Question

The national distribution of fatal work injuries in a country is shown in the table to the right under National ​%. You believe that the distribution of fatal work injuries is different in the western part of the country and randomly select 6231 fatal work injuries occurring in that region. At alpha equals 0.05 can you conclude that the distribution of fatal work injuries in the west is different from the national​ distribution? Complete parts a through d below. Cause National​ % Western Frequency Transportation 44​% 2894 Equipment 18​% 1159 Assaults 14​% 806 Falls 13​% 746 Harmful fumes 9​% 531 Fires 2​% 95 a. State Upper H 0 and Upper H Subscript a and identify the claim. What is the null​ hypothesis, Upper H 0​? A. The distribution of fatal work injuries in the west differs from the expected distribution. B. The distribution of fatal work injuries in the west is 2894 ​transportation, 1159 ​equipment, 806 ​assaults, 746 ​falls, 531 harmful​ fumes, and 95 fires. C. The distribution of fatal work injuries in the west is 44​% ​transportation, 18​% ​equipment, 14 % ​assaults, 13​% ​falls, 9​% harmful​ fumes, and 2​% fires. Your answer is correct. What is the alternate​ hypothesis, Upper H Subscript a​? A. The distribution of fatal work injuries in the west is the same as the expected distribution. B. The distribution of fatal work injuries in the west is 44​% ​transportation, 18​% ​equipment, 14 % ​assaults, 13​% ​falls, 9​% harmful​ fumes, and 2​% fires. C. The distribution of fatal work injuries in the west differs from the expected distribution. Your answer is correct. Which hypothesis is the​ claim? Upper H 0 Upper H Subscript a Your answer is correct. b. Determine the critical​ value, font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2​, and the rejection region. font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2equals 11.070 ​(Round to three decimal places as​ needed.) Determine the rejection region. A. font size increased by 1 font size increased by 1 font size increased by 1 chi squared less than font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2 B. font size increased by 1 font size increased by 1 font size increased by 1 chi squared greater than font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2 Your answer is correct.C. font size increased by 1 font size increased by 1 font size increased by 1 chi squared greater than or equals font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2 D. font size increased by 1 font size increased by 1 font size increased by 1 chi squared less than or equals font size increased by 1 font size increased by 1 font size increased by 1 chi Subscript 0 Superscript 2 c. Calculate the test statistic. font size increased by 1 font size increased by 1 font size increased by 1 chi squaredequals nothing ​(Round to three decimal places as​ needed.)

Answer 1

Solution:-

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

Null hypothesis: The distribution of fatal work injuries is not different in the western part of the country.

Alternative hypothesis: The distribution of fatal work injuries is different in the western part of the country.

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.

DF = k - 1 = 6 - 1
D.F = 5
(E_i) = n * p_i


X² = 28.44

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.

The P-value is the probability that a chi-square statistic having 5 degrees of freedom is more extreme than 28.44.

We use the Chi-Square Distribution Calculator to find P(X² > 28.44) = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that distribution of fatal work injuries in the west differs from the expected distribution.