For hypothesis tests, please give me the null and alternative hypotheses, the fo
ID: 3270216 • Letter: F
Question
For hypothesis tests, please give me the null and alternative hypotheses, the formula for the test statistic, the appropriate substitutions, the value of the test statistic, the rejection region (be careful on two-sided tests, don’t forget both tails) or p-value (Note: This means you can use either approach, unless specified), whether you accept or reject the null, and the interpretation of the results in the words of the problem.
For confidence and prediction intervals, give the formula used for the interval, the appropriate substitutions, the actual interval(s), and interpret it in the context of the problem.
1. A researcher wishes to test whether the proportion of college students who smoke is the same in four different colleges. She uses independent simple random samples to select 100 students from each college and records the number that smoke. The results are shown below, showing smoking status and college.
College: A B C D
Smoke: 17 26 11 34
Don’t Smoke: 83 74 89 66
Use a 1% significance level to test the claim that the distribution of smoking status is different at the four colleges.
Explanation / Answer
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H0: The proportion of college students who smoke is the same in four different colleges.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a chi-square test for homogeneity.
Analyze sample data. Applying the chi-square test for homogeneity 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.
Let DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, nr is the number of observations from population r, nc is the number of observations from level c of the categorical variable, n is the number of observations in the sample, Er,c is the expected frequency count in population r for level c, and Or,c is the observed frequency count in population r for level c.
DF = (r - 1) * (c - 1) = (2 - 1) * (4 - 1) = 3
The expected values are calculated in the below table , shown in brackets.
Er,c = (nr * nc) / n
Chisquare test statistic , 2 = [ (Or,c - Er,c)2 / Er,c ]
2 = (17 - 22)2/22 + (26 - 22)2/22 + (11 - 22)2/22 + (34 - 22)2/22 + (83 - 78)2/78 + (74 - 78)2/78 + (89 - 78)2/78 + (66 - 78)2/78 = 17.83
P-value = P(2 > 17.83) for DF = 3 is 0.0005
For two sided test, P-value = 2 * 0.0005 = 0.001
Interpret results. Since the P-value (0.001) is less than the significance level (0.01), we reject the null hypothesis and conclude that at least one of the proportion of college students who smoke is different from other colleges.
H0: The proportion of college students who smoke is the same in four different colleges.
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