full answer please 4. Are couch potatoes (people who watch a lot of television)
ID: 3171416 • Letter: F
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full answer please
4. Are couch potatoes (people who watch a lot of television) experts in sports? A survey was conducted to determine whether a person is an expert in a sport merely by watching a lot of it on television. Thus, the question is: Does watching a lot of sports on television mean that one is an expert in that/those sport(s)? Here is a table of the gathered information. To qualify as an expert in sports, a sports knowledge exam was given to all participants. Expert in sports Not an expert in sports Couch potato 22 38 Not a couch potato 15 25 a. The hypotheses are as follows: Ho: Sports expertise and watching a lot a television are independent variables Ha: Sports expertise and watching a lot a television are dependent variables Conduct the test for independence at the alpha 0.05 level of significance. Calculate the test statistic and the p-value. Make a decision. b. Conduct a two-proportion z-test, with alpha 0.05. Determine the critical region for the hypotheses Ho: TI m, vs. Ha 12, (the proportion of couch potatoes who are experts in sports is the same as the proportion of non-couch potatoes who are experts in sports) Calculate the test statistic and p-value for this test. Make a decision.Explanation / Answer
Degrees of freedom. The degrees of freedom (DF) is equal to:
DF = (r - 1) * (c - 1) = (2-1)*(2-1) = 1
where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable.
Expected frequencies. The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula.
Er,c = (nr * nc) / n
where Er,c is the expected frequency count for level r of Variable A and level c of Variable B, nr is the total number of sample observations at level r of Variable A, nc is the total number of sample observations at level c of Variable B, and n is the total sample size.
Test statistic. The test statistic is a chi-square random variable (2) defined by the following equation.
2 = [ (Or,c - Er,c)2 / Er,c ] = 0.00715
where Or,c is the observed frequency count at level r of Variable A and level c of Variable B, and Er,c is the expected frequency count at level r of Variable A and level c of Variable B.
Chi - Squared test Statistic = 0.00715.
We use the Chi-Square Distribution Calculator to find P(2 > 0.00715) = 0.932849
P- Value = 0.932849
Interpret results. Since the P-value (0.932849) is more than the significance level (0.05), we can accept the null hypothesis. Thus, we conclude that there is not a relationship between sport expert and watching TV.
E NE Total CP 22 38 60 NCP 15 25 40 Total 37 63 100 E11 22.2 E12 37.8 E21 14.8 E22 25.2Related Questions
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