Q1) Chi-Square Test for Goodness of Fit (7 points) Two employees at an ice cream

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Q1) Chi-Square Test for Goodness of Fit (7 points) Two employees at an ice cream store were discussing the popularity of four of the ice cream flavors they have: Vanilla, Chocolate, Mint, and Strawberry. Employee A thought that the four flavors are equally popular but employee B didn't think so. To settle their debate, they decided to tally the number of customers asking for those flavors for a whole day. There were 40 customers that day: o 8 asked for vanilla . 12 asked for chocolate 14 asked for Mint 6 asked for strawberry Conduct a Chi-Square Test for Goodness of Fit to answer the research question of whether the four flavors are equally popular, using a .05 as the significance level. a. What is the variable in this test? What type of variable is it (nominal, ordinal, or continuous)? (2 point total: 1 for each answer) b. State the null and alternative hypotheses in words (2 point total: 1 for each hypothesis) c. Calculate X2statistic (3 points total: give 2 if process is correct but final answer is wrong) d. Calculate the degree of freedom and then identify the critical value (1point total: .5 for df, .5 for critical value) e. Compare the X2statistic with the critical value, then report the hypothesis test result, using "reject" or "fail to reject" the null hypothesis in the answer (1 point total, 5 for each answer) f. Explain the conclusion in a sentence or two, to answer the research question. (1 point)

Explanation / Answer

here null hypothesis: three flavors are equally popular . p1=p2=p3 =1/3

alternate hypothesis: all flavours are not equall popular.

applying chi square goodness of fit test:

a) from above X2 statistic =1.40

b) degree of freedom =number of categories -1 =3-1 =2

c) for 2 degree of freedom and 0.05 level ; critical value =5.991

as test stat is lower then critical value ; fail to reject” the null hypothesis

d) we do not have sufficient evidence at 0.05 level that three flavors are differently popular,

observed Expected Chi square Type Probability O E=total*p =(O-E)^2/E Vanilla 1/3 13.000 10.00 0.90 Chocolate 1/3 8.000 10.00 0.40 strawberry 1/3 9.000 10.00 0.10 1 30 30 1.4000

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