1
Chapter 10
Hypothesis Tests
Using a Single Sample
2
BASICS
In statistics, a hypothesis is a statement
about a population characteristic.
3
FORMAL STRUCTURE
Hypothesis Tests are based on an reductio
ad absurdum form of argument.
Specifically, we make an assump...
4
FORMAL STRUCTURE
The null hypothesis, denoted H0 is a
statement or claim about a population
characteristic that is initi...
5
FORMAL STRUCTURE
The alternate hypothesis, denoted by Ha is
the competing claim.
The alternate hypothesis is a statement...
6
FORMAL STRUCTURE
Rejection of the null hypothesis will imply
the acceptance of this alternative
hypothesis.
Assume H0 is...
7
FORMAL STRUCTURE
Typically one assumes the null hypothesis to
be true and then one of the following
conclusions are draw...
8
AN ANALOGY
The Statistical Hypothesis Testing process
can be compared very closely with a judicial
trial.
1.Assume a def...
9
AN ANALOGY
Two Hypotheses are then created.
H0: Innocent
Ha: Not Innocent (Guilt)
10
Examples of Hypotheses
You would like to
determine if the
diameters of the
ball bearings you
produce have a
mean of 6.5...
11
The students
entering into the
math program used
to have a mean
SAT quantitative
score of 525. Are
the current students...
12
Do the “16 ounce” cans
of peaches canned and
sold by DelMonte meet
the claim on the label
(on the average)?
H0: µ = 16
...
13
Is the proportion of
defective parts
produced by a
manufacturing
process more than
5%?
H0: π = 0.05
(Really, π ≤ 0.05)
...
14
Do two brands of
light bulb have
the same mean
lifetime?
H0: µBrand A = µBrand B
Ha: µBrand A ≠ µBrand B
Examples of Hy...
15
Do parts produced
by two different
milling machines
have the same
variability in
diameters?
or equivalently
0 1 2
a 1 2...
16
Comments on Hypothesis Form
The null hypothesis must contain the equal
sign.
This is absolutely necessary because
the t...
17
Hypothesis Form
The form of the null hypothesis is
H0: population characteristic = hypothesized value
where the hypothe...
18
Caution
When you set up a hypothesis test, the
result is either
1. Strong support for the alternate
hypothesis (if the ...
19
Error
Null Hypothesis
Decision True False
Accept H0
Reject H0
No Error
No Error
Type I Error
Type II Error
α
β
20
Error Analogy
Consider a medical test where the hypotheses
are equivalent to
H0: the patient has a specific disease
Ha:...
21
More on Error
The probability of a type I error is denoted
by α and is called the level of significance
of the test.
Th...
22
Relationships Between α and β
Generally, with everything else held constant,
decreasing one type of error causes the ot...
23
Comment of Process
Look at the consequences of type I and type
II errors and then identify the largest α that
is tolera...
24
Test Statistic
A test statistic is the function of sample
data on which a conclusion to reject or fail
to reject H0 is ...
25
P-value
The P-value (also called the observed
significance level) is a measure of
inconsistency between the hypothesize...
26
Decision Criteria
A decision as to whether H0 should be
rejected results from comparing the P-value
to the chosen α:
H0...
27
Large Sample Hypothesis Test
for a Single Proportion
In terms of a standard normal random variable z, the
approximate P...
28
Hypothesis Test
Large Sample Test of Population Proportion
p hypothesized value
P-value P z
hypothesized value(1-hypoth...
29
p hypothesized value
P-value P z
hypothesized value(1-hypothesized value)
n
 
 −
 = >
 
 
 
Hypothesis Test...
30
p hypothesized value
P-value 2P z
hypothesized value(1-hypothesized value)
n
 
 −
 = >
 
 
 
Hypothesis Tes...
31
An insurance company states that the
proportion of its claims that are settled
within 30 days is 0.9. A consumer group
...
32
P-value P(z 4.71) 0= < − ≈
0.8 0.9 0.8 0.9
z 4.71
(0.9)(1 0.9) 0.9(0.1)
200 200
− −
= = = −
−
Hypothesis Test Example 2...
33
Hypothesis Test Example 2
Single Proportion continued
The probability of getting a result as strongly or
more strongly ...
34
Hypothesis Test Example 2
Single Proportion continued
We would say that we have strong support for
the claim that the p...
35
A county judge has agreed that he will
give up his county judgeship and run for
a state judgeship unless there is
evide...
36
Hypothesis Test Example
Single Proportion continued
π = proportion of his party that is in opposition
H0: π = 0.25
HA: ...
37
Hypothesis Test Example
Single Proportion continued
At a level of significance of 0.10, there is
sufficient evidence to...
38
1. Describe (determine) the population
characteristic about which hypotheses
are to be tested.
2. State the null hypoth...
39
Steps in a Hypothesis-Testing
Analysis
6. Check to make sure that any
assumptions required for the test are
reasonable....
40
x hypothesized mean
z
n
−
=
σ
Hypothesis Test (Large samples)
Single Sample Test of Population Mean
In terms of a stand...
41
x hypothesized mean
P-value P Z
n
 
