Question

In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him or herself talk. Let x1 be a random variable representing productive hours per week lost by peer employees of an intimidator. x1: 7 2 7 3 2 5 2 A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let x2 be a random variable representing productive hours per week lost by peer employees of a stressor. x2: 2 2 10 8 6 2 5 8 (i) Use a calculator with sample mean and sample standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to two decimal places.) x1 = s1 = x2 = s2 = (ii) Assuming the variables x1 and x2 are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a 5% level of significance. (Assume the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetric.) (a) What is the level of significance? State the null and alternate hypotheses. H0: μ1 < μ2; H1: μ1 = μ2 H0: μ1 = μ2; H1: μ1 ≠ μ2 H0: μ1 = μ2; H1: μ1 < μ2 H0: μ1 = μ2; H1: μ1 > μ2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.) (c) Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. Maple Generated Plot Maple Generated Plot Maple Generated Plot Maple Generated Plot (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence that the mean time lost due to stressors is greater than the mean time lost due to intimidators. Reject the null hypothesis, there is insufficient evidence that the mean time lost due to stressors is greater than the mean time lost due to intimidators. Fail to reject the null hypothesis, there is insufficient evidence that the mean time lost due to stressors is greater than the mean time lost due to intimidators. Reject the null hypothesis, there is sufficient evidence that the mean time lost due to stressors is greater than the mean time lost due to intimidators.

Answer 1

=4, s1 = 2.31, n1 = 7

= 5.38, s2 = 3.16, n2 = 8

a) The level of significance is 0.05

H₀:

H₁:

b) The student's t . We assume that both population distributions are approximately normal with unknown standard deviations.

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                            = (4 - 5.38)/sqrt((2.31)^2/7 + (3.16)^2/8)

                            = -0.97

c) df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n1 - 1))

        = ((2.31)^2/7 + (3.16)^2/8)^2/(((2.31)^2/7)^2/6 + ((3.16)^2/8)^2/7)

        = 13

P-value = P(T < -0.97)

            = 0.1749

0.050 < P-value < 0.125

d) Since the P-value is greater than (0.1749 > 0.05), so we should not reject the null hypothesis.

At the = 0.05 level, we fail to reject the null hypothesis and conclude that the data are not statistically significant.

e) Fail to reject the null hypothesis. There is insufficient evidence that there is a difference in mean time lost due to stressors is greater than the mean time lost due to intimidators.