Question

Scenario 2: Car crashes ~ Randomly selected deaths from car crashed were obtained and the results are shown in the table below:

Day	Sun	Mon	Tue	Wed	Thu	Fri	Sat
Fatalities	132	108	95	99	115	132	138

State officials wonder if car crash fatalities occur with unequal frequency on different days of the week. They represent this research question in terms of two competing hypotheses:

H0​: Car crashes occur with equal frequency throughout the week and that any uneven results in sample data are negligible and attributable to random chance.

Ha​: Car crashes do not occur with equal frequency throughout the week and that any uneven results in sample data are reflect true difference in crash rates across the days of the week.

1.If we assume the null hypothesis H_0H0​ to be true, what would be the expected number of fatalities each day of the week? ....the expected value of the test statistic?

2. To decide which of the hypotheses to continue believing, the researchers conducted a \chi^2χ2 Goodness of Fit test. The resulting test statistic representing departures from expected counts across the whole table was \chi^2 = 15.2479χ2=15.2479 with a corresponding p-value of 0.01840.0184. Which day of the week contributed the most toward this overall test statistic?

3. Based on the p-value given above, what would be your conclusion to this hypothesis test at the \alpha = 0.01α=0.01 level?

A)Reject the null hypothesis.

B)Fail to reject the null hypothesis.

Answer 1

1. If we assume the null hypothesis H₀ to be true, what would be the expected number of fatalities each day of the week? What is the expected value of the test statistic?

Solution:

Here, we have to use chi square test of goodness of fit.

Null hypothesis: H₀: Car crashes occur with equal frequency throughout the week and that any uneven results in sample data are negligible and attributable to random chance.

Alternative hypothesis: H_a: Car crashes do not occur with equal frequency throughout the week and that any uneven results in sample data are reflect true difference in crash rates across the days of the week.

We are given level of significance = α = 0.01

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

We are given

N = 7

Degrees of freedom = df = N – 1 = 7 – 1 = 6

α = 0.01

Critical value = 16.81189

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Day	O	E	(O - E)^2	(O - E)^2/E
Sun	132	117	225	1.923076923
Mon	108	117	81	0.692307692
Tue	95	117	484	4.136752137
Wed	99	117	324	2.769230769
Thu	115	117	4	0.034188034
Fri	132	117	225	1.923076923
Sat	138	117	441	3.769230769
Total	819	819		15.24786325

Expected number of fatalities = ∑O/N = 819/7 = 117

Chi square = ∑[(O – E)^2/E] = 15.24786325

Expected value of test statistic = 15.24786325

2. Which day of the week contributed the most toward this overall test statistic?

Tuesday, because the chi square contribution for this day is most among all given days.

(See above table)

3. Based on the p-value given above, what would be your conclusion to this hypothesis test at the α = 0.01 level?

Answer: B)Fail to reject the null hypothesis.

Explanation:

We have

P-value = 0.018414068

(By using Chi square table or excel)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that Car crashes occur with equal frequency throughout the week and that any uneven results in sample data are negligible and attributable to random chance.