Question

4. A researcher claims that the distribution of favorite pizza toppings among teenagers is as shown below. Topping Cheese Pepperoni Sausage Mushrooms Onions Frequency | 41% 25% 15% 10% 9%

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The distribution of favorite pizza toppings among teenagers is as per given distribution.

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 5 - 1
D.F = 4
(E_i) = n * p_i

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$
X² = 2.025

Observed E(x Expected Num ((Or r,c)21 Er 82 0.19512195121 Peperoni Sausage Mushrooms Onion Sum 78 52 30 25 15 200 0.41 0.25 0.15 0.10 0.09 0.08 0 1.25 0.5 200 2.0251219512195 50 30 20 18

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 2.025.

We use the Chi-Square Distribution Calculator to find P(X² > 2.025) = 0.7311.

Interpret results. Since the P-value (0.7371) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we can conclude that the distribution of favorite pizza toppings among teenagers is as per given distribution.