Statistical Inference: Confidence Intervals and Hypothesis Testing, Study notes of Mathematics

Download Statistical Inference: Confidence Intervals and Hypothesis Testing and more Study notes Mathematics in PDF only on Docsity! Statistical Testing 1. Vocabulary • A statistic is a statement of numerical information about a sample. • A parameter is a statement of numerical information about a popula- tion. • A point estimate is a single number, based on sample data, used to estimate a population parameter. • Usualy the point estimate is the mean of the data in the sample. • An confidence interval is an interval which contains the true value of the parameter to a high probability, called the confidence level. 2. Variables • µ is the mean for the entire population as a whole • σ is the standard deviation for the entire population • x is the mean over the sampling distribution • σx is the standard error of the mean or the standard deviation of the sampling distribution • n is the sample size 3. The Five Step Method • Step 1. Determine the level of significance you are using: 90%, 95%, or 99%. • Step 2. Compute z from the following table Level of confidence z-value 90% 1.645 95% 1.96 99% 2.575 • Step 3. Determine the mean x for the sample. • The Method Separates into two branches depending on whether – Case A: σ for the population is known – Case B: σ for the population is unknown 1 2 4. Branch A: Sigma is known • Goal: Compute the interval (x − zσx, x + zσx). • Step 4A. If σ for the population is known, compute σx by the formula on page 63 of the text: σx = σ√ n • Step 5A. Combine the numbers found above to calculate x − zσx and x + zσx 5. Two Remarks • Remark Number 1. In Step 2, if the significance level is not 90, 95, or 99 percent, compute the z-value as follows: First convert the significance level to a percentage and divide the percentage by two. Look up this percentage in the table on page 314 of the text and read off its corresponding z value. • Remark Number 2. If the standard deviation of the population is not known, or cannot be computed by the p–q rule, then it can be estimated by ajusting the standard deviation of one particular sample. • We will not consider this harder case. 6. Example. Popularity Polls • A recent survey of 500 Americans 185 approved of the way President Bush is handling the War in Iraq. • What is the associated confidence interval about 185 at a confidence level of .95? • Solution: – Step 1. We are given that the confidence level is .95 – Step 2. Find the z value. – From the table we see that the z–value corresponding to confidence level = .95 is z = 1.96 7. Step 3. Find x • Since 185 500 = .37, 5 13. Example from Health Field • According to national health statistics, 40% of patients in the United States that have full blown AIDS die within one year. • In a sample of 60 AIDS patients that are given the experimental drug AZT , 12 die within one year. • In order to determine if the drug is a medical breakthrough, the doc- tors wish to determine whether the percentage of deaths for the sample group falls within a 99% confidence interval about the national percent- age of 40%. • Hypothesis to be tested (or null hypothesis): The drug AZT did not significantly help the 60 AIDS patients. 14. Example from Health Field • Step 1. We are given that the confidence level is .99 • Step 2. Find the z value. • From the chart, we see that the z–value corresponding to confidence level .99 is z = 2.575 • Step 3. Find the mean of the sample distribution x • If the national average of a 40 percent death rate holds for the 60 AIDS patients, then we would expect x = (.40)(60) = 24 deaths per year 15. Step 4: Find σx • By the p − q rule, σ = √ (.60)(.40) = .49 • The standard error of the mean is σ√ n = .49√ 60 = .063 • Since the sample size is 60, it is helpful to rewrite 6.3 percent as a number out of 60. • So we compute 6.3 percent of 60: σx = .063 × 60 = 3.78 6 16. Step 5. Find the confidence interval • 99% of all 60 person samples should have death rates that fall within 2.575 standard deviations of the mean. • We calculate: • x − 2.575 σx = 24 − (2.575)(3.78) = 14.3 and • x + 2.575 σx = 24 + (2.575)(3.78) = 33.7 • This means that 99% of the time we would expect that in a group of 60 AIDS patients between 14 and 34 of them will die within one year. 17. Draw a conclustion Since the number of deaths in the sample given the drug AZT is 12 and 12 is outside of the confidence interval 14 to 34, we we reject the null hypothesis. The doctors should accept the alternative hyptothesis that the drug actually helped the patients in the sample. 