# Chapter 14

## Chapter 14

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##### Presentation Transcript

1. Chapter 14 Introduction to Inference Chapter 14

2. Statistical Inference • Provides methods for drawing conclusions about a population from sample data • Confidence Intervals • Tests of Significance Chapter 14

3. Inference about a MeanSimple Conditions • SRS from the population of interest • Variable has a Normal distribution N(m, s) in the population • Although the value of m is unknown, the value of the population standard deviation s is known Chapter 14

4. Confidence Interval A level C confidence interval has two parts • An interval calculated from the data, usually of the form:estimate ± margin of error • The confidence level C, which is the probability that the interval will capture the true parameter value in repeated samples; that is, C is the success rate for the method. Chapter 14

5. Case Study NAEP Quantitative Scores The NAEP survey includes a short test of quantitative skills, covering mainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to 500, with higher scores indicating greater numerical abilities. It is known that NAEP scores have standard deviation s = 60. Chapter 14

6. Case Study NAEP Quantitative Scores In a recent year, 840 men 21 to 25 years of age were in the NAEP sample. Their mean quantitative score was 272 (σ = 60). On the basis of this sample, estimate the mean score m in the population of all 9.5 million young men of these ages. Chapter 14

7. Case Study NAEP Quantitative Scores To estimate the unknown population mean m, use the sample mean = 272. The law of large numbers suggests that will be close to m, but there will be some error in the estimate. The sampling distribution of has the Normal distribution with mean m and standard deviation Chapter 14

8. The 68-95-99.7 rule indicates that and m are within two standard deviations (4.2) of each other in about 95% of all samples. Case Study NAEP Quantitative Scores Chapter 14

9. Example – Your Turn • A sample of 654 women in their 20s have an average Body Mass Index (BMI) of 26.8 with standard deviation σ = 7.5. • Use the 68-95-99.7 rule to compute the 95% confidence interval for this sample of women from the US population of 18 million women in this age group. Chapter 14

10. Confidence IntervalMean of a Normal Population Take an SRS of size n from a Normal population with unknown mean m and known standard deviation s. A level C confidence interval for m is: Chapter 14

11. Case Study NAEP Quantitative Scores Using the 68-95-99.7 rule gave an approximate 95% confidence interval. A more precise 95% confidence interval can be found using the appropriate value of z* (1.960) with the previous formula. We are 95% confident that the average NAEP quantitative score for all adult males is between 267.884 and 276.116. Chapter 14

12. Careful Interpretation of a Confidence Interval • “We are 95% confident that the mean NAEP score for the population of all adult males is between 267.884 and 276.116.” (We feel that plausible values for the population of males’ mean NAEP score are between 267.884 and 276.116.) • ** This does not mean that 95% of all males will have NAEP scores between 267.884 and 276.116. ** • Statistically: 95% of all samples of size 840 from the population of males should yield a sample mean within two standard errors of the population mean; i.e., in repeated samples, 95% of the C.I.s should contain the true population mean. Chapter 14

13. Looking Up Confidence Levels Find the appropriate z-value for the following confidence intervals: 90% 97% 98% 99.5% 2.17 2.33 2.81 1.65 Chapter 14

14. Homework • Read Pg 359 – 367 • Problems 1, 4, 5b, 34 - 36 Chapter 14

15. Stating HypothesesNull Hypothesis, H0 • The statement being tested in a statistical test is called the null hypothesis. • The test is designed to assess the strength of evidence against the null hypothesis. • Usually the null hypothesis is a statement of “no effect” or “no difference”, or it is a statement of equality. • When performing a hypothesis test, we assume that the null hypothesis is true until we have sufficient evidence against it. Chapter 14

16. Stating HypothesesAlternative Hypothesis, Ha • The statement we are trying to find evidence for is called the alternative hypothesis. • Usually the alternative hypothesis is a statement of “there is an effect” or “there is a difference”, or it is a statement of inequality. • The alternative hypothesis should express the hopes or suspicions we bring to the data. It is cheating to first look at the data and then frame Ha to fit what the data show. Chapter 14

17. Case Study I Sweetening Colas Diet colas use artificial sweeteners to avoid sugar. These sweeteners gradually lose their sweetness over time. Trained testers sip the cola and assign a “sweetness score” of 1 to 10. The cola is then retested after some time and the two scores are compared to determine the difference in sweetness after storage. Bigger differences indicate bigger loss of sweetness. Chapter 14

