96% of Americans report eating a diet that includes meat, 3.5% of Americans claim to be exclusively vegetarian, and 0.5% claim to be strictly vegan. Of a random sample of undergraduates in the University of Virginia dining hall (n = 175), 156 students report eating a diet that includes meat, 11 report being exclusively vegetarian, and 8 are strictly vegan.
5. State the test you would use to determine if the observed and expected frequencies of diets are consistent. List the conditions required for this test and verify that they are satisfied.
6. How many students would you expect to eat meat, be exclusively vegetarian, and be a strict vegan?
7. Calculate the chi-square statistic, the degrees of freedom associated with it, and the p-value.
8. Based on the p-value calculated in Question 7, what is the conclusion of the hypothesis test at Î± = 0.05? Interpret your conclusion in context of the question.

5) Chi-square statistic test , would use to determine if the observed and expected frequencies of diets are consistent. With null hypothesis as the observed and expected frequencies of diets are consistent.

6) The excepted students to eat

meat = 165

exclusively vegetarian= 6

strict vegetarian= 1

7) Chi-square statistic value is 62.755 and degree of freedom are 2 . The critical value for Chi-square statistic is 0.00001

8) P-value< 0.05, so reject the null hypothesis.

Therefore there is no sufficient evidence to support that the observed and expected frequencies of diets are consistent.

A random sample of undergraduates in the University of Virginia dining hall.

Sample size , n = 175

Three categories are given as shown in above table with percentage of report eating a diet.

Report also consists observed values about the number of students diet including the three different categories

5) The Chi-square test would use to determine the required result. Consider Null and alternative hypothesis as :

H₀ : the observed and expected frequencies of diets are consistent

Hₐ : the observed and expected frequencies of diets are not consistent

χ² =∑(Oᵢ - Eᵢ)²/Eᵢ

where, χ² --> chi-square cofficient or value

Oᵢ --> observed values which are provided

Eᵢ --> excepted values

Excepted values , Eᵢ = n× p

here we assume that,

These categories are independent
individuals are selected randomly

6) The number of students would you expect to eat meat = 168

The number of students would you expect to eat exclusively vegetarian

= 6.125 ~ 6

The number of students would you expect to eat strict vegetarian = 0.875 ~ 1

7) Degree of freedom,df = k - 1 = 2, where k denotes total number of categories here

Chi-square statistic, χ² = ∑(Oᵢ - Eᵢ)²/E

χ² = 62.755

Using chi-square table, the critical chi-square value for χ² = 62.755 and degree of freedom 2 is 0.00001 ..

8) Significance level, alpha = 0.05

As we see that p-value = 0.00001< alpha = 0.05

So, we reject the null hypothesis.

Thus, there is no sufficient evidence to support null hypothesis.

To learn more about Chi-square statistic, refer:

https://brainly.com/question/4543358

#SPJ4