A health club manager claims that young males spend more exercising than females
ID: 3224379 • Letter: A
Question
A health club manager claims that young males spend more exercising than females in the gym. A random sample students is selected and asked how many days per week they exercise in the gym. Is this a chi-square goodness-of-fit test, a chi-square test for independent variables, or a chi-square test for homogeneity of populations? Explain your answer in the context of the problem. Consider the females spending 2-3 days per week exercising in the gym. What is cell's expected frequency and its contribution to the chi-square test statistic? Identify clearly which is which. Calculate the test statistic and its P-value. Indicate clearly which is which. What are your conclusions about the null hypothesis and the manager's claim? State clearly whether you reject or fail to reject the null hypothesis and whether you support or do not, support the manager's claim.Explanation / Answer
17. The is chisquare test for independence of variables
since The test is applied when you have two categorical variables from a single population. It is used to determine whether there is a significant association between the two variables.
H0: Gender and Excercise are independent
H1: Gender and Excercise are not independent
18. The expected frequenicy of the particular cell = (150*121) / 275 = 66
The expced frequencies are calculated as follows
The chisquare contribution is
19. )
Degrees of freedom: 3
Test Statistic, X^2: 3.4934
Critical X^2: 7.814736
P-Value: 0.3216
20) Here P-value 0.3216 > alpha 0.05, we accept H0
Thus we conclude that Gender and Excercise are independent i.e. young males spent not more time excercising than female in the gym.
Gender 0-1 2-3 4-5 6-7 Total Males 33.636 55 28.636 7.727 124.999 Females 40.364 66 34.364 9.273 150.001 Total 74 121 63 17 275Related Questions
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