Question

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, 27 preferred the Half Pint lager, 29 preferred the XXX lager, and 24 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. State the value of the test statistic. (Round your answer to two decimal places.) $ \chi_{obt}^2 $

Answer 1

reject the null hypothesis

Explanation:

Total observation = 27+49+24 = 100

Expected Frequency :

Half Pint = 100 × 0.4 = 40

XXX = 100 × 0.4 = 40

Dark night = 100 × 0.2 = 20

The Chi-square goodness-of-fit test can be computed as follows:

Observed Expected
`X^2 = ((O- E)^2)/(E)`

Frequency(O) Frequency (E)

Half- 27 40 4.225

Pint

XXX 29 40 3.025

Dark- 24 20 0.8

night

Total 100 100 8.05

The test statistics
`X^2` = 8.05

8.05

degree of freedom = n -1

degree of freedom = 3 - 1

degree of freedom = 2

The p-value can be computed by using the EXCEL FORMULA (=CHIDIST(8.05,2)

p-value = 0.01786

Since the p-value is less than the level of significance, we reject the null hypothesis

Also using the critical value approach from the EXCEL FUNCTION;

critical chi-square (=CHIINV(0.05,2) = 5.99

`X^2 > 5.99`, therefore, we reject the null hypothesis