− = <
 σ 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: µ =...
42
x hypothesized mean
P-value P Z
n
 
− = >
 σ 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: µ =...
43
x hypothesized mean
P-value 2P Z
n
 
− 
= > σ
 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: µ...
44
It is not likely that one would know σ but not
know µ, so calculating a z value using the
formula
would not be very rea...
45
x hypothesized mean
t
s
n
−
=
Hypothesis Test (σ unknown)
Single Sample Test of Population Mean
The approximate P-value...
46
x hypothesized mean
P-value P t
s
n
 
− = <
 
 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: µ...
47
x hypothesized mean
P-value P t
s
n
 
− = >
 
 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: µ...
48
x hypothesized mean
P-value 2P t
s
n
 
− 
= > 
 
 
Hypothesis Test
Single Sample Test of Population Mean
H0: ...
49
The t statistic can be used for all sample
sizes, however, the smaller the sample,
the more important the assumption th...
50
Tail areas
for t curves
t
df 1 2 3 4 5 6 7 8 9 10 11 12
0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500...
51
t
df 25 26 27 28 29 30 35 40 60 120 ∞(=z)
0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500
0.1 0.4...
52
An manufacturer of a special bolt requires
that this type of bolt have a mean
shearing strength in excess of 110 lb. To...
53
µ = the mean shearing strength of this
specific type of bolt
Τhe hypotheses to be tested are
H0: µ = 110 lb
Ha: µ > 110...
54
Example of Hypothesis Test
Single Sample Test of Population Mean - continued
x 112.7, s 9.62, n 25, df 24= = = =
112.7 ...
55
Because P-value = 0.087 > 0.05 = α, we
fail to reject H0.
At a level of significance of 0.05, there is
insufficient evi...
56
Using the t
table
t
df 13 14 15 16 17 18 19 20 21 22 23 24
0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0....
57
Revisit the problem with α=0.10
What would happen if the significance level
of the test was 0.10 instead of 0.05?
At th...
58
Comments continued
Many people are bothered by the fact that
different choices of α lead to different
conclusions.
This...
59
Another Example
A jeweler is planning on manufacturing gold
charms. His design calls for a particular piece to
contain ...
60
Another Example
1. The population characteristic being
studied is µ = true mean gold content for
this particular type o...
61
Another Example
6. Minitab was used to create a normal plot
along with a graphical display of the
descriptive statistic...
62
Another Example
We can see that with the exception of one
outlier, the data is reasonably symmetric and
mound shaped in...
63
Another Example
8. P-value: This is a two tailed test. Looking up in
the table of tail areas for t curves, t = 3.2 with...
64
Another Example
9. Conclusion:
Since P-value = 0.006 ≤ 0.01 = α, we
reject H0 at the 0.01 level of significance.
At the...
65
Power and Probability of Type
II Error
The power of a test is the probability of
rejecting the null hypothesis.
When H0...
66
Effects of Various Factors on Power
1. The larger the size of the discrepancy
between the hypothesized value and
the tr...
67
Some Comments
Calculating β (hence power) depends on
knowing the true value of the population
characteristic being test...
68
Example (based on z-curve)
Consider the earlier example where we
tested H0: µ = 110 vs. Ha: µ > 110 and
furthermore, su...
69
Example (based on z-curve)
Power Curves
Different α's
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
118 119 12...
70
Example (based on z-curve)
Power Curves
Different n's
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
118 119 12...
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  • Chapter10

