1 answer

Optional Chapter 5: Smoking on campus. In Febrnuary 2013, SurveyUSA interviewed a random sample 800 adults...

Question:

Optional Chapter 5: Smoking on campus. In Febrnuary 2013, SurveyUSA interviewed a random sample 800 adults in the Tampa, Florida, area. Two of the ques- tions asked were Should smoking be allowed? Or banned on college campuses? and What is your smoking status The two-way table below summarizes the answers to these two questions.7 Smoking Status Former smoker Never a smoker Opinion Current smoker Allowed Banned Not sure 115 20 23 93 213 39 79 188 30 What do the data say about differences in opinion a smoking on colleg smoking habits? aboot r individuals with different irn

Optional Chapter 5: Smoking on campus. In Febrnuary 2013, SurveyUSA interviewed a random sample 800 adults in the Tampa, Florida, area. Two of the ques- tions asked were "Should smoking be allowed? Or banned on college campuses?" and "What is your smoking status" The two-way table below summarizes the answers to these two questions.7 Smoking Status Former smoker Never a smoker Opinion Current smoker Allowed Banned Not sure 115 20 23 93 213 39 79 188 30 What do the data say about differences in opinion a smoking on colleg smoking habits? aboot r individuals with different irn

Answers

Solution:-

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

Null hypothesis: The null hypothesis states that there is no differences in opinion about smoking on college campuses for individuals with different smoking habits.

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

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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.

DF = (r - 1) * (c - 1) = (3 - 1) * (3 - 1)
D.F = 4
Er,c = (nr * nc) / n

Current smoker Expected (O 115 Former rSmoker Never a smoker Expected -Erc)2E.</p><p>Allowed 57 60 93 Total 124 79 107 287 Banned Not sure Total 20 23 158 83 48 213 39 345 182 40 345 188 30 297 156 34 297 421 92 800 158 109
Er,c
Χ2 = 136

where 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.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 136.0.

We use the Chi-Square Distribution Calculator to find P(Χ2 > 136.00) = 0.0000.

Interpret results. Since the P-value (0.0000) is less than the significance level (0.05), we reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is differences in opinion about smoking on college campuses for individuals with different smoking habits.

.

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