Download 8-3C: Examples of hypothesis testing for population means

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Bootstrapping (statistics) wikipedia , lookup

Degrees of freedom (statistics) wikipedia , lookup

Psychometrics wikipedia , lookup

Resampling (statistics) wikipedia , lookup

Student's t-test wikipedia , lookup

Long tail wikipedia , lookup

Transcript
Section 8 – 3C
Testing a Claim About the Population Mean
Examples
Example 1 (One Tail Left Tail Test)
A local radio station claims that it plays 55 minutes of music every hour. (More Music Less Noise).
After several complaints the State Broadcasting Board conducts an investigation. The Board selects
41 random hours of the stations broadcasts and finds that the 41 broadcasts had an average of 53.4
minutes of music with a standard deviation of 3.8 minutes. Use a .01 significance level to test
the claim that the average minutes of music broadcast each hour is less than 55 minutes.
H 0: µ = 55
Sample Mean: x = 53.4
H1: µ < 55 (One Tail Test Left Tail)
Sample Standard Deviation: sx = 3.8
n = 41
DF = 40
α = .01
(all .01 in the left tail)
Graph of Critical information:
Test Statistic:
t=
t=
( x − µ)
sx
n
(53.4 − 55)
3.8
41
t = −2.70
Conclusion based on H 0 : Reject H0
Conclusion based on the problem:
There is sufficient evidence at the α = .01 level to support the claim that the average minutes of
music broadcast each hour is less than 55 minutes.
t
Distribution
Degrees of
Freedom
40
Area In One Tail (Right Tail)
0.100
0.050
0.025
0.010
0.005
1.303
1.684
2.021
2.423
2.704
If the Right Tail Critical Value is +2.423 then the the Left Tail Critical Value is –2.423
Section 8 – 3C
Page 1
©2013 Eitel
Testing a Claim About the Population Mean
Example 2 (One Tail Right Tail Test)
Avis Car Rentals reports that the average daily rental cost for a midsize car at airports used for national
flights is $ 47.56 per day base price without taxes, surcharges or other local costs. The Conde Nashʼs
travel magazine feels that this price is too low. They donʼt want travelers to read about the low cost of
car rentals if indeed the average cost is higher. A Conde Nash reporter checks the rates on mid sized
cars at 35 airport rentals selected at random. The sample shows that the average cost is $ 51.23 a
day with a standard deviation of $ 8.12. Test the claim at a .10 significance level that the
average daily rental cost for a midsize car at airports used for national flights is more than $ 47.56 per
day.
H 0: µ = 47.56
Sample Mean: x = 51.23
H1: µ > 47.56 (One Tail Teat Right Tail)
Sample Standard Deviation: sx = 8.12
n = 35
DF = 34
α = .10
(all .10 in the right tail)
Graph of Critical information:
Test Statistic:
t=
t=
(x − µ)
sx
n
(51.23− 47.56)
8.12
35
t = 2.67
Conclusion based on H 0 : Reject H0
Conclusion based on the problem:
There is sufficient evidence at the α = .10 level to support the claim that the average daily rental
cost for a midsize car at airports used for national flights is more than $ 47.56 per day
t
Distribution
Degrees of
Freedom
34
Section 8 – 3C
Area In One Tail (Right Tail)
0.100
0.050
0.025
0.010
0.005
1.307
1.691
2.032
2.441
2.728
Page 2
©2013 Eitel
Testing a Claim About the Population Mean
Example 3 (Two Tail Test)
The IRS Audit division reports that the average money deducted for donations to charities in 2008 tax
year was $ 245. The California Franchise Tax Board thinks that this amount is incorrect for the 2008
Californian returns. They select 16 California tax returns at random and find that the average
donation returns was $ 240 with a standard deviation of $ 14. Test the claim at a .05
significance level that the average California tax return for charitable donations was not equal to
$
245. Assume that the distribution of charitable contributions in the 2008 state tax returns is normally
distributed.
H 0: µ = 245
Sample Mean: x = 240
H1: µ ≠ 245 (Two Tail Test)
Sample Standard Deviation: sx = 14
n = 16
DF = 15
α = .05
( α 2 = .025in each tail)
Graph of Critical information:
Test Statistic:
t=
t=
(x − µ)
sx
n
(240 − 245)
14
16
t = −1.43
Conclusion based on H 0 :
Do not Reject H0
Conclusion based on the problem:
There is not sufficient evidence at the α = .05 level to reject the hypothesis that the average
2008 California tax return for charitable donations was equal to $ 245
Distribution
Degrees of
Freedom
15
Section 8 – 3C
Area In One Tail (Right Tail)
0.100
0.050
0.025
0.010
0.005
1.341
1.753
2.131
2.602
2.947
Page 3
©2013 Eitel