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BPS - 5th Ed. Chapter 141 Introduction to Inference.

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1 BPS - 5th Ed. Chapter 141 Introduction to Inference

2 BPS - 5th Ed. Chapter 142 u Provides methods for drawing conclusions about a population from sample data –Confidence Intervals v What is the population mean? –Tests of Significance v Is the population mean larger than 66.5? Statistical Inference

3 BPS - 5th Ed. Chapter SRS from the population of interest 2. Variable has a Normal distribution N( ,  ) in the population 3. Although the value of  is unknown, the value of the population standard deviation  is known Inference about a Mean Simple Conditions

4 BPS - 5th Ed. Chapter 144 A level C confidence interval has two parts 1. An interval calculated from the data, usually of the form: estimate ± margin of error 2. 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. Confidence Interval

5 BPS - 5th Ed. Chapter 145 Case Study NAEP Quantitative Scores (National Assessment of Educational Progress) Rivera-Batiz, F. L., “Quantitative literacy and the likelihood of employment among young adults,” Journal of Human Resources, 27 (1992), pp What is the average score for all young adult males?

6 BPS - 5th Ed. Chapter 146 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  = 60.

7 BPS - 5th Ed. Chapter 147 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. On the basis of this sample, estimate the mean score  in the population of all 9.5 million young men of these ages.

8 BPS - 5th Ed. Chapter 148 Case Study NAEP Quantitative Scores 1. To estimate the unknown population mean , use the sample mean = The law of large numbers suggests that will be close to , but there will be some error in the estimate. 3. The sampling distribution of has the Normal distribution with mean  and standard deviation

9 BPS - 5th Ed. Chapter 149 Case Study NAEP Quantitative Scores

10 BPS - 5th Ed. Chapter 1410 Case Study NAEP Quantitative Scores 4. The rule indicates that and  are within two standard deviations (4.2) of each other in about 95% of all samples.

11 BPS - 5th Ed. Chapter 1411 Case Study NAEP Quantitative Scores So, if we estimate that  lies within 4.2 of, we’ll be right about 95% of the time.

12 BPS - 5th Ed. Chapter 1412 Take an SRS of size n from a Normal population with unknown mean  and known standard deviation . A level C confidence interval for  is: Confidence Interval Mean of a Normal Population

13 BPS - 5th Ed. Chapter 1413 Confidence Interval Mean of a Normal Population

14 BPS - 5th Ed. Chapter 1414 Case Study NAEP Quantitative Scores Using the 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 and

15 BPS - 5th Ed. Chapter 1415 Careful Interpretation of a Confidence Interval u “We are 95% confident that the mean NAEP score for the population of all adult males is between and ” (We feel that plausible values for the population of males’ mean NAEP score are between and ) u ** This does not mean that 95% of all males will have NAEP scores between and ** u 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.

16 BPS - 5th Ed. Chapter 1416 u What would happen if we repeated the sample or experiment many times? u How likely would it be to see the results we saw if the claim of the test were true? u Do the data give evidence against the claim? Reasoning of Tests of Significance

17 BPS - 5th Ed. Chapter 1417 u The statement being tested in a statistical test is called the null hypothesis. u The test is designed to assess the strength of evidence against the null hypothesis. u Usually the null hypothesis is a statement of “no effect” or “no difference”, or it is a statement of equality. u When performing a hypothesis test, we assume that the null hypothesis is true until we have sufficient evidence against it. Stating Hypotheses Null Hypothesis, H 0

18 BPS - 5th Ed. Chapter 1418 u The statement we are trying to find evidence for is called the alternative hypothesis. u Usually the alternative hypothesis is a statement of “there is an effect” or “there is a difference”, or it is a statement of inequality. u 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 H a to fit what the data show. Stating Hypotheses Alternative Hypothesis, H a

19 BPS - 5th Ed. Chapter 1419 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.

20 BPS - 5th Ed. Chapter 1420 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  = 1. The mean  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 = Do the data provide sufficient evidence that the new cola lost sweetness in storage?

21 BPS - 5th Ed. Chapter 1421 Case Study I Sweetening Colas  If the claim that  = 0 is true (no loss of sweetness, on average), the sampling distribution of from 10 tasters is Normal with  = 0 and standard deviation  The data yielded = 1.02, which is more than three standard deviations from  = 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  = 0, there would be no evidence that the new cola lost sweetness in storage.

