Question

Technology has made telecommuting easier for workers, and most companies seem willing to let workers do their work remotely, at least on an occasional basis if the position allows for it. According to the Gallup's annual Work and Education poll, conducted Aug. 5-9, 2015, 37% of U.S. workers say they have telecommuted in 2015. Assume that 37% is an accurate representation of all U.S. workers. A company executive believes that in 2018, the proportion of all U.S. workers who telecommuted increased significantly from the rate reported in 2015; that is, to test H0: p = 0.37 vs Ha: p > 0.37, using a 5% significance level. The parameter p represents the population proportion of all U.S. workers who telecommuted in 2018. A random sample of 225 U.S. workers is taken, of which 93 workers said they telecommuted in 2018. (Challenge: Before looking at the questions, use the information provided in this background and work through the 4 steps of testing theories yourself, and then use your results to answer these questions ~ good practice.) Question 1 Subquestions 1.a 0.5 point(s) Assuming that the 225 U.S. workers can be considered a random sample, check the remaining condition that is needed in order to conduct a large sample z test. Make sure to include relevant formulas with correct notation and calculations. n = 225 ≥ 25. Because the sample size n is "large", i.e., at least 25, the distribution of all possible values of the sample proportion p̂ is approximately normal, and a large sample z test can be used. np̂ = 225(93/225) = 93 ≥ 10 and n(1- p̂) = 225(1 – 93/225) =132 ≥ 10. Because the observed number of successes, np̂ = 93, and the observed number of failures, n(1- p̂) = 132, are both at least 10, the distribution of all possible values of the sample proportion p̂ is approximately normal, and a large sample z test can be used. np = 225(0.37) = 83.25 ≥ 10 and n(1- p) = 225(1 – 0.37) = 141.75 ≥ 10. Because the expected number of successes, np=83.25, and the expected number of failures n(1-p)=141.75 are both at least 10, the distribution of all possible values of the sample proportion p̂ is approximately normal, and a large sample z test can be used. np0 = 225(0.37) = 83.25 ≥ 10 and n(1- p0) = 225(1 – 0.37) = 141.75 ≥ 10. Because the number of successes expected under the null hypothesis, np0=83.25, and the number of failures expected under the null hypothesis, n(1-p0)=141.75 are both at least 10, the distribution of all possible values of the sample proportion p̂ is approximately normal, and a large sample z test can be used. 1.b 0.5 point(s) Select the values that best completes the following statement: The observed sample proportion of U.S. workers who telecommuted for work in 2018 is __________ null standard deviations _______ the hypothesized population proportion of U.S. workers who telecommuted for work. 1.35, above 1.35, below 1.32, above 1.32, below 1.c 0.5 point(s) Select the value that best completes the following statement: Assuming the population proportion of all U.S. workers who telecommuted in 2018 is 37%, the probability of obtaining a z-test statistic as large or larger than the one we obtained is ___________. 0.05 0.10 0.01 0.0885 0.9115 1.d 0.5 point(s) Select the value that best completes the following statement: If this inference procedure were repeated many times, and the population proportion of all U.S. workers who telecommuted in 2017 is 37%, we would expect to see a z-test statistic value like the one we got or larger in about _______ % of the repetitions. 5 10 1 8.85 91.15 1.e 0.5 point(s) What is the smallest level of significance at which we can reject the null hypothesis? 0.05 0.10 0.01 0.0885 0.9115 1.f 0.5 point(s) At a 5% significance level, the results are not statistically significant. Which of the following is the appropriate conclusion? We fail to reject the null hypothesis that the population proportion of all U.S. workers who telecommuted for work in 2018 is 37%. There is not sufficient evidence to say that the population proportion of all U.S. workers who telecommuted for work is higher than of 37% There is not sufficient evidence to say that the population proportion of all U.S. workers who telecommuted for work in 2018 is higher than the 2015 rate of 37%. There is sufficient evidence to say that the population proportion of all U.S. workers who telecommuted for work in 2018 is higher than the 2015 rate of 37%. There is sufficient evidence to say that the population proportion of U.S. workers who telecommuted for work is higher than 37%. Next Top

Answer 1

1 (a) n * p = 225 * 0.37 = 83.5 10 and n * (1 -p) = 225 * (1 - 0.37) = 141.75 10,

(b) The Test Statistic: = 93/225 = 0.4133

The observed sample proportion of US workers who telecommuted for work in 2018 is 1.35 null standard deviations above the hypothesized population proportion..

(c) The probability of obtaining a Z statistic as large or larger than that obtained is 0.0885.

(d) If this inference procedure was repeated then we would expect to see a z test statistic like the one we got in 8.85% of the repetitions.

(e) There is not sufficient evidence to say that the population proportion of all US workers who telecommuted for work in 2018 is higher than the 2015 rate of 37%.