point estimate only and ignore sampling error (i.e., the margin of error in a confidence
interval). Suppose that a simple random sample of 51 stores this month shows mean
sales of 52 units of a small appliance with standard deviation 10 units. During the same
month last year, a sample of 75 stores gave mean sales of 48 units with standard
deviation 9 units. An increase from 48 to 52 is a rise of 8.33% so the manager is happy
because “Sales are up 8.33%,” he claims.
SOLUTION: Twosample ttest of two independent means
Use the version with the pooled estimate for the variance. Since the sample standard
deviations are so close to one another (10 and 9), we assume that the samples come
from populations with equal variances.
H
0
: µ
1
− µ
2
= 0, H
a
: µ
1
− µ
2
≠ 0, where µ
1
and µ
2
are the average sales of this year and
last year, respectively.
Test statistic: t = (52–48) / 9.416√[1/51+1/75] = 2.341, with 124 df.
Pvalue = TDIST(2.341,124,2) = 0.0208
At a 5% significance level, there is sufficient evidence to conclude that the change in
sales from last year to this is statistically significant. (But not at the 1% sig. level!)
Note: With the unpooled version, the test statistic = 2.294.
Explanation to Manager: The data come from only a sample of stores, not all stores in
the chain. It is probable that the true average sales—from all stores—changed by some
amount other than the 4 units, or 8.33%, observed in the sample.
***
95% confidence interval for the difference in the mean number of units sold at all retail
stores:
Pooled SD = 9.416
95% CI = (5248) ± 1.979 x 9.416√[1/51+1/75] = 4 ± 3.4
or
(0.6,7.4)
Note: The unpooled version would give an interval of (0.5,7.5)
Since 0 is not within the 95% confidence interval from part, a twosided test of H
0
, µ
1
− µ
2
= 0 at the 5% level of significance will necessarily be rejected. The difference between
the mean sales year over year is significantly different from 0.
c) Explain in language that the marketing manager can understand why he cannot be
certain that sales went up by 8.33%.
4
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Scenario 5. Hospital Admissions on Friday the 13th
This question is based on a research study quoted on CBC Radio 1 on Friday, February
13, 2009.
Is Friday the 13
th
really unlucky? Do hospital emergency wards see more admissions
(that is, acutely injured or sick patients) on a Friday the 13
th
than they see, for example,
on each previous Friday? A British study examined the number of emergency
admissions on various Fridays the 13
th
and on the previous Fridays the 6
th
between 1989
and 1995. Here are their results.
Admissions on
Admissions on
Friday the 6
th
Friday the 13
th
October 1989
29
41
July 1990
22
40
September 1991
40
46
December 1991
36
39
March 1992
15
14
November 1992
16
43
August 1993
19
19
May 1994
38
42
January 1995
28
27
October 1995
45
44
Assess the evidence that, in the long run, there are more emergency admissions on
Fridays the 13
th
than there are on the preceding Fridays by conducting a hypothesis test
at the 5% level.
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 Winter '10
 E.Fowler
 Statistics, Statistical hypothesis testing

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