(A) A particular report from 2004 classified 718 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table.January 38 March 43April 60May 79June 73July 97August 83September 63October 67November 44Febuary 31December 40 Use the given data to test the null hypothesis H0: ?1 = 1/12, ?2 = 1/12, . . . , ?12 = 1/12, where ?1 is the proportion of fatal bicycle accidents that occur in January, ?2 is the proportion for February, and so on. Use a significance level of 0.01. (Use 2 decimal places.)?2 = (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. Test the hypotheses proposed in H0: ?4 = ?6 = ?9 = ?11 = 30/366 ≈ 0.082, ?2 = 29/366 ≈ 0.085 using a 0.05 significance level. (Use 2 decimal places.)≈ 0.079, ?1 = ?3 = ?5 = ?7 = ?8 = ?10 = ?12 = 31/366

Name: (A) A particular report from 2004 classified 718 fatal bicycle acciden
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: (A) A particular report from 2004 classified 718 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table.January 38 March 43April 60May 79June 73July 97August 83September 63October 67Novembe

january 38 february 31 march 43 april 60 may 79 june 73 july 97 august 83 september 63 october 67 november 44 december 40 now, we have to test the null hypothesis that proportion of bicycle accidents are equally likely in every month For the data we use the chi-square test for goodness of fit. It is used to determine if a hypothesized probability distribution for a population provides a good fit. Acceptance or rejection of the hypothesized population distribution is based on the difference between observed frequencies and expected frequencies. χ2= ∑i=1n [(fi-ei)2]ei, (∑i=1n fi=∑i=1n ei) follows chi square with (n-1)df. The expected frequencies is the average of all the observed frequencies. We find the values from the above equation for each month. months obs freq exp freq (fi-ei)^2 (fi-ei)^2/ei rounded january 38 59.83 476.5489 7.965049 7.97 february 31 59.83 831.1689 13.89218 13.89 march 43 59.83 283.2489 4.734229 4.73 april 60 59.83 0.0289 0.000483 0 may 79 59.83 367.4889 6.142218 6.14 june 73 59.83 173.4489 2.899029 2.9 july 97 59.83 1381.609 23.09224 23.09 august 83 59.83 536.8489 8.972905 8.97 september 63 59.83 10.0489 0.167958 0.17 october 67 59.83 51.4089 0.85925 0.86 november 44 59.83 250.5889 4.188349 4.19 december 40 59.83 393.2289 6.572437 6.57 718 79.48 the tabulated value of chi square at 11 df at 0.01 level of significance is 24.72 Hence, observed value is less than of tabulated value , we reject null hypothesis and say that the fatal accidents are not equally likely to occur in any of the 12 months. now, we are asked to check for the months april , june , september and november as these months have equal no of days and that is 30 now, n=4 df = n-1 = 3 months obs freq exp freq fi-ei (fi-ei)^2 (fi-ei)^2/ei april 60 60 0 0 0 june 73 60 13 169 2.82 september 63 60 3 9 0.15 november 44 60 -16 256 4.267 240 240 7.23 The critical value of chi square at 3 df at 0.01 level of significance is 11.34. The calculated value of test statistic is less than the tabulated value of test statistic , hence we accept our null hypothesis and say the fatal accidents are equally likely to occur in these four months.

Answered: A) A particular report from 2004…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

(A) A particular report from 2004 classified 718 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table.

January 38

March 43

April 60

May 79

June 73

July 97

August 83

September 63

October 67

November 44

Febuary 31

December 40

Use the given data to test the null hypothesis H₀: ?₁ = 1/12, ?₂ = 1/12, . . . , ?₁₂ = 1/12, where ?₁ is the proportion of fatal bicycle accidents that occur in January, ?₂ is the proportion for February, and so on. Use a significance level of 0.01. (Use 2 decimal places.)
?² =

(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. Test the hypotheses proposed in H₀: ?₄ = ?₆ = ?₉ = ?₁₁ =

30/366 ≈ 0.082, ?₂ = 29/366 ≈ 0.085 using a 0.05 significance level. (Use 2 decimal places.)≈ 0.079, ?₁ = ?₃ = ?₅ = ?₇ = ?₈ = ?₁₀ = ?₁₂ = 31/366

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Hypothesis Tests and Confidence Intervals for Proportions$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.