Question

According to Detroit Red Wing statistics on season ticket holders in the private suites at Little Caesars’ Arena, 60% of fans drink beer, 30% prefer hard liquor, 5% order wine and 5% do not use alcohol.

You are asked to perform a Chi-Square Goodness-of- Fit Test based on the following sample of 300 fans at a recent Montreal Canadians’ game:

Beer = 210

Hard Liquor = 70

Wine = 10

No alcohol = 10.

1. If you set α = 0.05, what is the Critical Value?
2. Compute the Chi-Square test statistic, either by hand or using your calculator.
3. Yes or No…..Is this variation significant at the 0.05 level?
4. What is the p-value for this study?

Answer 1

a) Since there are 4 variety hence the degree of freedom=n-1=4-1=3

using chi-square table at 0.05 level of significance and 3 degrees of freedom

Critical value=7.815

b)

Since given as

VARIETY	OBSERVED VALUE	EXPECTED PERCENTAGE	EXPECTED VALUE
BEER	210	60	180
HARD LIQUOR	70	30	60
WINE	10	5	15
NO ALCOHOL	10	5	15

As we know that chi-square value is calculated as

using excel tool

chi square=10

c) The P-Value is 0.018566. The result is significant at p < .05.

d) the p-value calculated using the calculator as 0.0186