A local brewery produces three premium lagers named Half Pint, XXX, and Dark Night. Of its premium lagers, they bottle 40% Half Pint, 40% XXX, and 20% Dark Night lagers. In a marketing test of a sample of consumers, 36 preferred the Half Pint lager, 35 preferred the XXX lager, and 9 preferred the Dark Night lager. Using a chi-square goodness-of-fit test, decide to retain or reject the null hypothesis that production of the premium lagers matches these consumer preferences using a 0.05 level of significance.

a. State the value of the test statistic.

b. Retain or reject the null hypothesis?

(a) The test statistic value is, 5.382.

(b) Retain the null hypothesis.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

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 Ch-square table.

p-value = 0.1463 > α = 0.05.

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

Thus, concluding that the production of the premium lagers matches these consumer preferences.

yes because when there is a parenthesis next to a number like 5(6) it means to multiply both together just like f(x) is to multiply each other