Download 9.3 tests about a population mean

9.3 TESTS ABOUT A POPULATION MEAN
Carrying Out a Significance Test for μ
Conditions: Random, Normal, and Independent need to be met
Standard Deviation is Known
If the following cases are true
1. Population is Normal and n <30
or
2. N ≥ 30 (Central Limit Theorem)
We use z = (x­bar ­ μ)/(σ/√n)
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Ex: Better Batteries p566 and 567
Check Random, Normal and Independent
Test statistic = (statistic ­ parameter)/(standard deviation of statistic)
t = (xbar ­ μ0)/(sx/√n) = (33.9 ­ 30)/(9.8/√15) = 1.54
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Example: P568 Two­Sided Tests, Negative t values and more
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Technology Corner: Computing P­values from t distributions on the calculator
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Example: A consumer advocacy group suspects that a local supermarket's 10 ounce packages of cheddar cheese actually weighs less than 10 ounces. The group took a random sample of 20 such packages and found mean weight 9.9555 ounces. The population follows a Normal distribution with pop. standard deviation = .15 ounces
Will you Reject H0 at a significance level fo .01?
H0: μ =10
x­bar = 9.9555 n = 20 σ = .15 Normal
Ha: μ< 10
lower tail test
z = (x­bar ­ μ)/(σ/√n) = (9.955 ­10)/(.15/√20)
= ­1.34
P­value = P(z ≤ ­1.34) = .0901
Is .0901 ≤ Alpha
NO Fail To Reject H0
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Consider H0: μ=100
Ha: μ ≠ 100
A random sample of 64 observations produced a sample mean of 98. Assume that the pop. st. deviation is 12. Using a significance level of .01, would you reject the H0?
two­tail
x­bar = 98 n = 64 σ =12
z = (x­bar ­ μ)/(σ/√n) = (98 ­100)/(12/√64)
= ­1.33
P­value = 2P(z ≤ ­1.34) = 2(.0918) =.1836
Is .1836 ≤ Alpha
NO Fail To Reject H0
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STANDARD DEVIATION IS NOT KNOWN
1. (Hope) Population is Normal
sample standard deviation
2. n < 30 t = (x­bar ­ μ)/(s/√n)
where t has ν = n­1 degrees of freedom
Example: p567 Better Batteries
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One­Sample t Test
Choose an SRS of size n from a large population with unknown mean μ. To test the hypothesis H0: μ = μ0, compute the one­
sample t statistic
t = (x­bar ­ μ0)/(sx/√n)
Find the P­value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha in a t distribution with df = n­1
Ha: μ > μ0
Ha: μ < μ0
Ha: μ ≠ μ0
Use this test only when (1) the population distribution is Normal or the sample is large (n ≥ 30), and (2) the population is at least 10 times as large as the sample.
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Ex: Healthy Streams p571
Performing a significance test about μ
4.53 5.04 3.29 5.23 4.13 5.5 4.83 4.4
5.42 6.38 4.01 4.66 2.87 5.73 5.55
t = (xbar ­ μ0)/(sx/√n) = (.4771­5)/(.9396/√15) = ­.94
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Technology Corner p573
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Ex: According to a BB coach, the mean height of all female college BB players is 69.5 inches. A random sample of 25 such players produced a mean height of 70.25 inches with a standard deviation of 2.1 inches.
Assuming that the heights of all female players is Normally distributed test at .02 significance level whether the mean height is different from 69.5.
