Explanation Given info: The data represents smoking status and frequency of visits to bars. The significance level for the test is α = 0.05 . Calculation: To test whether the evidences suggest that smokers spend more time in bars than non-smokers it is to be checked whether the distribution of bar visits is the same for smokers and non-smokers. The test hypotheses are given below: Null hypothesis: H 0 : The distribution of bar visits is the same for smokers and non-smokers. Alternative hypothesis: H 1 : The distribution of bar visits is the different for smokers and non-smokers. Decision rule: If χ 0 2 > χ α 2 , then reject the null hypothesis. Test statistic: The row and column total is obtained below: Time spent in Bars Almost Daily Several Times a Week Several Times a Month Once a month Several Times a Year Once a Year Never Total Smoker 80 409 294 362 433 336 1,265 3,179 Non smoker 57 350 379 471 573 568 3,297 5,695 Total 137 759 673 833 1,006 904 4,562 8,874 Expected count: The formula for expected count is given below: ( row total ) ( column total ) ( table total ) The expected count for each category is tabulated below: Time spent in Bars Almost Daily Several Times a Week Several Times a Month Once a month Several Times a Year Once a Year Never Smoker ( 3179 × 137 ) 8874 = 49.079 ( 3179 × 759 ) 8874 = 271.902 ( 3179 × 673 ) 8874 = 241.094 ( 3179 × 833 ) 8874 = 298.412 ( 3179 × 1006 ) 8874 = 360.387 ( 3179 × 904 ) 8874 = 323.847 ( 3179 × 1265 ) 8874 = 1634.280 Non smoker ( 5695 × 137 ) 8874 = 87.921 ( 5695 × 759 ) 8874 = 487.098 ( 5695 × 673 ) 8874 = 431.906 ( 5695 × 833 ) 8874 = 534.588 ( 5695 × 1006 ) 8874 = 645.613 ( 5695 × 904 ) 8874 = 580.153 ( 5695 × 3297 ) 8874 = 2927.720 The requirements that must be satisfied to perform the test are, All the expected frequencies are greater than or equal to 1. That is, all E i ≥ 1 where E i represents the expected counts of category i. Not more than 20% of the expected frequencies are less than 5. From the table, it can be observed that none of the expected frequency is less than 5. Therefore, the requirement of the test is satisfied. The value of the test statistic is, χ 0 2 = ∑ ( O i − E i ) 2 E i = { ( 80 − 49.079 ) 2 49.079 + ( 409 − 271.902 ) 2 271.902 + ( 294 − 241.094 ) 2 241.094 + ( 362 − 298.412 ) 2 298.412 + ( 433 − 360

Fundamentals of Statistics (5th Edition)

5th Edition

ISBN: 9780134508306

Author: Michael Sullivan III

Publisher: PEARSON

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Fundamentals of Statistics (5th Edition)

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Whether smokers spend more time in bars than non-smokers.

Solution Summary: The author evaluates whether the evidences suggest that smokers spend more time in bars than non-smokers.

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