The Practice of Statistics, 4th edition For AP-STARNES, .ppt

1、The Practice of Statistics, 4th edition For AP* STARNES, YATES, MOORE,Unit 5: Hypothesis Testing,Unit 5: Hypothesis Testing,10.2 Significance Tests: The Basics 12.1 Tests about a Population Proportion 11.1 Tests about a Population Mean 10.4 Errors and the Power of a Test,Section 11.1 Tests About a P

2、opulation Mean,After this section, you should be able to CHECK conditions for carrying out a test about a population mean. CONDUCT a one-sample t test about a population mean. CONSTRUCT a confidence interval to draw a conclusion for a two-sided test about a population mean. PERFORM significance test

3、s for paired data.,Learning Objectives,Tests About a Population Mean,Introduction Confidence intervals and significance tests for a population proportion p are based on z-values from the standard Normal distribution. Inference about a population mean uses a t distribution with n - 1 degrees of freed

4、om, except in the rare case when the population standard deviation is known.,Carrying Out a Significance Test for ,Tests About a Population Mean,In an earlier example, a company claimed to have developed a new AAA battery that lasts longer than its regular AAA batteries. Based on years of experience

5、, the company knows that its regular AAA batteries last for 30 hours of continuous use, on average. An SRS of 15 new batteries lasted an average of 33.9 hours with a standard deviation of 9.8 hours. Do these data give convincing evidence that the new batteries last longer on average?,To find out, we

6、 must perform a significance test of H0: = 30 hours Ha: 30 hours where = the true mean lifetime of the new deluxe AAA batteries.,Check Conditions: Three conditions should be met before we perform inference for an unknown population mean: Random, Normal, and Independent. The Normal condition for mean

7、s is Population distribution is Normal or sample size is large (n 30) We often dont know whether the population distribution is Normal. But if the sample size is large (n 30), we can safely carry out a significance test (due to the central limit theorem). If the sample size is small, we should exami

8、ne the sample data for any obvious departures from Normality, such as skewness and outliers.,Carrying Out a Significance Test for ,Tests About a Population Mean,Check Conditions: Three conditions should be met before we perform inference for an unknown population mean: Random, Normal, and Independen

9、t.,Random The company tests an SRS of 15 new AAA batteries.,Independent Since the batteries are being sampled without replacement, we need to check the 10% condition: there must be at least 10(15) = 150 new AAA batteries. This seems reasonable to believe.,Normal We dont know if the population distri

10、bution of battery lifetimes for the companys new AAA batteries is Normal. With such a small sample size (n = 15), we need to inspect the data for any departures from Normality. The dotplot and boxplot show slight right-skewness but no outliers. The Normal probability plot is close to linear. We shou

11、ld be safe performing a test about the population mean lifetime .,Carrying Out a Significance Test,Test About a Population Mean,Calculations: Test statistic and P-value When performing a significance test, we do calculations assuming that the null hypothesis H0 is true. The test statistic measures h

12、ow far the sample result diverges from the parameter value specified by H0, in standardized units. As before,For a test of H0: = 0, our statistic is the sample mean. Its standard deviation is,Carrying Out a Hypothesis Test The battery company wants to test H0: = 30 versus Ha: 30 based on an SRS of 1

13、5 new AAA batteries with mean lifetime and standard deviation,Tests About a Population Mean,The P-value is the probability of getting a result this large or larger in the direction indicated by Ha, that is, P(t 1.54).,Go to the df = 14 row.Since the t statistic falls between the values 1.345 and 1.7

14、61, the “Upper-tail probability p” is between 0.10 and 0.05. The P-value for this test is between 0.05 and 0.10.,Because the P-value exceeds our default = 0.05 significance level, we cant conclude that the companys new AAA batteries last longer than 30 hours, on average.,Using Table B Wisely,Tests A

15、bout a Population Mean,Table B gives a range of possible P-values for a significance. We can still draw a conclusion from the test in much the same way as if we had a single probability by comparing the range of possible P-values to our desired significance level.Table B has other limitations for fi

16、nding P-values. It includes probabilities only for t distributions with degrees of freedom from 1 to 30 and then skips to df = 40, 50, 60, 80, 100, and 1000. (The bottom row gives probabilities for df = , which corresponds to the standard Normal curve.) Note: If the df you need isnt provided in Tabl

17、e B, use the next lower df that is available. Table B shows probabilities only for positive values of t. To find a P-value for a negative value of t, we use the symmetry of the t distributions.,Using Table B Wisely,Tests About a Population Mean,Suppose you were performing a test of H0: = 5 versus Ha