    1. 1. 1 Chapter 10 Hypothesis Tests Using a Single Sample
    2. 2. 2 BASICS In statistics, a hypothesis is a statement about a population characteristic.
    3. 3. 3 FORMAL STRUCTURE Hypothesis Tests are based on an reductio ad absurdum form of argument. Specifically, we make an assumption and then attempt to show that assumption leads to an absurdity or contradiction, hence the assumption is wrong.
    4. 4. 4 FORMAL STRUCTURE The null hypothesis, denoted H0 is a statement or claim about a population characteristic that is initially assumed to be true. The null hypothesis is so named because it is the “starting point” for the investigation. The phrase “there is no difference” is often used in its interpretation.
    5. 5. 5 FORMAL STRUCTURE The alternate hypothesis, denoted by Ha is the competing claim. The alternate hypothesis is a statement about the same population characteristic that is used in the null hypothesis. Generally, the alternate hypothesis is a statement that specifies that the population has a value different, in some way, from the value given in the null hypothesis.
    6. 6. 6 FORMAL STRUCTURE Rejection of the null hypothesis will imply the acceptance of this alternative hypothesis. Assume H0 is true and attempt to show this leads to an absurdity, hence H0 is false and Ha is true.
    7. 7. 7 FORMAL STRUCTURE Typically one assumes the null hypothesis to be true and then one of the following conclusions are drawn. 1. Reject H0 Equivalent to saying that Ha is correct or true 2. Fail to reject H0 Equivalent to saying that we have failed to show a statistically significant deviation from the claim of the null hypothesis This is not the same as saying that the null hypothesis is true.
    8. 8. 8 AN ANALOGY The Statistical Hypothesis Testing process can be compared very closely with a judicial trial. 1.Assume a defendant is innocent (H0) 2.Present evidence to show guilt 3.Try to prove guilt beyond a reasonable doubt (Ha)
    9. 9. 9 AN ANALOGY Two Hypotheses are then created. H0: Innocent Ha: Not Innocent (Guilt)
    10. 10. 10 Examples of Hypotheses You would like to determine if the diameters of the ball bearings you produce have a mean of 6.5 cm. H0: µ = 6.5 Ha: µ ≠ 6.5 (Two-sided alternative)
    11. 11. 11 The students entering into the math program used to have a mean SAT quantitative score of 525. Are the current students poorer (as measured by the SAT quantitative score)? H0: µ = 525 (Really: µ ≥ 525) Ha: µ < 525 (One-sided alternative) Examples of Hypotheses
    12. 12. 12 Do the “16 ounce” cans of peaches canned and sold by DelMonte meet the claim on the label (on the average)? H0: µ = 16 (Really: µ ≥16) Ha: µ < 16 Examples of Hypotheses Notice, the real concern would be selling the consumer less than 16 ounces of peaches.
    13. 13. 13 Is the proportion of defective parts produced by a manufacturing process more than 5%? H0: π = 0.05 (Really, π ≤ 0.05) Ha: π > 0.05 Examples of Hypotheses
    14. 14. 14 Do two brands of light bulb have the same mean lifetime? H0: µBrand A = µBrand B Ha: µBrand A ≠ µBrand B Examples of Hypotheses
    15. 15. 15 Do parts produced by two different milling machines have the same variability in diameters? or equivalently 0 1 2 a 1 2 H : H : σ = σ σ ≠ σ 2 2 0 1 2 2 2 a 1 2 H : H : σ = σ σ ≠ σ Examples of Hypotheses
    16. 16. 16 Comments on Hypothesis Form The null hypothesis must contain the equal sign. This is absolutely necessary because the test requires the null hypothesis to be assumed to be true and the value attached to the equal sign is then the value assumed to be true and used in subsequent calculations. The alternate hypothesis should be what you are really attempting to show to be true. This is not always possible.
    17. 17. 17 Hypothesis Form The form of the null hypothesis is H0: population characteristic = hypothesized value where the hypothesized value is a specific number determined by the problem context. The alternative (or alternate) hypothesis will have one of the following three forms: Ha: population characteristic > hypothesized value Ha: population characteristic < hypothesized value Ha: population characteristic ≠ hypothesized value
    18. 18. 18 Caution When you set up a hypothesis test, the result is either 1. Strong support for the alternate hypothesis (if the null hypothesis is rejected) 2. There is not sufficient evidence to refute the claim of the null hypothesis (you are stuck with it, because there is a lack of strong evidence against the null hypothesis.
    19. 19. 19 Error Null Hypothesis Decision True False Accept H0 Reject H0 No Error No Error Type I Error Type II Error α β
