HW5
Exercise 9-17 (p375)
Issue:
BusQues:
Data:
Sample of 40 Jones and Ryanemployees' bonuses with a sample mean, x-bar, = $118,000.
Analysis:
1. Hypotheses:
Ho: mu = $125,500
Note - the Bus Ques becomes the Ha.
Ha: mu <> $125,500
2. Test Statistic:
(1) x-bar, the sample mean.
(2) If Ho is true, x-bar is Normal
with mean =
$125,500
and stdev (or std error) =
$4,743.42
=30000/40^0.5
3. Significance Level:
alpha = 0.05
[prob that we will reject Ho when it is true.]
4. Reject Rule:
Reject Ho if x-bar <
$116,203.07
or if x-bar >
$134,796.93
=NORMINV(0.025,125500,4743.42)
=NORMINV(0.975,125500,4743.42)
5. Calculations:
x-bar =
$118,000 Sample mean is below $125,500 by quite a bit--but just a sample of 40.
p-value =
0.114
=2*NORMDIST(118000,125500,4743.42,TRUE)
"If Ho is true (mu = 125,500), the chance of a sample mean
from a random sample
of 40 being different than 125,500 by 7,500 (like this sample) or more is 0.114."
6. Conclusion:
DO NOT REJECT Ho.
Based on this sample of 40 bonuses at Jones and Ryan, there is insufficient evidence
to conclude that the average bonus of All Jones and Ryan employees is different than $125,500.
Could add the following:
Even though we obtained a sample mean $7500 below $125,500,
if mu=125,500,
sample means this far from 125,500 or further will occur more than 11% of the time.
In this test, we are using a "cutoff" of 5% to conclude that mu <> $125,500.
Wall Street record year-end bonuses of $125,500 per employee for 2005 (