155S8.3_3 Testing a Claim about a Proportion
November 04, 2011
MAT 155 Statistical Analysis
Dr. Claude Moore
Cape Fear Community College
Chapter 8
Hypothesis Testing
8­1 Review and Preview
8­2 Basics of Hypothesis Testing
8­3 Testing a Claim about a Proportion
8­4 Testing a Claim About a Mean: σ Known
8­5 Testing a Claim About a Mean: σ Not Known
8­6 Testing a Claim About a Standard Deviation or Variance
See the following lesson and document in Course Documents of CourseCompass:
S4.D2.Chapter 8 Hypothesis Testing
155 Chapter 8 Lesson ( Package file )
These notes cover the following topics: Basics of Hypothesis Testing; Testing Claim about Proportion; Testing Claim of Mean, σ known; Testing Claim of Mean, σ unknown; Testing Claim about Standard Deviation
S4.D3.P­value, find with calculator
Find P­Value ( 145920 Bytes )
This document shows screen shots and instructions for finding the P­value for the t­test using the TI calculator. You can use the normalcdf function to find the P­value for the z­test.
Click on the globe on the left to see tutorial for finding P­value for the z­distribution using the TI: http://cfcc.edu/faculty/cmoore/TI_P_Value_zTest.pdf
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Click on the globe on the left to see tutorial for finding P­value for the t­distribution using the TI: http://cfcc.edu/faculty/cmoore/TI_P_Value_tTest.pdf
Key Concept
This section presents complete procedures for testing a hypothesis (or claim) made about a population proportion. This section uses the components introduced in the previous section for the P­value method, the traditional method or the use of confidence intervals.
Two common methods for testing a claim about a population proportion are (1) to use a normal distribution as an approximation to the binomial distribution, and (2) to use an exact method based on the binomial probability distribution. Part 1 of this section uses the approximate method with the normal distribution, and Part 2 of this section briefly describes the exact method. Part 1:
Basic Methods of Testing Claims about a Population Proportion p
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155S8.3_3 Testing a Claim about a Proportion
Notation
n = number of trials
∧
p = population proportion (used in the null hypothesis)
q = 1 – p
Test Statistic for Testing a Claim About a Proportion
November 04, 2011
Requirements for Testing Claims About a Population Proportion p
1) The sample observations are a simple random sample.
2) The conditions for a binomial distribution are satisfied.
3) The conditions np ≥ 5 and nq ≥ 5 are both satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with µ = np and
σ = npq . Note: p is the assumed proportion not the sample proportion.
Caution
Don’t confuse a P­value with a proportion p.
P­values:
Use the standard normal distribution (Table A­2) and refer to Figure 8­5
P­value = probability of getting a test statistic at least as extreme as the one representing sample data
p = population proportion
Use the standard normal Critical Values: distribution (Table A­2).
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CAUTION
P­Value Method:
Use the same method as described in Section 8­2 and in Figure 8­8.
Use the standard normal distribution (Table A­2).
Traditional Method
Use the same method as described in Section 8­2 and in Figure 8­9.
Confidence Interval Method
Use the same method as described in Section 8­2 and in Table 8­2.
When testing claims about a population proportion, the traditional method and the P­value method are equivalent and will yield the same result since they use the same standard deviation based on the claimed proportion p. However, the confidence interval uses an estimated standard ∧
deviation based upon the sample proportion p. Consequently, it is possible that the traditional and P­value methods may yield a different conclusion than the confidence interval method.
A good strategy is to use a confidence interval to estimate a population proportion, but use the P­value or traditional method for testing a claim about the proportion.
Example:
Example:
The text refers to a study in which 57 out of 104 pregnant women correctly guessed the sex of their babies. Use these sample data to test the claim that the success rate of such guesses is no different from the 50% success rate expected with random chance guesses. Use a 0.05 significance level.
Requirements are satisfied: simple random sample; fixed number of trials (104) with two categories (guess correctly or do not); np = (104)(0.5) = 52 ≥ 5 and nq = (104)(0.5) = 52 ≥ 5
Step 1: original claim is that the success rate is no different from 50%: p = 0.50
Step 2: opposite of original claim is p ≠ 0.50
Step 3: p ≠ 0.50 does not contain equality so it is H1.
H0: p = 0.50 null hypothesis and original claim
H1: p ≠ 0.50 alternative hypothesis
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Example:
Example:
Step 4: significance level is α = 0.50
Table A­2: z = 0.98 has an area of 0.8365 to its left, so area to the right is 1 – 0.8365 = 0.1635, doubles yields 0.3270 (technology provides a more accurate P­value of 0.3268
Step 5: sample involves proportion so the relevant statistic is the sample proportion, Step 7: the P­value of 0.3270 is greater than the significance level of 0.50, so fail to reject the null hypothesis
Step 6: calculate z:
For a two­tailed test, P­value is twice the area to the right of test statistic.
