Answer to Question #267394 in Statistics and Probability for Maika Togavuto

Question

A study is conducted to determine the relationship between a driver’s age and the number of accidents he or she has over a 1-year period. The data are shown here.

Drivers Age x - 16, 24, 18, 17, 23, 27, 32

No. of accidents y - 3, 2, 5, 2, 0, 1, 1

a. Draw the scatter plot.

b. Compute the value of the correlation coefficient.

c. Test the significance of the correlation coefficient at alpha = 0.01.

d. Determine the regression line equation.

e. Predict the number of accidents of a driver who is 28.

Answer 1

a.

b.

correlation coefficient:

$r=\frac{\sum(x_i-\overline{x})(y_i-\overline{y})}{\sqrt{\sum(x_i-\overline{x})^2(y_i-\overline{y})^2}}$

$\overline{x}=22.4,\overline{y}=2$

$r=-0.61$

c.

Null Hypothesis: H₀: r=0

Alternate Hypothesis: H_a: r≠0

test statistic:

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}=\frac{-0.61\sqrt{5}}{\sqrt{1-0.61^2}}=-1.721$

$df=n-2=5$

critical value:

$t_{crit}=4.032$

Since $|t|<t_{crit}$ we accept Null Hypothesis. Correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship(correlation) between x and y.

d.

equation of regression line:

$y=ax+b$

$a=\frac{\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=-0.1701$

$b=\frac{\sum y\sum x^2-\sum x\sum xy}{n\sum x^2-(\sum x)^2}=5.816$

$y=5.816-0.1701x$

e.

number of accidents of a driver who is 28:

$y(28)=5.816-0.1701\cdot28=1.05\approx 1$ accident