18、: 5 based on a sample size of n = 37 and obtained t = -3.17. Since this is a two-sided test, you are interested in the probability of getting a value of t less than -3.17 or greater than 3.17.Due to the symmetric shape of the density curve, P(t -3.17) = P(t 3.17). Since Table B shows only positive t

19、-values, we must focus on t = 3.17.,Since df = 37 1 = 36 is not available on the table, move across the df = 30 row and notice that t = 3.17 falls between 3.030 and 3.385. The corresponding “Upper-tail probability p” is between 0.0025 and 0.001. For this two-sided test, the corresponding P-value wou

20、ld be between 2(0.001) = 0.002 and 2(0.0025) = 0.005.,The One-Sample t Test When the conditions are met, we can test a claim about a population mean using a one-sample t test.,Tests About a Population Mean,Choose an SRS of size n from a large population that contains an unknown mean . To test the hy

21、pothesis H0 : = 0, compute the one-sample t statisticFind the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha in a t-distribution with df = n - 1,One-Sample t Test,Use this test only when (1) the populat

22、ion distribution is Normal or the sample is large (n 30), and (2) the population is at least 10 times as large as the sample.,Example: Healthy Streams The level of dissolved oxygen (DO) in a stream or river is an important indicator of the waters ability to support aquatic life. A researcher measure

23、s the DO level at 15 randomly chosen locations along a stream. Here are the results in milligrams per liter:,Tests About a Population Mean,State: We want to perform a test at the = 0.05 significance level of H0: = 5 Ha: 5 where is the actual mean dissolved oxygen level in this stream.,Plan: If condi

24、tions are met, we should do a one-sample t test for . Random The researcher measured the DO level at 15 randomly chosen locations. Normal We dont know whether the population distribution of DO levels at all points along the stream is Normal. With such a small sample size (n = 15), we need to look at

25、 the data to see if its safe to use t procedures.,4.53 5.04 3.29 5.23 4.13 5.50 4.83 4.40 5.42 6.38 4.01 4.66 2.87 5.73 5.55 A dissolved oxygen level below 5 mg/l puts aquatic life at risk.,The histogram looks roughly symmetric; the boxplot shows no outliers; and the Normal probability plot is fairl

26、y linear. With no outliers or strong skewness, the t procedures should be pretty accurate even if the population distribution isnt Normal.,Independent There is an infinite number of possible locations along the stream, so it isnt necessary to check the 10% condition. We do need to assume that indivi

27、dual measurements are independent.,Example: Healthy Streams,Tests About a Population Mean,Conclude: The P-value, is between 0.15 and 0.20. Since this is greater than our = 0.05 significance level, we fail to reject H0. We dont have enough evidence to conclude that the mean DO level in the stream is

28、less than 5 mg/l.,Do: The sample mean and standard deviation are,P-value The P-value is the area to the left of t = -0.94 under the t distribution curve with df = 15 1 = 14.,Since we decided not to reject H0, we could have made a Type II error (failing to reject H0when H0 is false). If we did, then

29、the mean dissolved oxygen level in the stream is actually less than 5 mg/l, but we didnt detect that with our significance test.,Two-Sided Tests At the Hawaii Pineapple Company, managers are interested in the sizes of the pineapples grown in the companys fields. Last year, the mean weight of the pin

30、eapples harvested from one large field was 31 ounces. A new irrigation system was installed in this field after the growing season. Managers wonder whether this change will affect the mean weight of future pineapples grown in the field. To find out, they select and weigh a random sample of 50 pineap

31、ples from this years crop. The Minitab output below summarizes the data. Determine whether there are any outliers.,Tests About a Population Mean,IQR = Q3 Q1 = 34.115 29.990 = 4.125,Any data value greater than Q3 + 1.5(IQR) or less than Q1 1.5(IQR) is considered an outlier.,Q3 + 1.5(IQR) = 34.115 + 1

32、.5(4.125) = 40.3025 Q1 1.5(IQR) = 29.990 1.5(4.125) = 23.0825,Since the maximum value 35.547 is less than 40.3025 and the minimum value 26.491 is greater than 23.0825, there are no outliers.,Two-Sided Tests,Tests About a Population Mean,State: We want to test the hypotheses H0: = 31 Ha: 31 where = t

33、he mean weight (in ounces) of all pineapples grown in the field this year. Since no significance level is given, well use = 0.05.,Plan: If conditions are met, we should do a one-sample t test for . Random The data came from a random sample of 50 pineapples from this years crop. Normal We dont know w