∧
Obtaining p
∧
p sometimes is given directly
“10% of the observed sports cars are ∧
red” is expressed as p = 0.10
Here is the correct conclusion: There is not sufficient evidence to warrant rejection of the claim that women who guess the sex of their babies have a success rate equal to 50%. Part 2:
Exact Method for Testing Claims about a Proportion p
∧
p sometimes must be calculated
“96 surveyed households have cable TV
and 54 do not” is calculated using
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Testing Claims
Testing Claims
We can get exact results by using the binomial probability distribution. Binomial probabilities are a nuisance to calculate manually, but technology makes this approach quite simple. Also, this exact approach does not require that np ≥ 5 and nq ≥ 5 so we have a method that applies when that requirement is not satisfied. To test hypotheses using the exact binomial distribution, use the binomial probability distribution with the P­value method, use the value of p assumed in the null hypothesis, and find P­values as follows:
Left­tailed test:
The P­value is the probability of getting x or fewer successes among n trials.
Testing Claims
Recap
Right­tailed test:
The P­value is the probability of getting x or more successes among n trials.
Two­tailed test:
the P­value is twice the probability of getting x or more successes
the P­value is twice the probability of getting x or fewer successes
In this section we have discussed:
• Test statistics for claims about a proportion.
• P­value method.
• Confidence interval method.
• Obtaining p.
Click on the globe on the left to see tutorial for finding P­value for the z­distribution using the TI: http://cfcc.edu/faculty/cmoore/TI_P_Value_zTest.pdf
Click on the globe on the left to see tutorial for finding P­value for the t­distribution using the TI: http://cfcc.edu/faculty/cmoore/TI_P_Value_tTest.pdf
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427/10. Voting for the Winner In a presidential election, 308 out of 611 voters surveyed said that they voted for the candidate who won (based on data from ICR Survey Research Group). Use a 0.01 significance level to test the claim that among all voters, the percentage who believe that they voted for the winning candidate is equal to 43%, which is the actual percentage of votes for the winning candidate. What does the result suggest about voter perceptions?
426/7. Driving and Cell Phones In a survey, 1640 out of 2246 randomly selected adults in the United States said that they use cell phones while driving (based on data from Zogby International). When testing the claim that the proportion of adults who use cell phones while driving is equal to 75%, the TI­83/84 Plus calculator display in the margin is obtained. Use the results from the display with a 0.05 significance level to test the stated claim. 427/12. Screening for Marijuana Usage The company Drug Test Success provides a “ 1­
Panel­THC” test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than 10% of the test results are wrong. Does the test appear to be good for most purposes?
428/16. Predicting Sex of Baby Example 3 in this section included a hypothesis test involving pregnant women and their ability to predict the sex of their babies. In the same study, 45 of the pregnant women had more than 12 years of education, and 32 of them made correct pre­
dictions. Use these results to test the claim that women with more than 12 years of education have a proportion of correct predictions that is greater than the 0.5 proportion expected with random guesses. Use a 0.01 significance level. Do these women appear to have an ability to correctly predict the sex of their babies?
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155S8.3_3 Testing a Claim about a Proportion
428/18. Gender Selection for Boys The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability that a baby is a boy. As of this writing, among the babies born to parents using the YSORT method, 172 were boys and 39 were girls. Use the sample data with a 0.01 significance level to test the claim that with this method, the probability of a baby being a boy is greater than 0.5. Does the YSORT method of gender selection appear to work?
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429/22. New Sheriff in Town In recent years, the Town of Newport experienced an arrest rate of 25% for robberies (based on FBI data). The new sheriff compiles records showing that among 30 recent robberies, the arrest rate is 30%, so she claims that her arrest rate is greater than the 25% rate in the past. Is there sufficient evidence to support her claim that the arrest rate is greater than 25%?
The test statistic is z = 9.16.
The reject H0 region is the shaded region to the right of the C.V. z = 2.33.
Since the test statistic is z = 9.16 is in the "reject region," reject H0.
429/24. Smoking and College Education A survey showed that among 785 randomly selected subjects who completed four years of college, 18.3% smoke and 81.7% do not smoke (based on data from the American Medical Association). Use a 0.01 significance level to test the claim that the rate of smoking among those with four years of college is less than the 27% rate for the general population. Why would college graduates smoke at a lower rate than others? 429/29. Scream A survey of 61,647 people included several questions about office relationships. Of the respondents, 26% reported that bosses scream at employees. Use a 0.05 significance level to test the claim that more than 1/4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an Elle/MSNBC. COM survey in which Internet users chose whether to respond?
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429/31. Finding a Job Through Networking In a survey of 703 randomly selected workers, 61% got their jobs through networking (based on data from Taylor Nelson Sofres Research). Use the sample data with a 0.05 significance level to test the claim that most workers (more than 50%) get their jobs through networking. What does the result suggest about the strategy for finding a job after graduation?
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430/32. Mendel's Genetics Experiments When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in 580 offspring peas, with 26.2% of them having yellow pods. According to Mendel's theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level to test the claim that the proportion of peas with yellow pods is equal to 1/4.
P­value = 2P﴾z>0.671﴿ = 0.5021
Conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that p = 0.25. There is not sufficient evidence to reject the claim that the proportion of peas with yellow pods is equal to 1/4.
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