In: Statistics and Probability
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 | 6 | 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 | 4 | 5 | 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?
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.)
(b) Find a 95% confidence interval for μ1 − μ2.(Round your answers to two decimal places.)
lower limit ? | |
upper limit ? |
mean of sample 1, x̅1= 4.57
standard deviation of sample 1, s1 =
2.82
mean of sample 2, x̅2= 6.00
standard deviation of sample 2, s2 =
2.62
a)
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
Level of Significance , α =
0.05
Sample #1 ----> successful
mean of sample 1, x̅1= 4.57
standard deviation of sample 1, s1 =
2.82
size of sample 1, n1= 7
Sample #2 ----> unsuccessful
mean of sample 2, x̅2= 6.00
standard deviation of sample 2, s2 =
2.62
size of sample 2, n2= 8
difference in sample means = x̅1-x̅2 =
4.5714 - 6.0 =
-1.43
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 2.7134
std error , SE = Sp*√(1/n1+1/n2) =
1.4043
t-statistic = ((x̅1-x̅2)-µd)/SE = ( -1.4286
- 0 ) / 1.40
= -1.017
Degree of freedom, DF= n1+n2-2 =
13
t-critical value , t* =
2.1604 (excel formula =t.inv(α/2,df)
Decision: | t-stat | < | critical value |, so, Do
not Reject Ho
p-value = 0.327587
(excel function: =T.DIST.2T(t stat,df) )
Conclusion: p-value>α , Do not reject null
hypothesis
There is not enough evidence 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)
the level of significance = 0.05
c)
Degree of freedom, DF= n1+n2-2 =
13
t-critical value = t α/2 =
2.1604 (excel formula =t.inv(α/2,df)
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 2.7134
std error , SE = Sp*√(1/n1+1/n2) =
1.4043
margin of error, E = t*SE = 2.1604
* 1.4043 =
3.0339
difference of means = x̅1-x̅2 =
4.5714 - 6.000 =
-1.4286
confidence interval is
Interval Lower Limit= (x̅1-x̅2) - E =
-1.4286 - 3.0339 =
-4.4624
Interval Upper Limit= (x̅1-x̅2) + E =
-1.4286 + 3.0339 =
1.6053
Please revert in case of any doubt.
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