18. More Complicated Cases • The value of σ cannot always be determined by the p − q rule. • For example, if we are computing the mean grade point average of a Math 101 student, we are not in a “yes–no” situation. • If we perform one sample, say we compute the mean of the gpa’s of a random sample of 50 Math 101 students. • Then we use this value both for µ and for x. • How do we find σx? 19. Values for Standard Deviation There are three separate values for the standard deviation: • 1. σ = the standard deviation of the entire population • 2. σx = the error of the mean given by the formula σx = σ√ n 7 • 3. S = the standard deviation of the one particular sample we took of n = 50 students. 20. The Estimator • Without the p − q rule, one idea is to use S to obtain an estimate for σ. • One estimate commonly used for σ is • σ ≈ S √ n n − 1 21. Case B If σ for the population is not known, we compute the interval (x− zσ̂x, x + zσ̂x). Step 4B. If σ for the population is not known, compute σ̂x by the formula σ̂x = S √ n n−1√ n = S√ n − 1 where S is the standard deviation of the sample. Step 5B. Combine the numbers found above to calculate x−zσ̂x and x+zσ̂x. 22. Testing for Bias Racial bias in death penalty Death Sentence yes no White 19 141 Black 17 149 10 29. The Null Hypothesis Use Chi-square testing with alpha level .05 to test the Null Hypothesis: There is NO DIFFERENCE between the opinions of the religious and non-religious NIU students. 30. Computing the Expected Values If there was no bias among the 40 religious students, then we would expect that (.50) × (40) = 20 would approve (.40) × (40) = 16 would disapprove (.10) × (40) = 4 would have no opinion Of the 60 non-religious students, we would expect the following outcomes: (.50) × (60) = 30 would approve (.40) × (60) = 24 would disapprove (.10) × (60) = 6 would have no opinion 31. Expected Data Putting this information into a chart Approve Disapprove No opinion Total Religious 20 16 4 40 Not Relig. 30 24 6 60 Total 50 40 10 100 11 32. The χ2 Calculation We compute χ2 = (F1 − E1)2 E1 + (F2 − E2)2 E2 + (F3 − E3)2 E3 + · · · + (F6 − E6)2 E6 = (14 − 20)2 20 + (23 − 16)2 16 + (3 − 4)2 4 + (36 − 30)2 30 + (17 − 24)2 24 + (7 − 6)2 6 = 36 20 + 49 16 + 1 4 + 36 30 + 49 24 + 1 6 = 8.52 33. The Conclusion The degrees of freedom for this problem is d = 2, since you can determine all the other numbers once you know two of them. The table for one degree of freedom at a .05 significance level gives the value 5.991 In our calculation χ2 = 8.52 exceed 5.991 This means that there is a less than 5% chance that the observed frequencies occurred by chance. Conclusion: at a .05 significance level, the data indicates that reglious students are more likely to disapprove of gay marriage than non-religious stu- dents. Note that the table for a .01 significance level gives the value 9.21. So at a .01 significance level, we accept the null hypothesis that the observed frequencies lie within a 99% probability of occurring merely by chance. 12 34. Are you Psychic? [Compare with Homework Problem 27 on page 106.] • One test of physic powers is for one person to concentrate on the 52 cards in a standard deck. • The person whose physic powers are to be tested reads the mind of the card holder and says what suit she believes the card is, hearts, diamonds, clubs, or spades. • The card holder keeps track of the number of correct guesses the “po- tential psychic” makes. 35. Find the Expected Values • Null hypothesis to be tested: The person is not psychic. • Set the alpha level. This is the level at which we will reject the null hypothesis. Here let’s say α = .05 • For a non-physic person, you would expect 13 correct guesses (1 out of 4) and 39 incorrect guesses. 36. The Chi Square Calculation Suppose the data comes out: Observed Expected Correct 20 13 Incorrect 32 39 χ2 = (F1 − E1)2 E1 + (F2 − E2)2 E2 = (20 − 13)2 13 + (32 − 39)2 39 = 49 13 + 49 39 = 5.03