Page 342 - The p-value approach to hypothesis testing: Two-tailed tests

1. Page 342 - The p-value approach to hypothesis testing: (Two-tailed tests) In recent years? widely available statistical software? alternative approach to hypothesis testing. The p-value is the probability of obtaining a test statistic equal to or more extreme than the result obtained from the sample data, given that the null hypothesis Ho is true. Also called the observed level of significance It is the smallest level at which Ho can be rejected for a given set of data. p-value ? ?, fail to reject Ho p-value < ? , reject Ho Easy for the normal distribution ? but the computation of the p-value can be very difficult for other distributions. Excel routinely present the p-value as part of the output for many hypothesis-testing procedures. P343 Phstat | One-Sample Tests | Z test for the mean, ? known

3. Page 345 Confidence intervals estimation verse hypothesis testing Two major components of statistical inference Same set of concepts, use them for different purposes. In many situations we can use confidence intervals to do a test of a null hypothesis. If the hypothesized value of ? falls into the confidence interval, the null hypothesis would not be rejected. If the hypothesized value of ? does not fall into the confidence interval, the null hypothesis would be rejected, because it would then be considered an unusual value. Cereal box filling example: Ho = 368, ? = 15. n = 25, X-bar = 372.5, ? =.05, two tailed error confidence interval estimate of ?. ? Xbar?Z(?/n�) 372.5 ? 5.88 95% CI of ? ? 366.62 ?? ? 378.38 95% CI includes Ho ? fail to reject

4. E7.3 p343 Hypothesis test steps 1. State the null hypothesis, Ho. (stated in statistical terms) 2. State the alternative hypothesis, H1. 3. Choose the level of significance, ?. (This along with the sample size, determines �) 4. Choose the sample size, n. Sample size is determined after taking into account the specified risks of committing Type I and Type II errors i.e., selected levels of ? and � 5. Determine the appropriate statistical technique and corresponding test statistic to use. (? known ? Z test) (? unknown ? t test) 6. Collect the data and compute the sample value of the appropriate test statistic. 7. Calculate the p-value based on the test statistic. 8. Compare the p-value to ? 9. Make the statistical decision reject Ho if p-value <= ? fail to reject Ho if p-value > ? 10. Express the statistical decision in terms of the problem. 11. Put it all in a picture.

5. Book example p343. Is the population mean weight of cereal per box different from at 368 grams? 1) Ho: ? = 368 grams 2) H1: ? ? 368 grams 3) ? = .05 4) n = 25 5) ? = 15 grams, so ? is known, use Z-test, normal distribution. 6) Collect data and calculate test statistic X-bar = 372.5 grams. e7.1, p340 Z = (Xbar - ?)/(?/n�) = +1.50 7) Calculate the p-value Phstat | One-Sample Tests | Z test for the mean, ? known Null Hypothesis ?= 368 Level of Significance = .05 Population Standard Deviation = 15 Sample Size = 25 Sample Mean = 372.5 Two Tailed Test Output results p-Value = .1336 8) p-Value = .1336 > ? = .05 9) Barely fail to reject Ho. Borderline situation. There is no evidence that the mean is significantly different from the hypothesized value. 10) Do not fix the machine. We have no evidence that the population mean is significantly different from 368 grams. 11) Picture, f7.4 p343 Homework 7.30, 7.26, Solutions on the web site.

6. 7.3 One-tailed tests Ho = , H1 ? , ? contained two possibilities In some situations the alternative hypothesis focuses in a particular direction also called directional test ? entire rejection region is contained in one tail of the sampling distribution Upper tail - reject Ho when the sample mean is significantly above(greater than) ? Lower tail - reject Ho when the sample mean is significantly below (less than)? Chief financial officer (CFO) would be interested in filling above 368 grams Ho: ? ? 368, H1: ?>368 ? = 15, n = 25, X-bar = 372.5, ? =.05, one-tailed error, so CV=+1.645, Z still = +1.50 We would conclude that there is no evidence that the average fill per box is above 368 grams. The result from the sample is deemed due to chance or sampling error, it is not statistically significant. One tail with ? = .01 ? CV=2.33 One tail with ? = .10 ? CV=1.28

7. p-value for one tailed test Depending on the direction of the alternative hypothesis: Compute the probability of obtaining a value greater than the computed test statistic, or Compute the probability of obtaining a value less than the computed test statistic Ho: ? ? 368, H1: ?>368 X-bar = 372.5 Z = +1.50 1.50 to the normal distribution table ? .4332 p = .5000 - .4332 = .0668 So... the probability of obtaining a value greater than 372.5 is .0668 (or 6.68 percent of the time). In Class: GMAT example Homework: 7.40 and others on web page