18. Case Study I Sweetening Colas Suppose we know that for any cola, the sweetness loss scores vary from taster to taster according to a Normal distribution with standard deviation s = 1. The mean m for all tasters measures loss of sweetness. The sweetness losses for a new cola, as measured by 10 trained testers, yields an average sweetness loss of = 1.02. Do the data provide sufficient evidence that the new cola lost sweetness in storage? Chapter 14

19. Case Study I Sweetening Colas • If the claim that m = 0 is true (no loss of sweetness, on average), the sampling distribution of from 10 tasters is Normal with m = 0 and standard deviation • The data yielded = 1.02, which is more than three standard deviations from m = 0. This is strong evidence that the new cola lost sweetness in storage. • If the data yielded = 0.3, which is less than one standard deviations from m = 0, there would be no evidence that the new cola lost sweetness in storage. Chapter 14

20. Case Study I Sweetening Colas Chapter 14

21. The Hypotheses for Means • Null: H0:m= m0 • One sided alternatives Ha:m > m0 Ha:m < m0 • Two sided alternative Ha:m ¹ m0 Chapter 14

22. Case Study I Sweetening Colas The null hypothesis is no average sweetness loss occurs, while the alternative hypothesis (that which we want to show is likely to be true) is that an average sweetness loss does occur. H0: m = 0 Ha: m > 0 This is considered a one-sided test because we are interested only in determining if the cola lost sweetness (gaining sweetness is of no consequence in this study). Chapter 14

23. Example – Two Tailed • Situation 1: the hypothesis states that the sample mean (k) is the same as the population mean. So H0: µ = k and Ha: µ  k. Example: will a new drug raise or lower your normal heart rate? This is a two-tailed test because if the results are either up or down, the null-hypothesis doesn’t hold. Chapter 14

24. Example – Right Tailed • Situation 2: the hypothesis states that the sample mean (k) is larger than the population mean. So H0: µ  k and Ha: µ > k. Example: will a new chemical increase battery life? This test is right-tailed since we only care about an increase in battery life. Chapter 14

25. Example – Left Tailed • Situation 3: the hypothesis states that the sample mean (k) is smaller than the population mean. So H0: µ  k and Ha: µ < k. Example: will heating bills be reduced with a new type of insulation? This test is left-tailed since we only care about a decrease in energy costs. Chapter 14

26. Case Study II Studying Job Satisfaction Does the job satisfaction of assembly workers differ when their work is machine-paced rather than self-paced? A matched pairs study was performed on a sample of workers, and each worker’s satisfaction was assessed after working in each setting. The response variable is the difference in satisfaction scores, self-paced minus machine-paced. Chapter 14

27. Case Study II Studying Job Satisfaction The null hypothesis is no average difference in scores in the population of assembly workers, while the alternative hypothesis (that which we want to show is likely to be true) is there is an average difference in scores in the population of assembly workers. H0: m = 0 Ha: m ≠ 0 This is considered a two-sided test because we are interested determining if a difference exists (the direction of the difference is not of interest in this study). Chapter 14

28. You write the hypothesis conjectures • 1) A researcher thinks that if expectant mothers use vitamin pills, then the birth weight of their babies will increase. The average birth weight of the population is 8.6 pounds. H1: µ  8.6 and Ha: µ > 8.6 • 2) An engineer hypothesizes that defects can be decreased in a manufacturing process by using robots instead of humans for certain tasks. The mean number of defects per 1000 is 18. H1: µ  18 and Ha: µ < 18 • 3) A psychologist thinks that playing soft music during a test will change the results of the test. He is not sure if the scores will go up or down. The current average score is 73. H1: µ = 73 and Ha: µ  73 Chapter 14

29. Test StatisticTesting the Mean of a Normal Population Take an SRS of size n from a Normal population with unknown mean m and known standard deviation s. The test statistic for hypotheses about the mean (H0: m = m0) of a Normal distribution is the standardized version of : Chapter 14

30. Case Study I Sweetening Colas If the null hypothesis of no average sweetness loss is true, the test statistic would be: Because the sample result is more than 3 standard deviations above the hypothesized mean 0, it gives strong evidence that the mean sweetness loss is not 0, but positive. Chapter 14