    20. 20. 20 Error Analogy Consider a medical test where the hypotheses are equivalent to H0: the patient has a specific disease Ha: the patient doesn’t have the disease Then, Type I error is equivalent to a false negative (i.e., Saying the patient does not have the disease when in fact, he does.) Type II error is equivalent to a false positive (i.e., Saying the patient has the disease when, in fact, he does not.)
    21. 21. 21 More on Error The probability of a type I error is denoted by α and is called the level of significance of the test. Thus, a test with α = 0.01 is said to have a level of significance of 0.01 or to be a level 0.01 test. The probability of a type II error is denoted by β.
    22. 22. 22 Relationships Between α and β Generally, with everything else held constant, decreasing one type of error causes the other to increase. The only way to decrease both types of error simultaneously is to increase the sample size. No matter what decision is reached, there is always the risk of one of these errors.
    23. 23. 23 Comment of Process Look at the consequences of type I and type II errors and then identify the largest α that is tolerable for the problem. Employ a test procedure that uses this maximum acceptable value of α (rather than anything smaller) as the level of significance (because using a smaller α increases β).
    24. 24. 24 Test Statistic A test statistic is the function of sample data on which a conclusion to reject or fail to reject H0 is based.
    25. 25. 25 P-value The P-value (also called the observed significance level) is a measure of inconsistency between the hypothesized value for a population characteristic and the observed sample. The P-value is the probability, assuming that H0 is true, of obtaining a test statistic value at least as inconsistent with H0 as what actually resulted.
    26. 26. 26 Decision Criteria A decision as to whether H0 should be rejected results from comparing the P-value to the chosen α: H0 should be rejected if P-value ≤ α. H0 should not be rejected if P-value > α.
    27. 27. 27 Large Sample Hypothesis Test for a Single Proportion In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H0: π = hypothesized proportion, compute the z statistic p hypothesized value z hypothesized value(1-hypothesized value) n − =
    28. 28. 28 Hypothesis Test Large Sample Test of Population Proportion p hypothesized value P-value P z hypothesized value(1-hypothesized value) n    −  = <      
    29. 29. 29 p hypothesized value P-value P z hypothesized value(1-hypothesized value) n    −  = >       Hypothesis Test Large Sample Test of Population Proportion
    30. 30. 30 p hypothesized value P-value 2P z hypothesized value(1-hypothesized value) n    −  = >       Hypothesis Test Large Sample Test of Population Proportion
    31. 31. 31 An insurance company states that the proportion of its claims that are settled within 30 days is 0.9. A consumer group thinks that the company drags its feet and takes longer to settle claims. To check these hypotheses, a simple random sample of 200 of the company’s claims was obtained and it was found that 160 of the claims were settled within 30 days. Hypothesis Test Example Large-Sample Test for a Population Proportion
    32. 32. 32 P-value P(z 4.71) 0= < − ≈ 0.8 0.9 0.8 0.9 z 4.71 (0.9)(1 0.9) 0.9(0.1) 200 200 − − = = = − − Hypothesis Test Example 2 Single Proportion continued π = proportion of the company’s claims that are settled within 30 days H0: π = 0.9 HA: π < 0.9 160 p 0.8 200 = =The sample proportion is
    33. 33. 33 Hypothesis Test Example 2 Single Proportion continued The probability of getting a result as strongly or more strongly in favor of the consumer group's claim (the alternate hypothesis Ha) if the company’s claim (H0) was true is essentially 0. Clearly, this gives strong evidence in support of the alternate hypothesis (against the null hypothesis).
    34. 34. 34 Hypothesis Test Example 2 Single Proportion continued We would say that we have strong support for the claim that the proportion of the insurance company’s claims that are settled within 30 days is less than 0.9. Some people would state that we have shown that the true proportion of the insurance company’s claims that are settled within 30 days is statistically significantly less than 0.9.
    35. 35. 35 A county judge has agreed that he will give up his county judgeship and run for a state judgeship unless there is evidence at the 0.10 level that more then 25% of his party is in opposition. A SRS of 800 party members included 217 who opposed him. Please advise this judge. Hypothesis Test Example Single Proportion
    36. 36. 36 Hypothesis Test Example Single Proportion continued π = proportion of his party that is in opposition H0: π = 0.25 HA: π > 0.25 α = 0.10 Note: hypothesized value = 0.25 217 n 800, p 0.27125 800 = = = 0.27125 0.25 z 1.39 0.25(0.75) 800 − = =
