Question

1	11	1.771
2	9	1.392
3	10	1.495
4	16	4.561
5	14	3.136
6	11	1.606
7	15	2.835
8	10	1.317
9	9	0.925
10	10	1.761
11	9	0
12	19	5.902
13	17	4.624
14	9	0.84
15	12	2.802
16	15	3.789
17	8	1.334
18	7	1.244
19	12	1.578
20	8	1.231
21	9	1.693
22	3	0
23	11	2.035
24	11	1.885
25	12	1.482
26	14	3.719
27	14	1.333
28	15	2.244
29	7	0.572
30	9	1.924
31	9	1.413
32	9	0

nPlot the number of fatal accidents as a function of percentage of drivers under 21. Based on the plot, try to anticipate whether or not the following analysis will show that there is a significant increase or decrease in number of fatalities with increases in percentage of drivers under 21.

                                                                                                                                      {Example 27}

nRegression analysis, where one variable depends on another, can be used to predict levels of a dependent variable for specified levels of an independent variable. Use the EXCEL REGRESSION command to calculate the intercept and slope of the least‑squares line, as well as the analysis of variance associated with that line. Fill in the following table and use the results to answer the next few questions. Carefullychoose your independent and dependent variables and input them correctly using EXCEL’s regression command. In this example, the percentage of drivers under the age of 21 affects the number of Fatals/1000 licenses.

The regression equation (least‑squares line) is

            Fatals/1000 licenses =                  +                   % under 21

                                                 (intercept)       (slope)

Analysis of variance

Source                         DF          SS                 MS          F               P

Regression                   1         ________   _______ ________   _______                      

Residual (Error)           30        ________   _______

15. What are the degrees of freedom for the standard error of estimate (and the standard deviation of the slope); i.e. what are the error degrees of freedom?

                                                                                                                                      {Example 29}

nStandard methodology for a single sample mean can be used to calculate a confidence interval for the slope of the least‑squares line and to test hypotheses other than H₀: ß₁= 0. In both cases, one needs to have an estimate of the slope and of its standard deviation (sometimes called standard error). Furthermore, one needs to recognize that the degrees of freedom for the standard deviation is the same as the error degrees of freedom (n ‑ 2).

Note that the EXCEL gives the standard error of estimate directly, but correctly calls it the standard deviation of the slope. Therefore, you must not divide by the square root of sample sizeas in example 16.

Use the above information to calculate a 90% confidence interval for the slope of the true regression line. For 30 degrees of freedom and a= 0.1, the critical t‑value is 1.697.

16. What is the margin of error for calculating a 90% confidence interval for the slope of the regression line (i.e. 1.697 ´the standard deviation of the slope)?

17. What is the lower 90% confidence limit for the slope?        (i.e. slope – margin of error)

18. What is the upper 90% confidence limit for the slope?

       (i.e. slope + margin of error)

                                                                                                                                                            

nUse this same information to calculate a statistic to test the null hypothesis H₀: ß₁= 0.05 against a one‑sided alternative H₁: ß₁> 0.05. Use a 1 percent significance level (for which the critical value is 2.423).

            Reminder:  t = estimated value - hypothesized value  =   slope -  0.05

                                    standard error (deviation) of estimate        st dev of slope

19. What is the value of the test statistic for testing this hypothesis?

Answer 1

Plot:

The above plot shows the number of fatal accidents as a function of the percentage of drivers under 21. Clearly, the trend line in the above graph or plot shows that there is a significant increase in the number of fatalities (accidents) with increases in the percentage of drivers under 21.

The regression equation (least‑squares line) :

Fatals/1000 licenses =        6.9165     +     2.1247      % under 21

                                      (intercept)       (slope)

Analysis of Variance
	df	SS	MS	F	Significance F
Regression	1	263.2533888	263.2534	86.14937	3.4824E-10
Residual	29	88.61757893	3.055779
Total	30	351.8709677

15. The error degrees of freedom is (n-2) i.e. 30.

16. The margin of error for calculating a 90% confidence interval for the slope of the regression line (i.e. 1.697 ´the standard deviation of the slope) = 1.697 * Standard error = 1.697 * 1.7480 (From Excel) = 2.966