Reuben is a human resources executive for a technology company. He is deciding between two types of plans for vacation allowance for the employees of the company: Unlimited and Traditional. Reuben wants to estimate, for workers in the tech industry, the difference between the yearly mean number of vacation days taken by workers with an Unlimited plan and the yearly mean number of vacation days taken by workers with a Traditional plan. Reuben surveys a random sample of 17 workers who have the Unlimited plan and a random sample of 16 workers who have the Traditional plan. (These samples are chosen independently.) For each worker, he records the number of vacation days taken last year. For the workers with an Unlimited plan, the sample mean is 17.7 with a sample variance of 44.8 . For the workers with a Traditional plan, the sample mean is 14.3 with a sample variance of 5.1 . Assume that the two populations of vacation days taken are approximately normally distributed. Let \( \mu_{1} \) be the population mean number of vacation days taken by workers with an Unlimited plan. Let \( \mu_{2} \) be the population mean number of vacation days taken by workers with a Traditional plan. Construct a \( 99 \% \) confidence interval for the difference \( \mu_{1}-\mu_{2} \). Then find the lower and upper limit of the \( 99 \% \) confidence interval. Carry your intermediate computations to three or more decimal places. Round your answers to two or more decimal places. (If necessary, consult a list of formulas.)
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- For the Unlimited plan: - Sample mean (\( \bar{x}_1 \)) = 17.7 - Sample variance (\( s_1^2 \)) = 44.8 - Sample size (\( n_1 \)) = 17 - For the Traditional plan: - Sample mean (\( \bar{x}_2 \)) = 14.3 - Sample variance (\( s_2^2 \)) = 5.1 - Sample Show more…
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Chris is a human resources executive for a technology company. He is deciding between two types of plans for vacation allowance for the employees of the company: Unlimited and Traditional. Chris wants to determine, for workers in the tech industry, if the yearly mean number of vacation days taken by workers with an Unlimited plan is greater than the yearly mean number of vacation days taken by workers with a Traditional plan. Chris surveys a random sample of 17 workers who have the Unlimited plan and a random sample of 15 workers who have the Traditional plan. (These samples are chosen independently.) For each worker, he records the number of vacation days taken last year. These data are shown in the table. Vacation days taken Unlimited 14, 16,23 , 15,15 ,9 , 15,18 ,17 ,28 ,25 , 24, 19, 12, 28, 22, 21 Traditional 19, 16, 15, 19, 20,22 , 18, 19,17 , 19, ,16 , 14,14 , 14,19 Assume that the two populations of vacation days taken are approximately normally distributed. Can Chris conclude, at the 0.10 level of significance, that the population mean of the yearly number of vacation days taken by workers with an Unlimited plan is greater than the population mean of the yearly number of vacation days taken by workers with a Traditional plan? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternate hypothesis H1 . (b) Determine the type of test statistic to use. T Degrees of freedom: (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the p-value. (Round to three or more decimal places.) (e) At the 0.10 level of significance, can Chris conclude that the yearly mean number of vacation days taken by workers with an Unlimited plan is greater than the yearly mean number of vacation days taken by workers with a Traditional plan? Yes or No
Kenny M.
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.3 days per employee with a standard deviation of 1.9 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.4 day per employee with a standard deviation of 1.8 days. Suppose we randomly select a sample of 100 blue-collar workers. Based on Martocchio's estimates: (a) What is the probability that the average amount of paid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? (Use the rounded mean and standard error to compute the rounded Z-score used to find the probability. Round means to 1 decimal place, standard deviations to 2 decimal places, and probabilities to 4 decimal places. Round z-value to 2 decimal places.) μx̄ = μ 1.4 σ^x̄ 1.30 P(x̄ > 1.5) 0.1300 (b) What is the probability that the average amount of unpaid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? (Use the rounded mean and standard error to compute the rounded Z-score used to find the probability. Round standard deviations to 2 decimal places and probabilities to 4 decimal places. Round z-value to 2 decimal places.) μx̄ = μ σ^x̄ P(x̄ > 1.5) (c) Suppose we randomly select a sample of 100 blue-collar workers, and suppose the sample mean amount of unpaid time lost during a three-month period actually exceeds 1.5 days. Would it be reasonable to conclude that the mean amount of unpaid time lost has increased above the previously estimated 1.4 days? Yes, the probability of observing the sample is if the mean is actually 1.4.
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