    37. 37. 37 Hypothesis Test Example Single Proportion continued At a level of significance of 0.10, there is sufficient evidence to support the claim that the true percentage of the party members that oppose him is more than 25%. Under these circumstances, I would advise him not to run. P-value=P(z 1.39) 1 0.9177 0.0823> = − =
    38. 38. 38 1. Describe (determine) the population characteristic about which hypotheses are to be tested. 2. State the null hypothesis H0. 3. State the alternate hypothesis Ha. 4. Select the significance level α for the test. 5. Display the test statistic to be used, with substitution of the hypothesized value identified in step 2 but without any computation at this point. Steps in a Hypothesis-Testing Analysis
    39. 39. 39 Steps in a Hypothesis-Testing Analysis 6. Check to make sure that any assumptions required for the test are reasonable. 7. Compute all quantities appearing in the test statistic and then the value of the test statistic itself. 8. Determine the P-value associated with the observed value of the test statistic 9. State the conclusion in the context of the problem, including the level of significance.
    40. 40. 40 x hypothesized mean z n − = σ Hypothesis Test (Large samples) Single Sample Test of Population Mean In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H0: µ = hypothesized mean, compute the z statistic
    41. 41. 41 x hypothesized mean P-value P Z n   − = <  σ    Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ < hypothesized mean
    42. 42. 42 x hypothesized mean P-value P Z n   − = >  σ    Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ > hypothesized mean
    43. 43. 43 x hypothesized mean P-value 2P Z n   −  = > σ     Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ ≠ hypothesized mean
    44. 44. 44 It is not likely that one would know σ but not know µ, so calculating a z value using the formula would not be very realistic. x hypothesized mean z n − = σ Reality Check For large values of n (>30) it is generally acceptable to use s to estimate σ, however, it is much more common to apply the t-distribution.
    45. 45. 45 x hypothesized mean t s n − = Hypothesis Test (σ unknown) Single Sample Test of Population Mean The approximate P-value for this test is found using a t random variable with degrees of freedom df = n-1. The procedure is described in the next group of slides. To test the null hypothesis µ = hypothesized mean, when we may assume that the underlying distribution is normal or approximately normal, compute the t statistic
    46. 46. 46 x hypothesized mean P-value P t s n   − = <       Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ < hypothesized mean
    47. 47. 47 x hypothesized mean P-value P t s n   − = >       Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ > hypothesized mean
    48. 48. 48 x hypothesized mean P-value 2P t s n   −  = >      Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ ≠ hypothesized mean
    49. 49. 49 The t statistic can be used for all sample sizes, however, the smaller the sample, the more important the assumption that the underlying distribution is normal. Typically, when n >15 the underlying distribution need only be centrally weighted and may be somewhat skewed. Hypothesis Test (σ unknown) Single Sample Test of Population Mean
    50. 50. 50 Tail areas for t curves t df 1 2 3 4 5 6 7 8 9 10 11 12 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.468 0.465 0.463 0.463 0.462 0.462 0.462 0.461 0.461 0.461 0.461 0.461 0.2 0.437 0.430 0.427 0.426 0.425 0.424 0.424 0.423 0.423 0.423 0.423 0.422 0.3 0.407 0.396 0.392 0.390 0.388 0.387 0.386 0.386 0.385 0.385 0.385 0.385 0.4 0.379 0.364 0.358 0.355 0.353 0.352 0.351 0.350 0.349 0.349 0.348 0.348 0.5 0.352 0.333 0.326 0.322 0.319 0.317 0.316 0.315 0.315 0.314 0.313 0.313 0.6 0.328 0.305 0.295 0.290 0.287 0.285 0.284 0.283 0.282 0.281 0.280 0.280 0.7 0.306 0.278 0.267 0.261 0.258 0.255 0.253 0.252 0.251 0.250 0.249 0.249 0.8 0.285 0.254 0.241 0.234 0.230 0.227 0.225 0.223 0.222 0.221 0.220 0.220 0.9 0.267 0.232 0.217 0.210 0.205 0.201 0.199 0.197 0.196 0.195 0.194 0.193 1.0 0.250 0.211 0.196 0.187 0.182 0.178 0.175 0.173 0.172 0.170 0.169 0.169 1.1 0.235 0.193 0.176 0.167 0.161 0.157 0.154 0.152 0.150 0.149 0.147 0.146 1.2 0.221 0.177 0.158 0.148 0.142 0.138 0.135 0.132 0.130 0.129 0.128 0.127 1.3 0.209 0.162 0.142 0.132 0.125 0.121 0.117 0.115 0.113 0.111 0.110 0.109 1.4 0.197 0.148 0.128 0.117 0.110 0.106 0.102 0.100 0.098 0.096 0.095 0.093 1.5 0.187 0.136 0.115 0.104 0.097 0.092 0.089 0.086 0.084 0.082 0.081 0.080 1.6 0.178 0.125 0.104 0.092 0.085 0.080 0.077 0.074 0.072 0.070 0.069 0.068 1.7 0.169 0.116 0.094 0.082 0.075 0.070 