8. GMAT example. The Director of Admissions at MSU believes that their MBA students are above the national average. The population average GMAT score is 500 with a population standard deviation of 100. A sample of 12 MSU MBA students is selected at random. The sample mean is 537. Use a level of significance of .01. 1) Ho: ? <= 500 points 2) H1: ? > 500 points 3) ? = .01 4) n = 12 5) ? = 100 points, so ? is known, use Z-test, normal distribution. 6) Collect data and calculate test statistic X-bar = 537 grams. e7.1, p340 Z = (Xbar - ?)/(?/n�) = Z = (537 - 500) / (100 / 12�) = +1.28 7) Calculate the p-value Phstat | One-Sample Tests | Z test for the mean, ? known Null Hypothesis ?= 500 Level of Significance = .01 Population Standard Deviation = 100 Sample Size = 12 Sample Mean = 537 Upper Tailed Test Output results p-Value = .09997 8) p-Value = .09997 > ? = .01 9) Barely fail to reject Ho. Borderline situation. There is no evidence that the mean is significantly greater than the hypothesized value. 10) The Director of Admission is incorrect. We have no evidence that the population mean is significantly greater than 500 points. 11) Picture:

9. Review Critical Values - Normal Distribution t distribution, each one is different p-value analysis - very subjective p-value >.2 ? Not close to rejecting. Sample statistic is close to the hypothesized mean. A strong fail to reject. p-value < .009 ? Little chance of failing to reject. Sample statistic is far from the hypothesized mean. Sample statistic is way out in the tail. A strong rejection. .009 < p-value < .2 ? Boarder line situation. Barely rejecting or barely failing to reject. When p value and ? are similar.

10. 7.4 t-test of hypothesis for the mean (? Unknown) In most hypothesis-testing situations, the standard deviation of the population (?) is unknown. It is estimated by S, the standard deviation of the sample. If the population is assumed to be normally distributed ? the sampling distribution of the mean will follow a t distribution with n-1 degrees of freedom. In practice ? As long as the sample size is not very small and the population is not very skewed. E7.2 p350 ? t = (Xbar - ?)/(S/n�) ? is the test statistic for determining the difference between the sample mean X-bar and the population mean ?.

11. Assumptions of the one-sample t Test (review) Each of the rows in T.E3 corresponds to a particular t distribution. The one-sample t test is considered a classical parametric procedure. Assumptions required ? a random sample from a population that is normally distributed Need to determine how closely the actual data match the normal distribution�s theoretical properties. We can use a stem-and-leaf display, a box-and-whisker plot, and a normal probability plot (p357) If the sample size is small (n less than 30) and/or we can not meet the normality assumption ? Distribution-free testing procedures are likely to be more powerful Two excel files on web page to calculate p-values for the t-distribution: p-value_t_given_info.xls p-value_t_rawdata.xls In class 7.50 p356 Cost of Books, Coin-Op Laundry Homework 7.54 and others on web page.

14. 7.5 One sample Z test for the proportion The sample proportion ( ps=X/n ) is compared to the hypothesized value of the parameter, p Evaluate the magnitude of the difference between ps and p ? e7.3 p357 ? Z test statistic Evaluate the magnitude of the difference between the # of successes in the sample and the hypothesized # of successes in the population ? e7.4 p358 ? Z test statistic e7.3 and e7.4 will provide the same result Z test statistic will be approximately normally distributed if the sample size is large enough ? np?5 and n(1-p)?5 We will not work any problems of this type now, but we will work some in chapter 10, two sample with categorical data.

15. 7.6 Pitfalls and Ethical Issues Fundamental hypothesis testing methodology. It is used for analyzing differences between sample estimates (i.e., statistics) of hypothesized population characteristics (i.e., parameters). It is used to make decisions. Ensure proper methodology: 1. What is the goal? 2. Two-tailed or one-tailed? 3. Can a random sample be drawn? 4. Measurements - numerical or categorical? 5. Significance level - ? 6. Sample size vs. power given ? 7. What statistical test - normal or t 8. Conclusions and interpretations A person with substantial statistical training should be consulted and involved early in the process. Not after the data have been collected.

16. Data collection method - many issues, self-selecting subjects Human subjects - consent Type of test - Prior information leads to a specifically directed one-tailed test. Interested only in differences then use a two-tailed test. In the overwhelming majority of research studies, a two-tailed test should be employed. Level of significance - ? is selected in advance of data collection. Solution ? always report the p-value Data snooping - steps must be done first before the data are collected. Cleansing and discarding of data - flagging outlier observations. Look at the displays. Should an observation be removed from a study? Reporting of findings - document both good and bad results Next up�Exam 2, same format, 20 MC, problems Ethical Issues