determine the x2 test statistic the degrees of freedom, the p value, and test the hypothesis at the a

The null hypothesis will not be rejected at 5% significance level.

Step-by-step explanation:

Consider a Chi-square test for goodness of fit.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

The information provided is:

Observed values:

Half Pint: 36  

XXX: 35  

Dark Night: 9  

TOTAL: 80

The expected proportions are:  

Half Pint: 40%

XXX: 40%  

Dark Night: 20%  

Compute the expected values as follows:

E (Half Pint)  

E (XXX)  

E (Dark night)  

Compute the test statistic as follows:

The test statistic value is, 5.382.

The degrees of freedom of the test is:

n - 1 = 3 - 1 = 2

The significance level is, α = 0.05.

Compute the p-value of the test as follows:

p-value = 0.1463

*Use a Chi-square table.

p-value = 0.1463 > α = 0.05.

So, the null hypothesis will not be rejected at 5% significance level.

power property.