Question

Among drivers who have had a car crash in the last year, 170 were randomly selected and categorized by age, with the results listed in the table below. Age Under 25 25-44 45-64 Over 64 Drivers 70 41 22 37 If all ages have the same crash rate, we would expect (because of the age distribution of licensed drivers) the given categories to have 16%, 44%, 27%, 13% of the subjects, respectively. At the 0.025 significance level, test the claim that the distribution of crashes conforms to the distribution of ages.

The test statistic is ?2=

The p-value is 0

The conclusion is

A. There is not sufficient evidence to warrant the rejection of the claim that the distribution of crashes conforms to the distibuion of ages.

B. There is sufficient evidence to warrant the rejection of the claim that the distribution of crashes conforms to the distibuion of ages.

Answer 1

Null Hypothesis : H_o : The distribution of crashes conforms to the distribution of ages

Alternate Hypothesis : H_a:  The distribution of crashes does not conform to the distribution of ages

Where O : Observed Frequency

E : Expected Frequency

Given,

Age	Under 25	25-44	45-64	over 64	Total
Drivers: Observe Frequency :O	70	41	22	37	170
Age Distribution	16%	44%	27%	13%

Expected frequency : Age Distribution percentage x Total number drivers.

Age	Under 25	25-44	45-64	over 64	Total
Drivers: Observe Frequency :O	70	41	22	37	170
Age Distribution	16%	44%	27%	13%
Expected Frequency: E	16% of 170	44% of 170	27% of 170	13% of 170
Expected Frequency: E	27.2	74.8	45.9	22.1	170

O	E	O-E	(O-E)2	(O-E)2/E
70	27.2	42.8	1831.84	67.34706
41	74.8	-33.8	1142.44	15.27326
22	45.9	-23.9	571.21	12.44466
37	22.1	14.9	222.01	10.0457
			Total	105.1107

Degrees of freedom = Number of Categories -1 = 4-1 = 3

p-Value = P( > Test Statistic ) = P( > 105.1107) = 0.000

Level of significance : = 0.025

As p-value : 0.0297 < : 0.05 ; Reject null hypothesis ;

Therefore,

The conclusion is

B. There is sufficient evidence to warrant the rejection of the claim that the distribution of crashes conforms to the distibuion of ages.

p-value computation using excel:

CHISQ.DIST.RT function

Returns the right-tailed probability of the chi-squared distribution.

The ?2 distribution is associated with a ?2 test. Use the ?2 test to compare observed and expected values. For example, a genetic experiment might hypothesize that the next generation of plants will exhibit a certain set of colors. By comparing the observed results with the expected ones, you can decide whether your original hypothesis is valid.

Syntax

CHISQ.DIST.RT(x,deg_freedom)

The CHISQ.DIST.RT function syntax has the following arguments:

• X     Required. The value at which you want to evaluate the distribution.

• Deg_freedom     Required. The number of degrees of freedom.