[1%] CHI SQUARE TEST OF INDEPENDENCE: Of a sample of 80 single and married women surveyed, we obtained the following data as to the sizes of cars each one owns (assuming 1 car per person). Test whether car size is independent of marital status at the .01 significance level.
Compact Midsize Luxury TOTAL
Single 8 14 8 30
Married 32 10 8 50
TOTAL 40 24 16 80
7a) P-value = ________________
7b) REJECT THE NULL or DO NOT REJECT THE NULL? (circle one)
7c) Can we conclude that car size is dependent on marital status? YES or NO
(circle one)
To test whether car size is independent of marital status, we need to perform a chi-square test of independence.
To calculate the chi-square statistic, we can use the following formula:
chi-square = Σ ((O - E)^2 / E)
where O represents the observed frequency in each category and E represents the expected frequency in each category under the assumption of independence.
To calculate the expected frequencies, we can use the formula:
Expected frequency = (row total * column total) / grand total
Now, let's perform the necessary calculations:
First, let's calculate the expected frequencies for each category:
For the "Compact" category in the "Single" group:
Expected frequency = (row total for "Compact" * column total for "Single") / grand total
Expected frequency = (30 * 40) / 80 = 15
We can follow the same process to calculate the expected frequencies for the remaining categories and groups:
Expected frequency for "Midsize" in "Single" = 30 * 24 / 80 = 9
Expected frequency for "Luxury" in "Single" = 30 * 16 / 80 = 6
Expected frequency for "Compact" in "Married" = 50 * 40 / 80 = 25
Expected frequency for "Midsize" in "Married" = 50 * 24 / 80 = 15
Expected frequency for "Luxury" in "Married" = 50 * 16 / 80 = 10
Now, let's calculate the chi-square statistic:
chi-square = ((8 - 15)^2 / 15) + ((14 - 9)^2 / 9) + ((8 - 6)^2 / 6) + ((32 - 25)^2 / 25) + ((10 - 15)^2 / 15) + ((8 - 10)^2 / 10)
chi-square = (49/15) + (25/9) + (4/6) + (49/25) + (25/15) + (4/10)
chi-square = 3.2667 + 2.7778 + 0.6667 + 1.9600 + 1.6667 + 0.4000
chi-square = 10.7379
Now that we have the chi-square statistic, we need to compare it to the critical value from the chi-square distribution table. The degrees of freedom for this test is (number of rows - 1) * (number of columns - 1) = (2 - 1) * (3 - 1) = 1 * 2 = 2.
Using the chi-square distribution table, we find that the critical value (at the 0.01 significance level with 2 degrees of freedom) is approximately 9.2103.
Comparing the calculated chi-square statistic (10.7379) with the critical value (9.2103), we can conclude the following:
7a) The p-value is the probability of obtaining a test statistic as extreme as the observed one. To determine the p-value, we need to compare the chi-square statistic with the chi-square distribution table. In this case, the p-value is less than 0.01, which indicates strong evidence against the null hypothesis.
7b) We REJECT THE NULL hypothesis. Since the p-value is less than the significance level, we reject the null hypothesis that car size is independent of marital status.
7c) We can conclude that car size is dependent on marital status.