Question

In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable x1 measures manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions. x1: 3 5 8 2 2 4 10 The variable x2 measures manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable". x2: 8 5 6 7 9 4 10 3 (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) Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use α = 0.05. Assume that the two lost-time population distributions are 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 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. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. 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 > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 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 statistically significant. At the α = 0.05 level, we fail to 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 not statistically significant. (e) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is insufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Reject the null hypothesis, there is sufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes.

Answer 1

$\bar x1$ = 4.86, s1 = 3.08, n1 = 7

$\bar x2$ = 6.5, s2 = 2.45, n2 = 8

a) The level of significance is 0.05

H₀: $\mu1 = \mu2$

H₁: $\mu1 \neq \mu2$

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

The test statistic t = ($\bar x1 - \bar x2$)/sqrt(s1^2/n1 + s2^2/n2)

                            = (4.86 - 6.5)/sqrt((3.08)^2/7 + (2.45)^2/8)

                            = -1.13

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

        = ((3.08)^2/7 + (2.45)^2/8)^2/(((3.08)^2/7)^2/6 + ((2.45)^2/8)^2/7)

        = 11

P-value = 2 * P(T < -1.13)

            = 2 * 0.1413 = 0.2826

0.250 < P-value < 0.500

0.1413 0.1 13

d) Since the P-value is greater than $\alpha$ (0.2826 > 0.05), so we should not reject the null hypothesis.

At the $\alpha$ = 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 hot tempers technical workers ' attitudes.