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: 1 5 8 4 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 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 = (a) Does the information indicate that the population mean time lost due to hot tempers is different (either way) from the 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. (i) 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 . (ii) What sampling distribution will you use? What assumptions are you making? 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 standard normal. 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? Compute the corresponding z or t value as appropriate. (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.) (iii) 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 Correct: Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot Correct: (iv) Based on your answers in parts (i)−(iii), 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 statistically significant. 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 not statistically significant. (v) Interpret your conclusion in the context of the application. 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. 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. 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 insufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. (b) Find a 95% confidence interval for μ1 − μ2. (Round your answers to two decimal places.) lower limit upper limit Explain the meaning of the confidence interval in the context of the problem. Because the interval contains only positive numbers, this indicates that at the 95% confidence level, the mean amount of a manager's hours lost due to hot tempers is greater than those lost due to disputes arising from technical workers' superior attitudes. Because the interval contains both positive and negative numbers, this indicates that at the 95% confidence level, we cannot say that the mean amount of a manager's hours lost per week differs between the two categories. Because the interval contains both positive and negative numbers, this indicates that at the 95% confidence level, the mean amount of a manager's hours lost due to hot tempers is more than those lost due to disputes arising from technical workers' superior attitudes. Because the interval contains only negative numbers, this indicates that at the 95% confidence level, the mean amount of a manager's hours lost due to hot tempers is less than those lost due to disputes arising from technical workers' superior attitudes.

Answer 1

a.

Given that,
mean(x)=4.8571
standard deviation , s.d1=3.1848
number(n1)=7
y(mean)=6.5
standard deviation, s.d2 =2.4495
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.8571-6.5/sqrt((10.14295/7)+(6.00005/8))
to =-1.1079
| to | =1.1079
critical value
the value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 2.447
we got |to| = 1.10789 & | t α | = 2.447
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.1079 ) = 0.31
hence value of p0.05 < 0.31,here we do not reject Ho
ANSWERS
---------------
i.
level of significance =0.05
ii.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t.
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.1079
critical value: -2.447 , 2.447
iv.
decision: do not reject Ho
iii.
p-value: 0.31
p value is greater than alpha value
v.
we do not have enough evidence to support the claim that the population mean time lost due to hot tempers is different (either way)
from the population mean time lost due to disputes arising from technical workers' superior attitudes.
b.
TRADITIONAL METHOD
given that,
mean(x)=4.8571
standard deviation , s.d1=3.1848
number(n1)=7
y(mean)=6.5
standard deviation, s.d2 =2.4495
number(n2)=8
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((10.143/7)+(6/8))
= 1.483
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 2.447
margin of error = 2.447 * 1.483
= 3.629
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (4.8571-6.5) ± 3.629 ]
= [-5.272 , 1.986]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=4.8571
standard deviation , s.d1=3.1848
sample size, n1=7
y(mean)=6.5
standard deviation, s.d2 =2.4495
sample size,n2 =8
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 4.8571-6.5) ± t a/2 * sqrt((10.143/7)+(6/8)]
= [ (-1.643) ± t a/2 * 1.483]
= [-5.272 , 1.986]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-5.272 , 1.986] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion