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Hypothesis TestsUsing a single Sample
10.1: Hypotheses & Test Procedures
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Basics: In statistics, a hypothesis is a statement about a population characteristic.
FORMAL STRUCTURE
4
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.
FORMAL STRUCTURE
5
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.
FORMAL STRUCTURE
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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.
FORMAL STRUCTURE
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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.
FORMAL STRUCTURE
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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 hypothesisThis is not the same as saying that the null hypothesis is true.
AN ANALOGY
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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)
AN ANALOGY
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Two Hypotheses are then created.
H0: Innocent
Ha: Not Innocent (Guilt)
Examples of Hypotheses
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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)
Examples of Hypotheses
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The students entering into the math program used to have a mean SAT quantitative score of 525. Are the current students scoring less (as measured by the SAT quantitative score)?
H0: µ = 525
(Really: µ Ha: µ < 525 (One-sided alternative)
Examples of Hypotheses
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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: ≥
Ha: µ< 16•Notice, the real concern would be selling the consumer less than 16 ounces of peaches.
Examples of Hypotheses
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Is the proportion of defective parts produced by a manufacturing process more than 5%?
H0: = 0.05
(Really,
Ha: > 0.05
Examples of Hypotheses
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Do two brands of light bulb have the same mean lifetime?
H0: µBrand A = µBrand B
Ha: µBrand A µBrand B
Examples of Hypotheses
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Do parts produced by two different milling machines have the same variability in diameters?
or equivalently
0 1 2
a 1 2
H :
H :
2 2
0 1 2
2 2
a 1 2
H :
H :
Comments on Hypothesis Form
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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.
Hypothesis Form
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The form of the null hypothesis isH0: population characteristic = hypothesized value H0: population characteristic hypothesized value 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 valueHa: population characteristic < hypothesized valueHa: population characteristic hypothesized value
Caution
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When you set up a hypothesis test, the result is either1. 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.
10.2: Errors in Hypothesis Testing
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Definitions: Type I error: The error of rejecting Ho when Ho is
actually true. Type II error: The error of failing to reject Ho when
Ho is actually false.
Error
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Null Hypothesis
Decision True False
Accept H0
Reject H0
No Error
No ErrorType I Error
Type II Error
Error Analogy
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Consider a medical test where the hypotheses are equivalent toH0: the patient has a specific diseaseHa: 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.)
More on Error
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The probability of a type I error is denoted by and is called the level of significance of the test.
Thus, a test with = 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 .
Relationships Between and
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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.
Comment on Process
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So why not just set =.01
Look at the consequences of type I and type II errors and then identify the largest that is tolerable for the problem.
Employ a test procedure that uses this maximum acceptable value of (rather than anything smaller) as the level of significance (because using a smaller increases ).
10.3: Test Statistic
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A test statistic is the function of sample data on which a conclusion to reject or fail to reject H0 is based.
P-value
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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.
Decision Criteria
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A decision as to whether H0 should be rejected results from comparing the P-value to the chosen :
H0 should be rejected if P-value
H0 should not be rejected if P-value >
Large Sample Hypothesis Test for a Single Proportion
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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: = hypothesized proportion,
compute the z statistic
p hypothesized valuez
hypothesized value(1-hypothesized value)n
Hypothesis Test Large Sample Test of Population Proportion
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p hypothesized valueP-value P z
hypothesized value(1-hypothesized value)n
Hypothesis Test Large Sample Test of Population Proportion
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p hypothesized valueP-value P z
hypothesized value(1-hypothesized value)
n
Hypothesis Test Large Sample Test of Population Proportion
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p hypothesized valueP-value 2P z
hypothesized value(1-hypothesized value)
n
Hypothesis Test Example
Large-Sample Test for a Population Proportion
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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 ExampleSingle Proportion continued
34
P-value P(z 4.71) 0
0.8 0.9 0.8 0.9z 4.71
(0.9)(1 0.9) 0.9(0.1)200 200
= proportion of the company’s claims that are settled within 30 days
H0: = 0.9 or really 0.9HA: 0.9
160p 0.8
200 The sample proportion is
Hypothesis Test ExampleSingle Proportion continued
35
The probability of getting the company’s claim (Ho) has a probability of being true essentially 0% of the time. This result is strongly in favor of the consumer group's claim that they settle a smaller percent of claims within 30 days (the alternate hypothesis Ha).
Hypothesis Test ExampleSingle Proportion continued
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This means 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 the 0.9 that the company indicated. So we reject the null and accept the alternative.
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 the companies claimed value of 0.9.
Hypothesis Test Example Single Proportion
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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 than 25% of his party is in opposition. A SRS of 800 party members included 217 who opposed him. Please advise this judge.
Hypothesis Test ExampleSingle Proportion continued
38
= proportion of his party that is in opposition
H0: = 0.25
HA: > 0.25
= 0.10
Note: hypothesized value = 0.25
217n 800, p 0.27125
800
0.27125 0.25z 1.39
0.25(0.75)800
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
39
Steps in a Hypothesis-Testing Analysis
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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 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
41
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.
10.4: Hypothesis Test (Large samples)Single Sample Test of Population Mean
42
x hypothesized meanz
n
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
Hypothesis Test ( unknown)
Single Sample Test of Population Mean
47
x hypothesized meant
sn
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
Hypothesis Test (
unknown) Single Sample Test of Population Mean
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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.
Tail areas for t curves
52
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 0.5000.1 0.468 0.465 0.463 0.463 0.462 0.462 0.462 0.461 0.461 0.461 0.461 0.4610.2 0.437 0.430 0.427 0.426 0.425 0.424 0.424 0.423 0.423 0.423 0.423 0.4220.3 0.407 0.396 0.392 0.390 0.388 0.387 0.386 0.386 0.385 0.385 0.385 0.3850.4 0.379 0.364 0.358 0.355 0.353 0.352 0.351 0.350 0.349 0.349 0.348 0.3480.5 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 0.2800.7 0.306 0.278 0.267 0.261 0.258 0.255 0.253 0.252 0.251 0.250 0.249 0.2490.8 0.285 0.254 0.241 0.234 0.230 0.227 0.225 0.223 0.222 0.221 0.220 0.2200.9 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 0.1691.1 0.235 0.193 0.176 0.167 0.161 0.157 0.154 0.152 0.150 0.149 0.147 0.1461.2 0.221 0.177 0.158 0.148 0.142 0.138 0.135 0.132 0.130 0.129 0.128 0.1271.3 0.209 0.162 0.142 0.132 0.125 0.121 0.117 0.115 0.113 0.111 0.110 0.1091.4 0.197 0.148 0.128 0.117 0.110 0.106 0.102 0.100 0.098 0.096 0.095 0.0931.5 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 0.0681.7 0.169 0.116 0.094 0.082 0.075 0.070 0.066 0.064 0.062 0.060 0.059 0.0571.8 0.161 0.107 0.085 0.073 0.066 0.061 0.057 0.055 0.053 0.051 0.050 0.0491.9 0.154 0.099 0.077 0.065 0.058 0.053 0.050 0.047 0.045 0.043 0.042 0.0412.0 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 0.0292.2 0.136 0.079 0.058 0.046 0.040 0.035 0.032 0.029 0.028 0.026 0.025 0.0242.3 0.131 0.074 0.052 0.041 0.035 0.031 0.027 0.025 0.023 0.022 0.021 0.0202.4 0.126 0.069 0.048 0.037 0.031 0.027 0.024 0.022 0.020 0.019 0.018 0.0172.5 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 0.0122.7 0.113 0.057 0.037 0.027 0.021 0.018 0.015 0.014 0.012 0.011 0.010 0.0102.8 0.109 0.054 0.034 0.024 0.019 0.016 0.013 0.012 0.010 0.009 0.009 0.0082.9 0.106 0.051 0.031 0.022 0.017 0.014 0.011 0.010 0.009 0.008 0.007 0.0073.0 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 0.0053.2 0.096 0.043 0.025 0.016 0.012 0.009 0.008 0.006 0.005 0.005 0.004 0.0043.3 0.094 0.040 0.023 0.015 0.011 0.008 0.007 0.005 0.005 0.004 0.004 0.0033.4 0.091 0.038 0.021 0.014 0.010 0.007 0.006 0.005 0.004 0.003 0.003 0.0033.5 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 0.0023.7 0.084 0.033 0.017 0.010 0.007 0.005 0.004 0.003 0.002 0.002 0.002 0.0023.8 0.082 0.031 0.016 0.010 0.006 0.004 0.003 0.003 0.002 0.002 0.001 0.0013.9 0.080 0.030 0.015 0.009 0.006 0.004 0.003 0.002 0.002 0.001 0.001 0.0014.0 0.078 0.029 0.014 0.008 0.005 0.004 0.003 0.002 0.002 0.001 0.001 0.001
Tail areas for t curves
53
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 0.5000.1 0.461 0.461 0.461 0.461 0.461 0.461 0.460 0.460 0.460 0.460 0.4600.2 0.422 0.422 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.4210.3 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.382 0.3820.4 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.345 0.345 0.3450.5 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 0.2740.7 0.245 0.245 0.245 0.245 0.245 0.245 0.244 0.244 0.243 0.243 0.2420.8 0.216 0.215 0.215 0.215 0.215 0.215 0.215 0.214 0.213 0.213 0.2120.9 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 0.1591.1 0.141 0.141 0.141 0.140 0.140 0.140 0.139 0.139 0.138 0.137 0.1361.2 0.121 0.120 0.120 0.120 0.120 0.120 0.119 0.119 0.117 0.116 0.1151.3 0.103 0.103 0.102 0.102 0.102 0.102 0.101 0.101 0.099 0.098 0.0971.4 0.087 0.087 0.086 0.086 0.086 0.086 0.085 0.085 0.083 0.082 0.0811.5 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 0.0551.7 0.051 0.051 0.050 0.050 0.050 0.050 0.049 0.048 0.047 0.046 0.0451.8 0.042 0.042 0.042 0.041 0.041 0.041 0.040 0.040 0.038 0.037 0.0361.9 0.035 0.034 0.034 0.034 0.034 0.034 0.033 0.032 0.031 0.030 0.0292.0 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 0.0182.2 0.019 0.018 0.018 0.018 0.018 0.018 0.017 0.017 0.016 0.015 0.0142.3 0.015 0.015 0.015 0.015 0.014 0.014 0.014 0.013 0.012 0.012 0.0112.4 0.012 0.012 0.012 0.012 0.012 0.011 0.011 0.011 0.010 0.009 0.0082.5 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 0.0052.7 0.006 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.004 0.004 0.0032.8 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.003 0.003 0.0032.9 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.002 0.0023.0 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 0.0013.2 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.0013.3 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.0003.4 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.0003.5 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 0.0003.7 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0003.8 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0003.9 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0004.0 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
Example of Hypothesis TestSingle Sample Test of Population Mean
54
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 TestSingle Sample Test of Population Mean continued
55
= the mean shearing strength of this specific type of bolt
he hypotheses to be tested are
H0: =110 lb
Ha: 110 lb
The significance level to be used for the test is = 0.05.
x 110t
sn
The test statistic is
Example of Hypothesis TestSingle Sample Test of Population Mean
continued
56
x 112.7, s 9.62, n 25, df 24
112.7 110P-value P t
9.6225
P(t 1.4) 0.087
Example of Hypothesis TestSingle Sample Test of Population Mean
conclusion
57
Because P-value = 0.087 > 0.05 = 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.
Looking at the t table in the back of the book, it is for 2 sided test so use .90 instead of .95 & our crit t for 24 df = 1.71: again we fail to reject
Using the t table
58
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 0.5000.1 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.4610.2 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.4220.3 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.383 0.383 0.3830.4 0.348 0.348 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.346 0.3460.5 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 0.2770.7 0.248 0.248 0.247 0.247 0.247 0.246 0.246 0.246 0.246 0.246 0.245 0.2450.8 0.219 0.219 0.218 0.218 0.217 0.217 0.217 0.217 0.216 0.216 0.216 0.2160.9 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 0.1641.1 0.146 0.145 0.144 0.144 0.143 0.143 0.143 0.142 0.142 0.142 0.141 0.1411.2 0.126 0.125 0.124 0.124 0.123 0.123 0.122 0.122 0.122 0.121 0.121 0.1211.3 0.108 0.107 0.107 0.106 0.105 0.105 0.105 0.104 0.104 0.104 0.103 0.1031.4 0.092 0.092 0.091 0.090 0.090 0.089 0.089 0.088 0.088 0.088 0.087 0.0871.5 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 0.0611.7 0.056 0.056 0.055 0.054 0.054 0.053 0.053 0.052 0.052 0.052 0.051 0.0511.8 0.048 0.047 0.046 0.045 0.045 0.044 0.044 0.043 0.043 0.043 0.042 0.0421.9 0.040 0.039 0.038 0.038 0.037 0.037 0.036 0.036 0.036 0.035 0.035 0.0352.0 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 0.0232.2 0.023 0.023 0.022 0.021 0.021 0.021 0.020 0.020 0.020 0.019 0.019 0.0192.3 0.019 0.019 0.018 0.018 0.017 0.017 0.016 0.016 0.016 0.016 0.015 0.0152.4 0.016 0.015 0.015 0.014 0.014 0.014 0.013 0.013 0.013 0.013 0.012 0.0122.5 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 0.0082.7 0.009 0.009 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.0062.8 0.008 0.007 0.007 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.0052.9 0.006 0.006 0.005 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.0043.0 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 0.0023.2 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.0023.3 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.0023.4 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.0013.5 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 0.0013.7 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.0013.8 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.0003.9 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.0004.0 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
Revisit the problem with =0.10
What would happen if the significance level of the test was 0.10 instead of 0.05?