H0: μ =69.5
Ha: μ≠ 69.5 (two tail)
x­bar = 70.25
sx = 2.1
n = 25 ν = 24
t = (xbar ­ μ)/(s/√n) = (70.25 ­69.5)/(2.1/√25)
= 1.786
1.786 does not fall in shaded region
Fail to Reject H0
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Ex: H0: μ =18
Ha: μ > 18 (upper tail)
x­bar = 22.5
n = 40 ν = 39
sx = 8
Test at .01
t = (xbar ­ μ)/(s/√n) = (22.5 ­18)/(8/√40)
= 3.558
Refer to chart to find t on graph
3.558 falls in the critical/shaded region
Reject H0
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EX: Juicy Pineapple p574 and p576
var
N
Weig
50
ht`
se medi
stdev min Q1
mean
an Q3
31.93
26.49 29.99 31.73 34.115
.339 2.394
5
1
0
9
mean
a) Q3 +1.5(IQR) = 40.3025>35.547
Q1 ­ 1.5(IQR) = 23.8025<26.491
Max
35.547
IQR = 4.125
b) State: H0: μ = 31 Ha: μ≠31 no significance level is given so use a standard such as α = .05
Plan: Conditions Random = yes, Normal 50>30, Independent ­ 10%
Do: t = (xbar ­ μ0)/(sx/√n) = (31.935 ­ 31)/(2.394/√50) = 2.762
Pvalue df = 49 calculator = 2tcdf(2.762, 100, 49) = .0081
Conclude: Pvalue .0081 is less than α = .05 so we have enough evidence to reject H0 and conclude that the mean weight of the pineapples in this year's crop is not 31 ounces.
c) No. This was not a comparative experiment, so we cannot infer causation.
Continue Example on P576 Juicy Pineapples
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TWO SIDED TESTS AND CONFIDENCE INTERVALS
Think About It. p. 576
Is there a connection between one­sided tests and confidence intervals for a population mean? Additional Examples p188 and 189 in Black Book
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Inference for Means: Paired Data
Definition: Study designs that involve making two observations on the same individual or one observation on each of two similar individuals, result in paired data
Ex: Is Caffeine Dependence Real?
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Data Exploration: Nitrogen In Tires?
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Using Tests Wisely:
1. Statistical Significance and Practical Importance
2. Don't Ignore Lack of Significance
3. Statistical Inference is Not Valid for All Sets of Data
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Ex: Wound Healing Time
p582
H0: μ = 7.6 days
Ha: μ < 7.6 days
!!! Remember: Statistical Significance is not the same thing as practical importance.
The foolish user of statistics who feeds the data to a calculator or computer without exploratory analysis will often be embarrassed.
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Example: Reducing HIV Transmission P583
!! When planning a study, verify that the test you plan to use has a high probability (power) of detecting a difference of the size you hope to find.
Example: Does Music Increase Worker Productivity? p584
Example: Cell Phones and Brain Cancer p584
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Paired Data Example:
Blood Samples during sleep on men
(paired b/c drawn on same men)
Day 1 No Activity (Control)
Day 2 Activity
Based on the data, test at a sig. level of .05 that there is a significant diff. between the control and the strenuous activity in Growth hormone level.
Subject
1
Post exercise 13.6
Control 8.5
difference 5.1
H0: μd = 0
Ha: μd ≠ 0
2
14.7
12.6
2.1
3
42.8
21.6
21.2
4
20
19.4
.6
5
19.2
14.7
4.5
Σd = 37.2 Σd2 = 514.16
sd = 7.5302
6
17.3
13.6
3.7
d­bar = 6.2
t = (6.2)/(7.5302/√6) = 2.017 ν = 6­1 = 5
Chart t does not fall in critical region
Fail to Reject H0
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Hypnotism example Paired Data
Study on effectiveness of hypnotism on pain. Using a sig. level of .05 test the claim that the sensory measurements are lower after hypnotism.
subject A
before 6.6
after 6.8
find diff. .2
B
6.5
2.4
C
4.0
2.4
D
E
F
10.3 11.3 8.1
8.5 8.1 6.1
­4.1 ­2.6 ­1.8 ­2.2
G
6.3
3.4
H
11.6
2.0
­2
­2.9 ­9.6
dbar = ­3.125
sd = 2.911
Σd = ­25
Σd2 = 137.46
H0: μd = 0
Ha: μd < 0
ν = 8 ­ 1 = 7 t = (­3.125)/(2.911/√8) = ­3.036
Calculate t value on chart
Reject H0
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Homework: p 564 and 587 57­60, 71, 73
p587 75,77,89,94­97, 99­104
Unit 9 Review
Unit 9 Practice Test
Unit 9 Strive for 5 10 MC and Frappy
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