Question

A researcher planned a study in which a crucial step was offering participants a food reward. It was important that three food rewards were equal in appeal. Thus, a pilot study was designed in which participants were asked which of the rewards they preferred. The observed frequencies are as follows:
Of the 60 participants, 16 preferred cupcakes, 26 preferred candy bars, and 18 favored dried apricots.

1. The appropriate statistical test for this problem is (be specific): ________.

2. What are the expected frequencies for this question?

3. What is the cutoff on the comparison distribution (step 3)?

4. What is the correct calculation for chi-square (step 4)?

Answer 1

Answer:

Step-by-step explanation:

Hello!

A pilot study was conducted to test if three food rewards are equally appealing to the participants.

Of 60 participants surveyed:

16 preferred cupcakes (CC)

26 preferred candy bars (CB)

18 preferred dried apricots (DA)

If the three types of food are equally appealing for the participants, you'd expect that their proportions will be equal: P(CC)=P(CB)=P(DA)= 1/3

1.

The objective of this pilot study is to test if the observed frequencies follow a theoretical model/ distribution. To analyze this, you have to apply a Chi Square Goodness to Fit test. $X^2=sum \frac{(O_i-E_i)^2}{E_i} ~~X^2_{k-1}$ Where k= number of categories of the variable.

For this example the statistical hypotheses are:

H₀: P(CC)=P(CB)=P(DA)= 1/3

H₁: At least one of the proportions isn't equal to the others.

2.

To calculate the expected frequencies for each category you have to use the formula: $E_i= n* P_i$ where Pi represents the theoretical proportion for the i category, stated in the null hypothesis.

$E_{CC}= n* P(CC)= 60*1/3= 20$

$E_{CB}= n*P(CB)= 60*1/3= 20$

$E_{DA}= n* P(DA)= 60* 1/3= 20$

3.

The cutoff or critical value indicates the beginning of the rejection region for the hypothesis test. For the Chi-Square tests, the rejection region is always one-tailed to the right, meaning that you'll reject the null hypothesis if the value of the statistic is big. For the goodness to fit test you have k-1 degrees of freedom, so the critical value will be:

Assuming α: 0.05

$X^2_{k-1;1-\alpha /2}= X^2_{2;0.975}= 7.378$

The rejection region is then X²₂ ≥ 7.378

4.

$X^2_{H_0}= \frac{(O_{CC}-E_{CC})^2}{E_{CC}} + \frac{(O_{CB}-E_{CB})^2}{E_{CB}} + \frac{(O_{DA}-E_{DA})^2}{E_{DA}} = \frac{(16-20)^2}{20} +\frac{(26-20)^2}{20} +\frac{(18-20)^2}{20}= \frac{14}{5}= 2.8$

I hope this helps!