Question

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 170170 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month?

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Fatal Accidents	1515	1010	1212	99	1515	2020	2020	2222	99	1414	1010	1414

Step 1 of 10:

State the null and alternative hypothesis.

Step 2 of 10:

What does the null hypothesis indicate about the proportions of fatal accidents during each month?

Step 3 of 10:

State the null and alternative hypothesis in terms of the expected proportions for each category.

Step 4 of 10:

Find the expected value for the number of fatal accidents that occurred in January. Round your answer to two decimal places.

Step 5 of 10:

Find the expected value for the number of fatal accidents that occurred in February. Round your answer to two decimal places

Step 6 of 10:

Find the value of the test statistic. Round your answer to three decimal places.

Step 7 of 10:

Find the degrees of freedom associated with the test statistic for this problem.

Step 8 of 10:

Find the critical value of the test at the 0.050.05 level of significance. Round your answer to three decimal places

Step 9 of 10:

Make the decision to reject or fail to reject the null hypothesis at the 0.050.05 level of significance.

Step 10 of 10:

State the conclusion of the hypothesis test at the 0.050.05 level of significance.

Answer 1

Solution:-

Step 1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The number of fatal accidents which occur in her state does not vary from month to month.

Alternative hypothesis: The number of fatal accidents which occur in her state does vary from month to month.

Step 2)

The null hypothesis indicate about the proportions of fatal accidents during each month is same and is equal is 0.0833.

Step 3)

Null hypothesis: p₁ = p₂ = p₃ = p₄ = .................................................=p₁₀ = p₁₁ =p₁₂ = 0.0833

Alternative hypothesis: At least one of the null hypothesis statements is false..

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

Step 4)
(E_i) = n * p_i

Step 6)

X² = 15.788

Step 7)

DF = k - 1 = 12 - 1
D.F = 11

Step 8)

X² = 19.68

Rejection region is X² > 19.68

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

Step 9)

Interpret results. Since the X² -value does not lies in the rejection region, hence we failed to reject the null hypothesis.

Step 10)

From the above test we have sufficient evidence in the favor of the claim that the number of fatal accidents which occur in her state does not vary from month to month.