Question

For each of the following questions, report 1) the appropriate statistical test or estimation procedure to use, 2) the null and alternate hypotheses, 3) the test statistic, 4) the P value, 5) whether you accept or reject the null, then 6) a sentence or two about what the results mean.

A student wanted to know if there was a difference between the sexes in basic food preference. Final results were: 65 females preferred salty and 35 preferred sweet. 46 males preferred salty and 54 preferred sweet.

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: There was no difference between the sexes in basic food preference.

Alternative hypothesis: H_a: There was a difference between the sexes in basic food preference.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1

α = 0.05

Critical value = 3.841459

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Sex
Row variable	Male	Female	Total
Salty	46	65	111
Sweet	54	35	89
Total	100	100	200

Expected Frequencies
	Sex
Row variable	Male	Female	Total
Salty	55.5	55.5	111
Sweet	44.5	44.5	89
Total	100	100	200

Calculations
(O - E)
-9.5	9.5
9.5	-9.5

(O - E)^2/E
1.626126	1.626126
2.02809	2.02809

Test Statistic = Chi square = ∑[(O – E)^2/E] = 7.308432

χ² statistic = 7.308432

P-value = 0.006863

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there was a difference between the sexes in basic food preference.