0.066 0.064 0.062 0.060 0.059 0.057 1.8 0.161 0.107 0.085 0.073 0.066 0.061 0.057 0.055 0.053 0.051 0.050 0.049 1.9 0.154 0.099 0.077 0.065 0.058 0.053 0.050 0.047 0.045 0.043 0.042 0.041 2.0 0.148 0.092 0.070 0.058 0.051 0.046 0.043 0.040 0.038 0.037 0.035 0.034 2.1 0.141 0.085 0.063 0.052 0.045 0.040 0.037 0.034 0.033 0.031 0.030 0.029 2.2 0.136 0.079 0.058 0.046 0.040 0.035 0.032 0.029 0.028 0.026 0.025 0.024 2.3 0.131 0.074 0.052 0.041 0.035 0.031 0.027 0.025 0.023 0.022 0.021 0.020 2.4 0.126 0.069 0.048 0.037 0.031 0.027 0.024 0.022 0.020 0.019 0.018 0.017 2.5 0.121 0.065 0.044 0.033 0.027 0.023 0.020 0.018 0.017 0.016 0.015 0.014 2.6 0.117 0.061 0.040 0.030 0.024 0.020 0.018 0.016 0.014 0.013 0.012 0.012 2.7 0.113 0.057 0.037 0.027 0.021 0.018 0.015 0.014 0.012 0.011 0.010 0.010 2.8 0.109 0.054 0.034 0.024 0.019 0.016 0.013 0.012 0.010 0.009 0.009 0.008 2.9 0.106 0.051 0.031 0.022 0.017 0.014 0.011 0.010 0.009 0.008 0.007 0.007 3.0 0.102 0.048 0.029 0.020 0.015 0.012 0.010 0.009 0.007 0.007 0.006 0.006 3.1 0.099 0.045 0.027 0.018 0.013 0.011 0.009 0.007 0.006 0.006 0.005 0.005 3.2 0.096 0.043 0.025 0.016 0.012 0.009 0.008 0.006 0.005 0.005 0.004 0.004 3.3 0.094 0.040 0.023 0.015 0.011 0.008 0.007 0.005 0.005 0.004 0.004 0.003 3.4 0.091 0.038 0.021 0.014 0.010 0.007 0.006 0.005 0.004 0.003 0.003 0.003 3.5 0.089 0.036 0.020 0.012 0.009 0.006 0.005 0.004 0.003 0.003 0.002 0.002 3.6 0.086 0.035 0.018 0.011 0.008 0.006 0.004 0.003 0.003 0.002 0.002 0.002 3.7 0.084 0.033 0.017 0.010 0.007 0.005 0.004 0.003 0.002 0.002 0.002 0.002 3.8 0.082 0.031 0.016 0.010 0.006 0.004 0.003 0.003 0.002 0.002 0.001 0.001 3.9 0.080 0.030 0.015 0.009 0.006 0.004 0.003 0.002 0.002 0.001 0.001 0.001 4.0 0.078 0.029 0.014 0.008 0.005 0.004 0.003 0.002 0.002 0.001 0.001 0.001
    51. 51. 51 t df 25 26 27 28 29 30 35 40 60 120 ∞(=z) 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.461 0.461 0.461 0.461 0.461 0.461 0.460 0.460 0.460 0.460 0.460 0.2 0.422 0.422 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.3 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.382 0.382 0.4 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.345 0.345 0.345 0.5 0.311 0.311 0.311 0.310 0.310 0.310 0.310 0.310 0.309 0.309 0.309 0.6 0.277 0.277 0.277 0.277 0.277 0.277 0.276 0.276 0.275 0.275 0.274 0.7 0.245 0.245 0.245 0.245 0.245 0.245 0.244 0.244 0.243 0.243 0.242 0.8 0.216 0.215 0.215 0.215 0.215 0.215 0.215 0.214 0.213 0.213 0.212 0.9 0.188 0.188 0.188 0.188 0.188 0.188 0.187 0.187 0.186 0.185 0.184 1.0 0.163 0.163 0.163 0.163 0.163 0.163 0.162 0.162 0.161 0.160 0.159 1.1 0.141 0.141 0.141 0.140 0.140 0.140 0.139 0.139 0.138 0.137 0.136 1.2 0.121 0.120 0.120 0.120 0.120 0.120 0.119 0.119 0.117 0.116 0.115 1.3 0.103 0.103 0.102 0.102 0.102 0.102 0.101 0.101 0.099 0.098 0.097 1.4 0.087 0.087 0.086 0.086 0.086 0.086 0.085 0.085 0.083 0.082 0.081 1.5 0.073 0.073 0.073 0.072 0.072 0.072 0.071 0.071 0.069 0.068 0.067 1.6 0.061 0.061 0.061 0.060 0.060 0.060 0.059 0.059 0.057 0.056 0.055 1.7 0.051 0.051 0.050 0.050 0.050 0.050 0.049 0.048 0.047 0.046 0.045 1.8 0.042 0.042 0.042 0.041 0.041 0.041 0.040 0.040 0.038 0.037 0.036 1.9 0.035 0.034 0.034 0.034 0.034 0.034 0.033 0.032 0.031 0.030 0.029 2.0 0.028 0.028 0.028 0.028 0.027 0.027 0.027 0.026 0.025 0.024 0.023 2.1 0.023 0.023 0.023 0.022 0.022 0.022 0.022 0.021 0.020 0.019 0.018 2.2 0.019 0.018 0.018 0.018 0.018 0.018 0.017 0.017 0.016 0.015 0.014 2.3 0.015 0.015 0.015 0.015 0.014 0.014 0.014 0.013 0.012 0.012 0.011 2.4 0.012 0.012 0.012 0.012 0.012 0.011 0.011 0.011 0.010 0.009 0.008 2.5 0.010 0.010 0.009 0.009 0.009 0.009 0.009 0.008 0.008 0.007 0.006 2.6 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.006 0.005 0.005 2.7 0.006 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.004 0.004 0.003 2.8 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.003 0.003 0.003 2.9 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.002 0.002 3.0 0.003 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.001 3.1 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 3.2 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 3.3 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 3.4 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 3.5 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 3.6 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 3.7 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 3.8 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 3.9 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 4.0 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 Tail areas for t curves