59
At the 0.10 level of significance there is sufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 110 lbs. Looking at the t table in the back of the book, it is for 2 sided test so use .80 instead of .90 & our crit t for 24 df = 1.32: so we reject
Now P-value = 0.087 < 0.10 = and we reject H0 at the 0.10 level of significance and conclude
Comments continued
60
Many people are bothered by the fact that different choices of 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.
Another Example
61
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 0.07770.0713 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.
Another Example
62
6. Minitab was used to create a normal plot along with a graphical display of the descriptive statistics for the sample data.
P-Value: 0.396A-Squared: 0.363
Anderson-Darling Normality Test
N: 16StDev: 0.0025143Average: 0.0779813
0.0820.0770.072
.999
.99
.95
.80
.50
.20
.05
.01
.001
Pro
bab
ility
Gold
Normal Probability Plot
Another Example
63
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
Maximum3rd QuartileMedian1st QuartileMinimum
NKurtosisSkewnessVarianceStDevMean
P-Value:A-Squared:
7.95E-02
3.89E-03
7.93E-02
8.18E-028.00E-027.82E-027.66E-027.13E-02
162.23191
-1.109226.32E-062.51E-037.80E-02
0.3960.363
95% Confidence Interval for Median
95% Confidence Interval for Sigma
95% Confidence Interval for Mu
Anderson-Darling Normality Test
Descriptive Statistics
Another Example
64
1. The population characteristic being studied is = 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: = 0.01
5. Test statistic:
x hypothesized mean x 0.08t
s sn n
Another Example
65
n 16, x 0.077981, s 0.0025143
0.077981 0.08t 3.2
0.002514316
7. Computations:
n 16, x 0.077981, s 0.0025143
0.077981 0.08t 3.2
0.002514316
7. Computations:
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
Crit t with 15 df & an of .01 = 2.95
Another Example
66
9. Conclusion: Since P-value = 0.006 0.01 = , (or Crit t with
15 df & an of .01 = 2.95) 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.
10.5: Power and Probability of Type II Error
67
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 – .
Effects of Various Factors on Power
68
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, , the higher the power of the test.
3. The larger the sample size, the higher the power of the test.
Some Comments
69
Calculating (hence power) depends on knowing the true value of the population characteristic being tested. Since the true value is not known, generally, one calculates for a number of possible “true” values of the characteristic under study and then sketches a power curve.
Example (based on z-curve)
72
Power CurvesDifferent 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
Po
we
r (1
-
)
n = 45
n = 90
n = 180
n = 360
H0: = 120
Ha: > 120 = 10
Proportion Example
73
The city council is concerned about a trend where apartment owners won’t rent to people with children. They selected a random sample of 125 apartments buildings & determined if children were allowed. Let be the true proportion of apartment buildings that prohibit children. If exceeds .75, city council will consider legislation.
A. If 102 of the 125 sampled exclude kids, would a level .05 test lead to legislation? B. What is the power of the test when = .8 and = .05?
Proportion Example
74
Let be the true proportion of apartments which prohibit children.
Ho: = 0.75 Ha: > 0.75 = 0.05 p = 102/125 = 0.816Since n = 125(0.816) = 102 10, and n(1 ) = 125(0.184)
= 23 10, the large sample z test for may be used.z = .816-.75/(.75)(.25)/125) = 1.71 P-value = area under the z curve to the right of 1.71 = 1
0.9564 = 0.0436.Since the P-value is less than , Ho is rejected. This 0.05 level
test does lead to the conclusion that more than 75% of the apartments exclude children.
10.6: Communicating & Interpreting Results of Analysis
76
Hypotheses – Important to state them whether in symbolic or sentence form.
Test Procedure – Be clear about what test you use & verify any assumptions you made (normal, etc.)
Test Statistic- Include the value of the statistic & the p value.
Conclusions in context – Don’t just say that you rejected, put it in the context of the problem.
value – whether you state it at the beginning or in conclusions, you must state your level.
Look For's in Published Data
77
What hypotheses were being tested & what characteristic(s) was(were) being looked at.
Was the appropriate test used? What were the assumptions & were they met?
What was the p – value associated with the test & what was the ? Was reasonable?
Were conclusions drawn consistent with the results of the hypothesis test?
Cautions
78
Remember, a hypothesis test can NEVER show support of any hypothesis, only support against.
That’s why we use a null hypothesis
If you can collect data on the entire population, don’t do a sample, use the population data.
There is a difference in statistical & practical/clinical significance.
Your test might show stat sig differ, but lowering systolic BP by 2 pts is not clinically significant
