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fAbout 13% of employed adults in the United States held multiple jobs. A random sample of 74 employed adults is chosen. Use the TI-84 Plus

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\fAbout 13% of employed adults in the United States held multiple jobs. A random sample of 74 employed adults is chosen. Use the TI-84 Plus calculator as needed. Part 1 of 5 (a) Is it appropriate to use the normal approximation to find the probability that less than 7.1% of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not. It is not appropriate to use the normal curve, since np = 9.62 10. Part 2 of 5 (b) A new sample of 223 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places. The probability that less than 7.1% of the individuals in this sample hold multiple jobs is .0042 Alternate Answer: 0.0042 Part: 2 / 5 Part 3 of 5 (c) Find the probability that more than 6.6% of the individuals in the sample of 225 hold multiple jobs. Round the answer to at least four decimal places. The probability that more than 6.6% of individuals in the sample of 225 hold multiple jobs isFind the critical value to /2 needed to construct a confidence interval of the given level with the given sample size. Round the answers to three decimal places. Part 1 of 4 (a) For level 80% and sample size 7 Critical value = [ X 5 Part 2 of 4 (b) For level 98% and sample size 10 Critical value = Part 3 of 4 (c) For level 90% and sample size 30 Critical value = Part 4 of 4 (d) For level 99% and sample size 14 Critical value =A sample of size n = 50 has sample mean x = 35.2 and sample standard deviation $ = 9.3. Part: 0 / 2 Part 1 of 2 Construct an 80% confidence interval for the population mean L. Round the answers to one decimal place. An 80% confidence interval for the population mean isOnline courses: A sample of 261 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the most favorable impression. The average score was 5.45, and the standard deviation was 0.90. Part: 0 / 2 Part 1 of 2 Construct a 99.5% interval for the mean score. Round the answers to two decimal places. A 99.5% confidence interval for the mean score is 4 The alternate hypothesis is (Choose one) eft-tailed X right-tailed two-tailedDetermine whether the alternate hypothesis is left-tailed, right-tailed, or two-tailed. Ho : H= 9 H1 := 9 The alternate hypothesis is (Choose one) left-tailed X 5 right-tailed two-tailed

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