump can be explained by the age of the fetus through the fitted regression line [latex]\hat y=b_0+b_1 x=-48.12838+11.65328x.[/latex] - We assume all assumptions for inference on simple linear regression are satisfied.

Step 1: Hypotheses. [latex]H_0: \beta_1=0[/latex] versus [latex]H_a: \beta_1\ne 0.[/latex]

Step 2: Significance level [latex]\alpha=0.01.[/latex]

Step 3: Test statistic [latex]t_o=\frac{b_1}{(\frac{s_e}{\sqrt{S_{xx}}})}=\frac{11.65328}{(\frac{5.518}{\sqrt{353.6}})}=39.71208[/latex] with [latex]df=n-2=10-2=8.[/latex]

Step 4: P-value. It is a two-tailed test, [latex]\text{p-value}= 2P(t\ge |t_o|)=2P(t\ge 39.71208)<2\times 0.0005=0.001.[/latex] Step 5: Decision. We reject [latex]H_0[/latex] since [latex]\text{p-value}<0.001<0.01(\alpha).[/latex] Step 6: Conclusion. At the 1% significant level, we have sufficient evidence that the age of fetuses is a useful predictor for the crown-rump length. - [latex]\hat y=b_0+b_1 x=-48.12838+11.65328x=-48.12838+11.65328\times 19=173.2839[/latex]

The predicted crown-rump length of a 19-week-old fetus is 173.2839 mm. - Residual [latex]e=y-\hat y=y-(b_0+b_1x)=280-(-48.12838+11.65328\times 28)=280-278.1635=1.8365.[/latex]