34、hether the population distribution of pineapple weights this year is Normally distributed. But n = 50 30, so the large sample size (and the fact that there are no outliers) makes it OK to use t procedures. Independent There need to be at least 10(50) = 500 pineapples in the field because managers ar

35、e sampling without replacement (10% condition). We would expect many more than 500 pineapples in a “large field.”,Two-Sided Tests,Tests About a Population Mean,Conclude: Since the P-value is between 0.005 and 0.01, it is less than our = 0.05 significance level, so we have enough evidence to reject H

36、0 and conclude that the mean weight of the pineapples in this years crop is not 31 ounces.,Do: The sample mean and standard deviation are,P-value The P-value for this two-sided test is the area under the t distribution curve with 50 - 1 = 49 degrees of freedom. Since Table B does not have an entry f

37、or df = 49, we use the more conservative df = 40. The upper tail probability is between 0.005 and 0.0025 so the desired P-value is between 0.01 and 0.005.,Confidence Intervals Give More Information,Tests About a Population Mean,Minitab output for a significance test and confidence interval based on

38、the pineapple data is shown below. The test statistic and P-value match what we got earlier (up to rounding).,As with proportions, there is a link between a two-sided test at significance level and a 100(1 )% confidence interval for a population mean . For the pineapples, the two-sided test at =0.05

39、 rejects H0: = 31 in favor of Ha: 31. The corresponding 95% confidence interval does not include 31 as a plausible value of the parameter . In other words, the test and interval lead to the same conclusion about H0. But the confidence interval provides much more information: a set of plausible value

40、s for the population mean.,Confidence Intervals and Two-Sided Tests,Tests About a Population Mean,The connection between two-sided tests and confidence intervals is even stronger for means than it was for proportions. Thats because both inference methods for means use the standard error of the sampl

41、e mean in the calculations.,A two-sided test at significance level (say, = 0.05) and a 100(1 )% confidence interval (a 95% confidence interval if = 0.05) give similar information about the population parameter.,When the two-sided significance test at level rejects H0: = 0, the 100(1 )% confidence in

42、terval for will not contain the hypothesized value 0 .,When the two-sided significance test at level fails to reject the null hypothesis, the confidence interval for will contain 0 .,Inference for Means: Paired Data,Test About a Population Mean,Comparative studies are more convincing than single-sam

43、ple investigations. For that reason, one-sample inference is less common than comparative inference. Study designs that involve making two observations on the same individual, or one observation on each of two similar individuals, result in paired data.,When paired data result from measuring the sam

44、e quantitative variable twice, as in the job satisfaction study, we can make comparisons by analyzing the differences in each pair. If the conditions for inference are met, we can use one-sample t procedures to perform inference about the mean difference d. These methods are sometimes called paired

45、t procedures.,Paired t Test Researchers designed an experiment to study the effects of caffeine withdrawal. They recruited 11 volunteers who were diagnosed as being caffeine dependent to serve as subjects. Each subject was barred from coffee, colas, and other substances with caffeine for the duratio

46、n of the experiment. During one two-day period, subjects took capsules containing their normal caffeine intake. During another two-day period, they took placebo capsules. The order in which subjects took caffeine and the placebo was randomized. At the end of each two-day period, a test for depressio

47、n was given to all 11 subjects. Researchers wanted to know whether being deprived of caffeine would lead to an increase in depression.,Tests About a Population Mean,State: If caffeine deprivation has no effect on depression, then we would expect the actual mean difference in depression scores to be

48、0. We want to test the hypotheses H0: d = 0 Ha: d 0 where d = the true mean difference (placebo caffeine) in depression score. Since no significance level is given, well use = 0.05.,Paired t Test,Tests About a Population Mean,Plan: If conditions are met, we should do a paired t test for d. Random re

49、searchers randomly assigned the treatment orderplacebo then caffeine, caffeine then placeboto the subjects. Normal We dont know whether the actual distribution of difference in depression scores (placebo - caffeine) is Normal. With such a small sample size (n = 11), we need to examine the data to se

50、e if its safe to use t procedures. The histogram has an irregular shape with so few values; the boxplot shows some right-skewness but not outliers; and the Normal probability plot looks fairly linear. With no outliers or strong skewness, the t procedures should be pretty accurate. Independent We arent sampling, so it isnt necessary to check the 10% condition. We will assume that the changes in depression scores for individual subjects are independent. This is reasonable if the experiment is conducted properly.,