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Lecture Slides

Lecture Slides. Elementary Statistics Twelfth Edition and the Triola Statistics Series by Mario F. Triola. Chapter 8 Hypothesis Testing. 8-1 Review and Preview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim about a Proportion 8-4 Testing a Claim About a Mean

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Lecture Slides

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  1. Lecture Slides Elementary StatisticsTwelfth Edition and the Triola Statistics Series by Mario F. Triola

  2. Chapter 8Hypothesis Testing 8-1 Review and Preview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim about a Proportion 8-4 Testing a Claim About a Mean 8-5 Testing a Claim About a Standard Deviation or Variance

  3. 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 P-value method, the iitional method or the use of confidence intervals.

  4. Key Concept 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.

  5. Part 1: Basic Methods of Testing Claims about a Population Proportion p

  6. Notation n = sample size or number of trials p = population proportion q = 1 – p

  7. 1) The sample observations are a simple random sample. 2) The conditions for a binomial distribution are satisfied. 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 = npand . Note: p is the assumed proportion not the sample proportion. Requirements for Testing Claims About a Population Proportion p

  8. Test Statistic for Testing a Claim About a Proportion Use the standard normal distribution (Table A-2) and refer to Figure 8-1. Use the standard normal distribution (Table A-2). P-values: Critical Values:

  9. Caution Don’t confuse a P-value with a proportion p. P-value = probability of getting a test statistic at least as extreme as the one representing sample data p = population proportion

  10. P-Value Method Computer programs and calculators usually provide a P-value, so the P-value method is used. If technology is not available, see Figure 8-1 in the text.

  11. Critical Value Method Use the same method as described in Figure 8-2 in Section 8-2.

  12. Confidence Interval Method In general, for two-tailed hypothesis tests, construct a confidence interval with a confidence level corresponding to the significance level, as in Table 8-1 in the text.

  13. Caution When testing claims about a population proportion, the Critical-value method and the P-value 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 . Consequently, it is possible that the Critical-value and P-value 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 P-value or Critical-value method for testing a claim about the proportion.

  14. Example Based on information from the National Cyber Security Alliance, 93% of computer owners believe they have antivirus programs installed on their computers. In a random sample of 400 scanned computers, it is found that 380 of them (or 95%) actually have antivirus software programs. Use the sample data from the scanned computers to test the claim that 93% of computers have antivirus software.

  15. Example - Continued • Requirement check: • The 400 computers are randomly selected. • There is a fixed number of independent trials with two categories (computer has an antivirus program or does not). • The requirements np ≥ 5 and nq ≥ 5 are both satisfied with n = 400

  16. Example - Continued • P-Value Method: • The original claim that 93% of computers have antivirus software can be expressed as p = 0.93. • The opposite of the original claim is p ≠ 0.93. • The hypotheses are written as:

  17. Example - Continued • P-Value Method: • For the significance level, we select α = 0.05. • Because we are testing a claim about a population proportion, the sample statistic relevant to this test is:

  18. Example - Continued • P-Value Method: • The test statistic is calculated as:

  19. Example - Continued • P-Value Method: • Because the hypothesis test is two-tailed with a test statistic of • z = 1.57, the P-value is twice the area to the right of z = 1.57. • The P-value is twice 0.0582, or 0.1164.

  20. Example - Continued • P-Value Method: • Because the P-value of 0.1164 is greater than the significance level of α = 0.05, we fail to reject the null hypothesis. • We fail to reject the claim that 93% computers have antivirus software. We conclude that there is not sufficient sample evidence to warrant rejection of the claim that 93% of computers have antivirus programs.

  21. Example - Continued • Critical Value Method: Steps 1 – 5 are the same as for the P-value method. • The test statistic is computed to be z = 1.57. We now find the critical values, with the critical region having an area of α = 0.05, split equally in both tails.

  22. Example - Continued • Critical Value Method: • Because the test statistic does not fall in the critical region, we fail to reject the null hypothesis. • We fail to reject the claim that 93% computers have antivirus software. We conclude that there is not sufficient sample evidence to warrant rejection of the claim that 93% of computers have antivirus programs.

  23. Example - Continued Confidence Interval Method: The claim of p = 0.93 can be tested at the α = 0.05 level of significance with a 95% confidence interval. Using the methods of Section 7-2, we get: 0.929 < p < 0.971 This interval contains p = 0.93, so we do not have sufficient evidence to warrant the rejection of the claim that 93% of computers have antivirus programs.

  24. Part 2 Exact Method for Testing Claims about a Proportion

  25. Testing Claims Using the Exact Method 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 P-value method, use the value of p assumed in the null hypothesis, and find P-values as follows:

  26. Testing Claims Using the Exact Method Left-tailed test: The P-value is the probability of getting x or fewer successes among n trials. Right-tailed test: The P-value is the probability of getting x or more successes among n trials.

  27. Testing Claims Using the Exact Method Two-tailed test: , the P-value is twice the probability of getting x or more successes , the P-value is twice the probability of getting x or fewer successes

  28. Example In testing a method of gender selection, 10 randomly selected couples are treated with the method, and 9 of the babies are girls. Use a 0.05 significance level to test the claim that with this method, the probability of a baby being a girl is greater than 0.75.

  29. Example - Continued We will test using technology to find probabilities in a binomial distribution with p = 0.75. Because it is a right-tailed test, the P-value is the probability of 9 or more successes among 10 trials, assuming p = 0.75.

  30. Example - Continued The accompanying STATDISK display shows exact binomial probabilities.

  31. Example - Continued The probability of 9 or more successes is 0.2440252, which is the P-value of the hypothesis test. The P-value is high (greater than 0.05), so we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that with the gender selection method, the probability of a girl is greater than 0.75.

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