22 BPS - 5th Ed. Chapter 1422 Case Study I Sweetening Colas

23 BPS - 5th Ed. Chapter 1423 The Hypotheses for Means  Null: H 0 :  =  0 u One sided alternatives H a :   H a :   u Two sided alternative H a :  0

24 BPS - 5th Ed. Chapter 1424 Sweetening Colas Case Study I 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. H 0 :  = 0H a :  > 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).

25 BPS - 5th Ed. Chapter 1425 Studying Job Satisfaction Case Study II 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.

26 BPS - 5th Ed. Chapter 1426 Studying Job Satisfaction Case Study II 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. H 0 :  = 0H a :  ≠ 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).

27 BPS - 5th Ed. Chapter 1427 Take an SRS of size n from a Normal population with unknown mean  and known standard deviation . The test statistic for hypotheses about the mean (H 0 :  =  0 ) of a Normal distribution is the standardized version of : Test Statistic Testing the Mean of a Normal Population

28 BPS - 5th Ed. Chapter 1428 Sweetening Colas Case Study I 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.

29 BPS - 5th Ed. Chapter 1429 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. P-value

30 BPS - 5th Ed. Chapter 1430 P-value for Testing Means  H a :  >  0 v P-value is the probability of getting a value as large or larger than the observed test statistic (z) value.  H a :  <  0 v P-value is the probability of getting a value as small or smaller than the observed test statistic (z) value.  H a :  0 v P-value is two times the probability of getting a value as large or larger than the absolute value of the observed test statistic (z) value.

31 BPS - 5th Ed. Chapter 1431 Sweetening Colas Case Study I For test statistic z = 3.23 and alternative hypothesis H a :  > 0, the P-value would be: P-value = P(Z > 3.23) = 1 – = If H 0 is true, there is only a (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 H 0 is true, we therefore have evidence against H 0 and in favor of H a.

32 BPS - 5th Ed. Chapter 1432 Sweetening Colas Case Study I

33 BPS - 5th Ed. Chapter 1433 Studying Job Satisfaction Case Study II Suppose job satisfaction scores follow a Normal distribution with standard deviation  = 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:

34 BPS - 5th Ed. Chapter 1434 Studying Job Satisfaction Case Study II For test statistic z = 1.20 and alternative hypothesis H a :  ≠ 0, the P-value would be: P-value= P(Z 1.20) = 2 P(Z 1.20) = (2)(0.1151) = If H 0 is true, there is a (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 H 0 is true, we therefore do not have good evidence against H 0 and in favor of H a.

35 BPS - 5th Ed. Chapter 1435 Studying Job Satisfaction Case Study II

36 BPS - 5th Ed. Chapter 1436  If the P-value is as small as or smaller than the significance level  (i.e., P-value ≤  ), then we say that the data give results that are statistically significant at level .  If we choose  = 0.05, we are requiring that the data give evidence against H 0 so strong that it would occur no more than 5% of the time when H 0 is true.  If we choose  = 0.01, we are insisting on stronger evidence against H 0, evidence so strong that it would occur only 1% of the time when H 0 is true. Statistical Significance

37 BPS - 5th Ed. Chapter 1437 The four steps in carrying out a significance test: 1. State the null and alternative hypotheses. 2. Calculate the test statistic. 3. Find the P-value. 4. 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. Tests for a Population Mean

38 BPS - 5th Ed. Chapter 1438

39 BPS - 5th Ed. Chapter 1439 Sweetening Colas Case Study I 1. Hypotheses:H 0 :  = 0 H a :  > 0 2. Test Statistic: 3. P-value:P-value = P(Z > 3.23) = 1 – = Conclusion: Since the P-value is smaller than  = 0.01, there is very strong evidence that the new cola loses sweetness on average during storage at room temperature.

40 BPS - 5th Ed. Chapter 1440 Studying Job Satisfaction Case Study II 1. Hypotheses:H 0 :  = 0 H a :  ≠ 0 2. Test Statistic: 3. P-value:P-value = 2P(Z > 1.20) = (2)(1 – ) = Conclusion: Since the P-value is larger than  = 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.

41 BPS - 5th Ed. Chapter 1441 Confidence Intervals & Two-Sided Tests A level  two-sided significance test rejects the null hypothesis H 0 :  =  0 exactly when the value  0 falls outside a level (1 –  ) confidence interval for .

42 BPS - 5th Ed. Chapter 1442 Case Study II A 90% confidence interval for  is: Since  0 = 0 is in this confidence interval, it is plausible that the true value of  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. Studying Job Satisfaction


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