31. Homework • Read Pg 368 – 372 • Problems 8 – 11, 40ab, 41ab, 42ab Chapter 14

32. P-value Assuming that the null hypothesis is true, the probability that the test statistic would take a value as extreme or more extreme than the value actually observed is called the P-value of the test. The smaller the P-value, the stronger the evidence the data provide against the null hypothesis. That is, a small P-value indicates a small likelihood of observing the sampled results if the null hypothesis were true. Chapter 14

33. P-value for Testing Means • Ha: m > m0 • P-value is the probability of getting a value as large or larger than the observed test statistic (z) value. • Ha: m < m0 • P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value. • Ha: m ¹ m0 • P-value is two times theprobability of getting a value as large or larger than the absolute value of the observed test statistic (z) value. Chapter 14

34. Case Study I Sweetening Colas For test statistic z = 3.23 and alternative hypothesisHa: m > 0, the P-value would be: P-value = P(Z > 3.23) = 1 – 0.9994 = 0.0006 If H0 is true, there is only a 0.0006 (0.06%) chance that we would see results at least as extreme as those in the sample; thus, since we saw results that are unlikely if H0 is true, we therefore have evidence against H0 and in favor of Ha. Chapter 14

35. Case Study I Sweetening Colas Chapter 14

36. Case Study II Studying Job Satisfaction Suppose job satisfaction scores follow a Normal distribution with standard deviation s = 60. Data from 18 workers gave a sample mean score of 17. If the null hypothesis of no average difference in job satisfaction is true, the test statistic would be: Chapter 14

37. Case Study II Studying Job Satisfaction For test statistic z = 1.20 and alternative hypothesisHa: m ≠ 0, the P-value would be: P-value = P(Z < -1.20 or Z > 1.20) = 2 P(Z < -1.20) = 2 P(Z > 1.20) = (2)(0.1151) = 0.2302 If H0 is true, there is a 0.2302 (23.02%) chance that we would see results at least as extreme as those in the sample; thus, since we saw results that are likely if H0 is true, we therefore do not have good evidence against H0 and in favor of Ha. Chapter 14

38. Case Study II Studying Job Satisfaction Chapter 14

39. Statistical Significance • If the P-value is as small as or smaller than the significance level a (i.e., P-value ≤ a), then we say that the data give results that are statistically significant at level a. • If we choose a = 0.05, we are requiring that the data give evidence against H0 so strong that it would occur no more than 5% of the time when H0 is true. • If we choose a = 0.01, we are insisting on stronger evidence against H0, evidence so strong that it would occur only 1% of the time when H0 is true. Chapter 14

40. Tests for a Population Mean The four steps in carrying out a significance test: • State the null and alternative hypotheses. • Calculate the test statistic. • Find the P-value. • State your conclusion in the context of the specific setting of the test. The procedure for Steps 2 and 3 is on the next page. Chapter 14

41. Chapter 14

42. Case Study I Sweetening Colas • Hypotheses: H0: m = 0 Ha: m > 0 • Test Statistic: • CRITICAL VALUE FOR 95% = 1.96 • Conclusion: Since the P-value is smaller than a = 0.01, there is very strong evidence that the new cola loses sweetness on average during storage at room temperature. Chapter 14

43. Case Study II Studying Job Satisfaction • Hypotheses: H0: m = 0 Ha: m ≠ 0 • Test Statistic: • P-value: P-value = 2P(Z > 1.20) = (2)(1 – 0.8849) = 0.2302 • Conclusion: Since the P-value is larger than a = 0.10, there is not sufficient evidence that mean job satisfaction of assembly workers differs when their work is machine-paced rather than self-paced. Chapter 14

44. Confidence Intervals & Two-Sided Tests A level a two-sided significance test rejects the null hypothesis H0: m = m0 exactly when the value m0 falls outside a level (1 – a) confidence interval for m. Chapter 14

45. Case Study II Studying Job Satisfaction A 90% confidence interval for m is: Since m0 = 0 is in this confidence interval, it is plausible that the true value of m is 0; thus, there is not sufficient evidence(at  = 0.10) that the mean job satisfaction of assembly workers differs when their work is machine-paced rather than self-paced. Chapter 14