A particular report classified 717 fatal bicycle accidents according to the mont
ID: 3363189 • Letter: A
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 anuary February March April May June July August September October November December 37 42 59 78 73 97 84 64 65 41 (a) Use the given data to test the null hypothesis H0: p1 1/12, P2 = 1/12, , p12-1/12, where pl 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 x value to two decimal places, and round your P-value to three decimal places.) x2 P-value What can you conclude? There is sufficient evidence to reject Ho There is insufficient evidence to reject Ho.Explanation / Answer
a)
Expected accidents each month = 717/12 = 59.75
2 = [ (Observed - Expected)2 / Expected ]
2 = (37 - 59.75)2/59.75 + (33 - 59.75)2/59.75 + (42 - 59.75)2/59.75 + (59 - 59.75)2/59.75 + (78 - 59.75)2/59.75 + (73 - 59.75)2/59.75 + (97 - 59.75)2/59.75 + (84 - 59.75)2/59.75 + (64 - 59.75)2/59.75 + (65 - 59.75)2/59.75 + (44 - 59.75)2/59.75 + (41 - 59.75)2/59.75
2 = 78.30
Degree of freedom = 12 - 1 = 11
P-value = P[2 > 78.30, 11] = 0
As, P-value is less than the significance level of 0.01,
There is sufficient evidence to reject H0
(b)
The correct option is 3rd one.
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)
Expected accidents for month with 31 days = 717*31 /365 = 60.7
Expected accidents for month with 30 days = 717*30 /365 = 58.93
Expected accidents for month with 28 days (Feb month) = 717*28 /365 = 55
2 = (37 - 60.7)2/60.7 + (33 - 55)2/55 + (42 - 60.7)2/60.7 + (59 - 58.93)2/58.93 + (78 - 60.7)2/60.7 + (73 - 58.93)2/58.93 + (97 - 60.7)2/60.7 + (84 - 60.7)2/60.7 + (64 - 58.93)2/58.93 + (65 - 60.7)2/60.7 + (44 - 58.93)2/58.93 + (41 - 60.7)2/60.7
= 73.67
P-value = P[2 > 73.67, 11] = 0
As, P-value is less than the significance level of 0.05,
There is sufficient evidence to reject H0
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