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Unformatted text preview: dent random samples of weekly proﬁts from the two populations
(one population is ﬂights to the proposed airport, and the other population is ﬂights to an alternative airport). Data are
as follows. For the proposed route, ̅ = $96,540 per week and = $12,522. For the alternative route, ̅ = $85,991 and
= $19,548. Test the hypothesis that the proposed route is more proﬁtable than the alternative route. Use a
signiﬁcance level of your choice.
Proposed Route
(1)
=6
̅ = 96,540 = 12,522 SOLUTION: Alternative Route
(2)
=9
̅ = 85,991 = 19,548 We assume that two populations are normally distributed.
: − ≤0
:−
>0
The test statistic value:
( ̅ − ̅ )−( − )
(96,540 − 85,991) − 0
=
=
≈ 1.2737
(12,522)
(19,548)
+
6
9
+
At = 0.05 and
= ( ⁄ (⁄+⁄)
) /( − 1) + ( ⁄ ) /( We have the critical value: = ( ,) = ( ,. ) − 1) = (12,522 ⁄6 + 19,548 ⁄9)
≈ 12.9993 ≈ 12
(12,522 ⁄6)
(19,548 ⁄9 )
+
(6 − 1)
(9 − 1) = 1.782 Thus, at 0.05 level of significance, we cannot reject
since
< . It means that with the hypothesis testing we do not
have sufficient evidence to prove that the proposed route is more profitable than the alternative route. Powered by statisticsforbusinessiuba.blogspot.com Statistics for Business  Chapter 08: The Comparison of Two Populations International University IU 15 PROBLEM I02D:
(Situation IV) The photography department of a fashion magazine needs to choose a camera. Of the two models the department is
considering, one is made by Nikon and one by Minolta. The department contracts with an agency to determine if one of
the two models gets a higher average performance rating by professional photographers, or whether the average
performance ratings of these two cameras are not statistically different. The agency asks 60 different professional
photographers to rate one of the cameras (30 photographers rate each model). The ratings are on a scale of 1 to 10. The
average sample rating for Nikon is 8.5, and the sample standard deviation is 2.1. For the Minolta sample, the average
sample rating is 7.8, and the standard deviation is 1.8. Is there a difference between the average population ratings of
the two cameras? If so, which one is rated higher?
Nikon
(1)
= 30
̅ = 8.5 = 2.1 SOLUTION: Minolta
(2)
= 30
̅ = 7.8 = 1.8 We assume that two populations are normally distributed
: − =0
:−
≠0
The test statistic value:
( ̅ − ̅ )−( − )
(8.5 − 7.8) − 0
=
=
≈ 1.3862
(2.1)
(1.8)
30 + 30
+
At = 0.05, the critical value(s): ± = ± / = ±1.96
Thus, at 0.05 level of significance, we cannot reject
since ∈ [− , ]. It means that with the hypothesis testing we
do not have sufficient evidence to prove the difference between the average population ratings of two cameras. Powered by statisticsforbusinessiuba.blogspot.com Statistics for Business  Chapter 08: The Comparison of Two Populations International University IU 16 International University IU
PART II HYPOTHESIS TESTING
PROCESS Two – tailed Testing Right – tailed Testing Step 01
: /
ℎ (1 −
Condition
Step 02
: − =( − )
: − ≤( −
Determine the null and
alternative
hypotheses
: − ≠( − )
: − >( −
( and )
Step 03 and
Compute the test statistic For all instances, we always use −
value ( / ) and t...
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This document was uploaded on 03/01/2014 for the course ACCT 404 at Indiana State University .
 Winter '09

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