Question

Questions 19-22 are based on the following data: A sample of randomly selected non-fatal work related injuries categorized according to the weekday they occurred in is given below.

Day	Mon	Tues	Wed	Thurs	Fri
Number	23	23	21	21	19

20. What is the expected value for Mondays if injuries were thrice as likely on Mondays and twice as likely on Fridays as they were in all other days of the working week?

• 13.375

• 23

• 26.75

• 40.125

Use a .05 significance level to test the claim that such injuries occur with equal frequency throughout the week. What is the p-value?

• Same frequency, since p = .01

• Different frequencies, since p = .01

• Same frequency, since p = .99

• Different frequencies, since p = .99

Answer 1

Solution:-

19)

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

Null hypothesis: The non-fatal work related injuries categorized according to the weekday occurred as per given distribution.

Alternative hypothesis: At least one of the proportions in the null hypothesis 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.

DF = k - 1 = 5 - 1
D.F = 4
(E_i) = n * p_i

X² = 25.174

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 4 degrees of freedom is more extreme than 25.174.

We use the Chi-Square Distribution Calculator to find P(X² > 25.174) = 0.00.

Interpret results. Since the P-value (0.00) is less than the significance level (0.05), we cannot accept the null hypothesis.

20)

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

Null hypothesis: Non-fatal work related injuries occur with equal frequency throughout the week.

Alternative hypothesis: At least one of the proportions in the null hypothesis 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.

DF = k - 1 = 5 - 1
D.F = 4
(E_i) = n * p_i

X² = 0.5234

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 4 degrees of freedom is more extreme than 0.5234.

We use the Chi-Square Distribution Calculator to find P(X² > 0.5234) = 0.97

Interpret results. Since the P-value (0.97) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject the null hypothesis, From the above test we have sufficient evidence in the favor of the claim that such injuries occur with equal frequency throughout the week.

Same frequency, since p = 0.97.