13.11 Review Questions

Wanhua Su

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 ( $x$ ) in weeks and the length of crown-rump ( $y$ ) in millimeters.

Age ( $x$ )	10	10	13	13	18	19	19	23	25	28
Length ( $y$ )	66	66	108	106	161	166	177	228	235	280

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

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 $r$ , the correlation coefficient between $y$ and $x$ . Interpret the number.
Calculate the coefficient of determination $r^2$ . 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 $s_e = 5.518.$
Predict the crown-rump length of a 19-week-old fetus.
What is the residual for the last observation with response $y = 280$ and $x = 28$ ?

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$\begin{align*} S_{xx}&=\sum x_i^2-\frac{(\sum x_i)^2}{n} = 3522-\frac{(178)^2}{10} =353.6; \\ S_{xy}&=\sum x_iy_i-\frac{(\sum x_i)(\sum y_i)}{n} =32476- \frac{(178)(1593)}{10}=4120.6; \\ S_{yy}&=\sum y_i^2-\frac{(\sum y_i)^2}{n} = 302027- \frac{(1593)^2}{10} = 48262.1; \\ b_1&=\frac{S_{xy}}{S_{xx}}=\frac{4120.6}{353.6}=11.65328;\\ b_0&=\frac{\sum y_i}{n}-b_1\times \frac{\sum x_i}{n}=\frac{1593}{10}-11.65328\times \frac{178}{10}=-48.12838. \end{align*}$
Therefore, the least-squares regression equation $\hat y=b_0+b_1 x=-48.12838+11.65328x$ or $\widehat {length}=-48.12838+11.65328age$ .
Left as an exercise for the reader.
The slope is $b_1=11.65328$ .
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 $r$ is given by $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{4120.6}{\sqrt{(353.6)(48262.1)}}=0.9974732.$
Interpretation: there is a very strong, positive, linear association between the length of crown-rump $(y)$ and the age $(x)$ of the fetus.
The coefficient of determination is $r^2=0.9974732^2=0.9949528.$
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 $\hat y=b_0+b_1 x=-48.12838+11.65328x.$
We assume all assumptions for inference on simple linear regression are satisfied.
Step 1: Hypotheses. $H_0: \beta_1=0$ versus $H_a: \beta_1\ne 0.$
Step 2: Significance level $\alpha=0.01.$
Step 3: Test statistic $t_o=\frac{b_1}{(\frac{s_e}{\sqrt{S_{xx}}})}=\frac{11.65328}{(\frac{5.518}{\sqrt{353.6}})}=39.71208$ with $df=n-2=10-2=8.$
Step 4: P-value. It is a two-tailed test, $\text{p-value}= 2P(t\ge |t_o|)=2P(t\ge 39.71208)<2\times 0.0005=0.001.$ Step 5: Decision. We reject $H_0$ since $\text{p-value}<0.001<0.01(\alpha).$ 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.
$\hat y=b_0+b_1 x=-48.12838+11.65328x=-48.12838+11.65328\times 19=173.2839$
The predicted crown-rump length of a 19-week-old fetus is 173.2839 mm.
Residual $e=y-\hat y=y-(b_0+b_1x)=280-(-48.12838+11.65328\times 28)=280-278.1635=1.8365.$

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