Question

Automobile insurance companies take many factors into consideration when setting rates. These factors include age, marital status and miles driven per year. To determine the effect of gender, a random sample of young (under 25, with at least 2 years of driving experience) male and female drivers was surveyed. Each was asked how many miles he or she had driven in the past year. The distances (in thousands of miles) are stored in stacked format (column 1= driving distances and column 2 identifies gender where 1=male and 2=female). (Assume equal variances.) (a) Can we conclude that male & female drivers differ in the numbers of miles driven per year? (you need to compute the sample means and sample standard deviations for each gender in the Excel Workbook that comes with this exercise. Use ”=AVERAGE()” and ”=STDEV()” formulas. This also applies to exercises further below.) (b) Estimate with 95% confidence the difference in mean distance driven by male and female drivers.

Men	Women	Diff
75.45	155.84	-80.39
1869.44	1420.88	448.56
487.22	267.56	219.66
1529.57	1843.48	-313.91
423.12	338.49	84.63
279.68	757.35	-477.67
794.43	442.36	352.07
1.13	329.31	-328.18
56.78	305.57	-248.79
699.41	514.4	185.01
1278.74	1048.98	229.76
395.54	526.7	-131.16
2217.96	2404.58	-186.62
996.27	622.56	373.71
640.77	459.78	180.99
1866.03	1777.64	88.3899999999999
587.89	807.67	-219.78
520.63	726.32	-205.69
1477.49	1609.63	-132.14
392.41	1298.86	-906.45
1724.05	1350.07	373.98
506.07	608.41	-102.34
1357.56	1155.45	202.11
259.9	406.43	-146.53
432.8	-12.14	444.94
3033.65	3450	-416.35
978.01	792.47	185.54
1953.09	1828.4	124.69
722.98	1069.65	-346.67
1806.8	2237.03	-430.23
1031.63	1184.15	-152.52
-86.52	319.43	-405.95
1828.28	1845.77	-17.49
2408.31	2734.14	-325.83
2676.72	2523.3	153.42
1870.92	1235.51	635.41
2751.37	3183.55	-432.18
1405.73	1746.62	-340.89
1530.83	1114.66	416.17
1796.1	2091.21	-295.11
2537.42	2844.19	-306.77
1291.7	1216.35	75.3500000000001
1013.79	870.75	143.04
1443.57	1509.97	-66.4000000000001
1822.24	1656.05	166.19

Answer 1

We have already assumed equal variances.

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

alpha= 0.05

Since Variances are equal we need to find pooled variance

Pooled variance=

The following formula is used to compute the pooled variance:

sp^2​=(n1​−1)s1^2​+(n2​−1)s2^2​​/n1​+n2​−2

Plugging in the corresponding values in the formula above, we get that

sp^2=(45−1)(814.2299)^2+(45−1)(653.2246)^2/(​45+45−2)=544836.354

Now that we have the pooled variance, we can compute the pooled standard deviation by simply taking the squared root of the value obtained for the variance.

The following pooled standard deviation is obtained:

sp​=sqrt(544836.354​)=738.1303

standard error of mean difference= 738.1303*sqrt(1/45+1/45)= 155.61

difference in means= (1215.266-1258.208)=-42.9427

t= -42.9427/155.61= -0.27596

t=-0.276

degrees of freedom= 45+45-2= 88

The P-Value is .783195.

P value>0.05 hence not significant.

Thefore fail to reject null hypothesis. We dont have sufficient evidence to show that there is difference in means.

b) 95% confidence interval

Confidence Interval
μ1 - μ2 = (M1 - M2) ± ts(M1 - M2) = -42.942 ± (1.99 * 155.61) = -42.942 ± 309.664

95% CI [-352.606,266.722].

You can be 95% confident that the difference between your two population means (μ1 - μ2) lies between -352.606, 266.722.