Respuesta :
Answer:
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([tex]\chi^2[/tex] < 4.12) < 0.9
Chi-square calculator, P([tex]\chi^2[/tex] < 4.12) = 0.87
Step-by-step explanation:
The given data is presented as follows;
[tex]\begin{array}{ccc}Day&&Number\\Fri&&21\\Sat&&19\\Sun&&10\end{array}[/tex]
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.[tex]\overline 6[/tex]
We get;
[tex]\begin{array}{ccc}Day & Observed \ Number& Expected \ Number \\Fri&21& 16.\overline 6\\Sat&19&16.\overline 6\\Sun&10&16.\overline 6\end{array}[/tex]
The chi-square test statistics
[tex]\chi^2[/tex] = (21 - 50/3)²/(50/3) + (19 - 50/3)²/(50/3) + (10 - 50/3)²/(50/3) = 4.12
The [tex]\chi^2_u[/tex] with (3 - 1) = 2 degrees of freedom = 5.991
The decision rule, [tex]\chi^2[/tex] > 5.991, we reject the null hypothesis
Given that we have [tex]\chi^2[/tex] = 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([tex]\chi^2[/tex] < 4.12) = 0.87.
