Question

A particular report classified 717 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table.

Month	Number of Accidents
January	37
February	31
March	44
April	59
May	79
June	75
July	97
August	83
September	63
October	66
November	43
December	40

(a) Use the given data to test the null hypothesis H₀: p₁ = 1/12, p₂ = 1/12, ... , p₁₂ = 1/12, where p₁ is the proportion of fatal bicycle accidents that occur in January, p₂ is the proportion for February, and so on. Use a significance level of .01. (Round your ?² value to two decimal places, and round your P-value to three decimal places.)

?2	=	1
P-value	=	2

What can you conclude?

There is sufficient evidence to reject H₀. There is insufficient evidence to reject H₀.    

(b) The null hypothesis in Part (a) specifies that fatal accidents were equally likely to occur in any of the 12 months. But not all months have the same number of days. What null and alternative hypotheses would you test to determine if some months are riskier than others if you wanted to take differing month lengths into account?

H₀: p₁ = 31/365, p₂ = 28/365, ... , p₁₂ = 31/365
H_a: All of the true category proportions differ from the hypothesized values. H₀: p₁ = 30/365, p₂ = 30/365, ... , p₁₂ = 30/365
H_a: At least one of the true category proportions differs from the hypothesized value.     H₀: p₁ = 30/365, p₂ = 30/365, ... , p₁₂ = 30/365
H_a: All of the true category proportions differ from the hypothesized values. H₀: p₁ = 31/365, p₂ = 28/365, ... , p₁₂ = 31/365
H_a: At least one of the true category proportions differs from the hypothesized value.

(c) Test the hypothesis proposed in Part (b) using a .05 significance level. (Round your ?² value to two decimal places, and round your P-value to three decimal places.)

?2	=	5
P-value	=	6

What can you conclude?

There is sufficient evidence to reject H₀. There is insufficient evidence to reject H₀.   

Answer 1

(a)

Following table shows the calculations for chi square test statistics:

Month	Number of Accidents, O	p	E=p*717	(O-E)^2/E
January	37	0.083333333	59.75	8.662133891
February	31	0.083333333	59.75	13.83368201
March	44	0.083333333	59.75	4.15167364
April	59	0.083333333	59.75	0.009414226
May	79	0.083333333	59.75	6.201882845
June	75	0.083333333	59.75	3.892259414
July	97	0.083333333	59.75	23.22280335
August	83	0.083333333	59.75	9.04707113
September	63	0.083333333	59.75	0.176778243
October	66	0.083333333	59.75	0.65376569
November	43	0.083333333	59.75	4.695606695
December	40	0.083333333	59.75	6.528242678
Total	717		717	81.07531381

The test statistics is:

Degree of freedom: df=n-1=11

The p-value is: 0.000

Since p-value is less than 0.05 so we reject the null hypothesis.

There is sufficient evidence to reject H0.

(b)

H0: p1 = 31/365, p2 = 28/365, ... , p12 = 31/365

Ha: At least one of the true category proportions differs from the hypothesized value.

(c)

Following table shows the calculations for chi square test statistics:

Month	Number of Accidents, O	p	E=p*717	(O-E)^2/E
January	37	0.084931507	60.89589041	9.376881998
February	31	0.076712329	55.00273973	10.47459667
March	44	0.084931507	60.89589041	4.687855139
April	59	0.082191781	58.93150685	7.96062E-05
May	79	0.084931507	60.89589041	5.382280837
June	75	0.082191781	58.93150685	4.381297644
July	97	0.084931507	60.89589041	21.40549585
August	83	0.084931507	60.89589041	8.023392998
September	63	0.082191781	58.93150685	0.280879234
October	66	0.084931507	60.89589041	0.427811048
November	43	0.082191781	58.93150685	4.306913637
December	40	0.084931507	60.89589041	7.170241425
Total	717	1	717	75.91772608

The test statistics is:

Degree of freedom: df=n-1=11

The p-value is: 0.000

Since p-value is less than 0.05 so we reject the null hypothesis.

There is sufficient evidence to reject H0.