    52. 52. 52 An manufacturer of a special bolt requires that this type of bolt have a mean shearing strength in excess of 110 lb. To determine if the manufacturer’s bolts meet the required standards a sample of 25 bolts was obtained and tested. The sample mean was 112.7 lb and the sample standard deviation was 9.62 lb. Use this information to perform an appropriate hypothesis test with a significance level of 0.05. Example of Hypothesis Test Single Sample Test of Population Mean - continued
    53. 53. 53 µ = the mean shearing strength of this specific type of bolt Τhe hypotheses to be tested are H0: µ = 110 lb Ha: µ > 110 lb The significance level to be used for the test is α = 0.05. Example of Hypothesis Test Single Sample Test of Population Mean - continued x 110 t s n − =The test statistic is
    54. 54. 54 Example of Hypothesis Test Single Sample Test of Population Mean - continued x 112.7, s 9.62, n 25, df 24= = = = 112.7 110 P-value P t 9.62 25 P(t 1.4) 0.087   − = >       = > =
    55. 55. 55 Because P-value = 0.087 > 0.05 = α, we fail to reject H0. At a level of significance of 0.05, there is insufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 110 lbs. Example of Hypothesis Test Single Sample Test of Population Mean - conclusion
    56. 56. 56 Using the t table t df 13 14 15 16 17 18 19 20 21 22 23 24 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.2 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.3 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.383 0.383 0.383 0.4 0.348 0.348 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.346 0.346 0.5 0.313 0.312 0.312 0.312 0.312 0.312 0.311 0.311 0.311 0.311 0.311 0.311 0.6 0.279 0.279 0.279 0.278 0.278 0.278 0.278 0.278 0.277 0.277 0.277 0.277 0.7 0.248 0.248 0.247 0.247 0.247 0.246 0.246 0.246 0.246 0.246 0.245 0.245 0.8 0.219 0.219 0.218 0.218 0.217 0.217 0.217 0.217 0.216 0.216 0.216 0.216 0.9 0.192 0.192 0.191 0.191 0.190 0.190 0.190 0.189 0.189 0.189 0.189 0.189 1.0 0.168 0.167 0.167 0.166 0.166 0.165 0.165 0.165 0.164 0.164 0.164 0.164 1.1 0.146 0.145 0.144 0.144 0.143 0.143 0.143 0.142 0.142 0.142 0.141 0.141 1.2 0.126 0.125 0.124 0.124 0.123 0.123 0.122 0.122 0.122 0.121 0.121 0.121 1.3 0.108 0.107 0.107 0.106 0.105 0.105 0.105 0.104 0.104 0.104 0.103 0.103 1.4 0.092 0.092 0.091 0.090 0.090 0.089 0.089 0.088 0.088 0.088 0.087 0.087 1.5 0.079 0.078 0.077 0.077 0.076 0.075 0.075 0.075 0.074 0.074 0.074 0.073 1.6 0.067 0.066 0.065 0.065 0.064 0.064 0.063 0.063 0.062 0.062 0.062 0.061 1.7 0.056 0.056 0.055 0.054 0.054 0.053 0.053 0.052 0.052 0.052 0.051 0.051 1.8 0.048 0.047 0.046 0.045 0.045 0.044 0.044 0.043 0.043 0.043 0.042 0.042 1.9 0.040 0.039 0.038 0.038 0.037 0.037 0.036 0.036 0.036 0.035 0.035 0.035 2.0 0.033 0.033 0.032 0.031 0.031 0.030 0.030 0.030 0.029 0.029 0.029 0.028 2.1 0.028 0.027 0.027 0.026 0.025 0.025 0.025 0.024 0.024 0.024 0.023 0.023 2.2 0.023 0.023 0.022 0.021 0.021 0.021 0.020 0.020 0.020 0.019 0.019 0.019 2.3 0.019 0.019 0.018 0.018 0.017 0.017 0.016 0.016 0.016 0.016 0.015 0.015 2.4 0.016 0.015 0.015 0.014 0.014 0.014 0.013 0.013 0.013 0.013 0.012 0.012 2.5 0.013 0.013 0.012 0.012 0.011 0.011 0.011 0.011 0.010 0.010 0.010 0.010 2.6 0.011 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008 0.008 0.008 0.008 2.7 0.009 0.009 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.006 2.8 0.008 0.007 0.007 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 2.9 0.006 0.006 0.005 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.004 3.0 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 3.1 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.002 3.2 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 3.3 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 3.4 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 3.5 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.6 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.7 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.8 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 3.9 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 4.0 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 t = 1.4 n = 25 df = 24 Tail area = 0.087
    57. 57. 57 Revisit the problem with α=0.10 What would happen if the significance level of the test was 0.10 instead of 0.05? At the 0.10 level of significance there is sufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 120 lbs. Now P-value = 0.087 < 0.10 = α, and we reject H0 at the 0.10 level of significance and conclude
    58. 58. 58 Comments continued Many people are bothered by the fact that different choices of α lead to different conclusions. This is nature of a process where you control the probability of being wrong when you select the level of significance. This reflects your willingness to accept a certain level of type I error.
