13. Randomly selected nonfatal occupational injuries and illnesses are categorized according to the day of week that they first occurred, and the results are listed below. Use a .05 significance level and that such injuries and illnesses occur with equal frequency on the weekend. State the null and alternative hypotheses, the pvalue, and conclusion. Day Fri Sat Sun Number 21 19 10

Part A

The null hypothesis H₀; The injuries and illnesses have no relationship in the weekend

The alternative hypothesis Hₐ; The injuries and illnesses occur with equal frequency on the weekend

Part B

P-value from;

Tables, 0.1 < P( < 4.12) < 0.9

Chi-square calculator, P( < 4.12) = 0.87

Step-by-step explanation:

The given data is presented as follows;

The significance level, α = 0.05

Part A

The hypothesis is to test that the given injuries and illnesses occur with equal frequency of the weekend

The null hypothesis H₀; The injuries and illnesses have no relationship in the weekend

The alternative hypothesis Hₐ; The injuries and illnesses occur with equal frequency on the weekend

Let the expected frequency, each day of the weekend = (21 + 19 + 10)/3 = 50/3 = 16.

We get;

The chi-square test statistics

= (21 - 50/3)²/(50/3) + (19 - 50/3)²/(50/3) + (10 - 50/3)²/(50/3) = 4.12

The with (3 - 1) = 2 degrees of freedom = 5.991

The decision rule, > 5.991, we reject the null hypothesis

Given that we have = 4.12 < 5.991, we fail to reject H₀, therefore, there is not enough statistical evidence to suggest that the meal plan are related at α = 0.05

Part B

From the chi-square table, we have the p-value at 2  degrees of freedom for the test statistic of 4.12 is between 0.9 and 0.1

Using the chi-square calculator, we have;

P( < 4.12) = 0.87.

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square