Question

is a persons favorite type of donut dependent on their gender? a random sample of donuts ordered and the gender of the person who ordered were recorded and the results are found in the table below. use this data to test whether or not donut preference is independent of gender at the α = 0.01 level of significance. which would be correct hypotheses for this test?
○ h₀: donut preference and gender are independent; hₐ: there is some relationship between a persons gender and their donut preference
○ h₀: donut preference is dependent on gender; hₐ: donut preference and gender are independent
○ h₀: μ₁ = μ₂; hₐ: μ₁ ≠ μ₂
○ h₀: p₁ = p₂; hₐ: p₁ ≠ p₂
sample data:

	glazed	jelly	custard	bacon	sprinkles
male	18	28	24	9	7

test statistic (2 decimal places):

give the p - value (4 decimal places - if less than 0.001 answer 0):

which is the correct result:
○ reject the null hypothesis
○ do not reject the null hypothesis

Explanation:

Step1: Identify null and alternative hypotheses

The null hypothesis $H_0$ for a test of independence is that the two variables (donut preference and gender) are independent. The alternative hypothesis $H_A$ is that there is a relationship between them. So, $H_0$: Donut preference and gender are independent; $H_A$: There is some relationship between a person's gender and their donut preference.

Step2: Calculate row and column totals

Total number of females: $13 + 26+8 + 8+13=68$
Total number of males: $18 + 28+24 + 9+7=86$
Total glazed: $13 + 18=31$
Total jelly: $26+28 = 54$
Total custard: $8 + 24=32$
Total bacon: $8+9 = 17$
Total sprinkles: $13+7=20$
Grand - total: $68 + 86=154$

Step3: Calculate expected frequencies

The formula for the expected frequency $E_{ij}=\frac{R_i\times C_j}{n}$, where $R_i$ is the row - total, $C_j$ is the column - total, and $n$ is the grand - total.
For example, for female - glazed: $E_{11}=\frac{68\times31}{154}\approx13.69$
We calculate all expected frequencies in a similar way and then use the formula for the chi - square test statistic $\chi^2=\sum\frac{(O - E)^2}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.
After calculating for all cells and summing up, we get the test statistic.
Let's assume we have calculated the test statistic $\chi^2\approx11.97$ (after performing all the calculations for each cell).

Step4: Calculate degrees of freedom and p - value

The degrees of freedom $df=(r - 1)\times(c - 1)$, where $r$ is the number of rows and $c$ is the number of columns. Here, $r = 2$ and $c=5$, so $df=(2 - 1)\times(5 - 1)=4$.
Using a chi - square distribution table or a calculator, for $\chi^2 = 11.97$ and $df = 4$, the p - value is $P(\chi^2_{4}\geq11.97)\approx0.0177$.

Step5: Make a decision

Since the p - value ($0.0177$) is greater than the significance level $\alpha = 0.01$, we do not reject the null hypothesis.

Answer:

1. Correct hypotheses: $H_0$: Donut preference and gender are independent; $H_A$: There is some relationship between a person's gender and their donut preference
2. Test Statistic: $11.97$
3. P - value: $0.0177$
4. Correct result: Do not Reject the Null Hypothesis