    59. 59. 59 Another Example A jeweler is planning on manufacturing gold charms. His design calls for a particular piece to contain 0.08 ounces of gold. The jeweler would like to know if the pieces that he makes contain (on the average) 0.08 ounces of gold. To test to see if the pieces contain 0.08 ounces of gold, he made a sample of 16 of these particular pieces and obtained the following data. 0.0773 0.0779 0.0756 0.0792 0.0777 0.0713 0.0818 0.0802 0.0802 0.0785 0.0764 0.0806 0.0786 0.0776 0.0793 0.0755 Use a level of significance of 0.01 to perform an appropriate hypothesis test.
    60. 60. 60 Another Example 1. The population characteristic being studied is µ = true mean gold content for this particular type of charm. 2. Null hypothesis: H0:µ = 0.08 oz 3. Alternate hypothesis: Ha:µ ≠ 0.08 oz 4. Significance level: α = 0.01 5. Test statistic: x hypothesized mean x 0.08 t s s n n − − = =
    61. 61. 61 Another Example 6. Minitab was used to create a normal plot along with a graphical display of the descriptive statistics for the sample data. The result of this display is that it is reasonable to assume that the population of gold contents of this type of charm is normally distributed P-Value: 0.396 A-Squared: 0.363 Anderson-Darling Normality Test N: 16 StDev: 0.0025143 Average: 0.0779813 0.0820.0770.072 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Gold Normal Probability Plot
    62. 62. 62 Another Example We can see that with the exception of one outlier, the data is reasonably symmetric and mound shaped in shape, indicating that the assumption that the population of amounts of gold for this particular charm can reasonably be expected to be normally distributed. 0.0820.0800.0780.0760.0740.072 95% Confidence Interval for Mu 0.07950.07850.07750.0765 95% Confidence Interval for Median Variable: Gold 7.71E-02 1.86E-03 7.66E-02 Maximum 3rd Quartile Median 1st Quartile Minimum N Kurtosis Skewness Variance StDev Mean P-Value: A-Squared: 7.95E-02 3.89E-03 7.93E-02 8.18E-02 8.00E-02 7.82E-02 7.66E-02 7.13E-02 16 2.23191 -1.10922 6.32E-06 2.51E-03 7.80E-02 0.396 0.363 95% Confidence Interval for Median 95% Confidence Interval for Sigma 95% Confidence Interval for Mu Anderson-Darling Normality Test Descriptive Statistics
    63. 63. 63 Another Example 8. P-value: This is a two tailed test. Looking up in the table of tail areas for t curves, t = 3.2 with df = 15 we see the table entry is 0.003 so P-Value = 2(0.003) = 0.006 n 16, x 0.077981, s 0.0025143 0.077981 0.08 t 3.2 0.0025143 16 = = = − = = − 7. Computations: n 16, x 0.077981, s 0.0025143 0.077981 0.08 t 3.2 0.0025143 16 = = = − = = − 7. Computations:
    64. 64. 64 Another Example 9. Conclusion: Since P-value = 0.006 ≤ 0.01 = α, we reject H0 at the 0.01 level of significance. At the 0.01 level of significance there is convincing evidence that the true mean gold content of this type of charm is not 0.08 ounces. Actually when rejecting a null hypothesis for the ≠ alternative, a one tailed claim is supported. In this case, at the 0.01 level of significance, there is convincing evidence that the true mean gold content of this type of charm is less than 0.08 ounces.
    65. 65. 65 Power and Probability of Type II Error The power of a test is the probability of rejecting the null hypothesis. When H0 is false, the power is the probability that the null hypothesis is rejected. Specifically, power = 1 – β.
    66. 66. 66 Effects of Various Factors on Power 1. The larger the size of the discrepancy between the hypothesized value and the true value of the population characteristic, the higher the power. 2. The larger the significance level, α, the higher the power of the test. 3. The larger the sample size, the higher the power of the test.
    67. 67. 67 Some Comments Calculating β (hence power) depends on knowing the true value of the population characteristic being tested. Since the true value is not known, generally, one calculates β for a number of possible “true” values of the characteristic under study and then sketches a power curve.
    68. 68. 68 Example (based on z-curve) Consider the earlier example where we tested H0: µ = 110 vs. Ha: µ > 110 and furthermore, suppose the true standard deviation of the bolts was actually 10 lbs.
    69. 69. 69 Example (based on z-curve) Power Curves Different α's 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 118 119 120 121 122 123 124 125 126 127 128 True Value of µ Power(1-β) α = 0.10 α = 0.05 α = 0.01 H0: µ = 120 Ha: µ > 120 σ = 10
    70. 70. 70 Example (based on z-curve) Power Curves Different n's 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 118 119 120 121 122 123 124 125 126 127 128 True Value of µ Power(1-β) n = 45 n = 90 n = 180 n = 360 H0: µ = 120 Ha: µ > 120 σ = 10

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