#### Transcript LGM #12 - Penn State Statistics Department

Presentation 11 More About Significance Tests Inference Techniques Confidence Intervals: 1. 1-Proportion 2. 1-Mean 3. Difference between 2-Proportions 4. Difference between 2-Means Hypothesis Tests: 1. 1-Proportion 2. 1-Mean 3. Difference between 2-Means 4. Difference between 2-Proportions 5. Difference between Paired Means. 6. Chi-Square (Relationship between 2 Categorical Variables) 7. Regression (Relationship between 2 Quantitative Variables) 8. ANOVA (Difference between 3 or More Means) 9. Median Tests General Steps for Hypothesis Tests Step 1: Determine the H0 and Ha. The alternative hypothesis, Ha, is the claim regarding the population parameter that we want to test. There are three possible Ha's - the parameter is not equal to a null value (two-sided), less than a null value (one-sided) or greater than a null value (one-sided). The null hypothesis, H0, claims that nothing is happening. H0 can be the opposite of Ha or just that the parameter is equal to the null value. Step 2: Verify necessary data conditions, if met summarize the data into an appropriate test statistic. The conditions to be verified are the same conditions we have seen in Chapter 12 for creating C.I's. The test statistic is a standardized statistic, i.e. is of the form Sample statistic - Null value Null s.e (sample statistic) Under H0 the test statistic follows either a normal distribution (proportion cases) or a t-distribution with some d.f (mean cases). General Steps for Hypothesis Tests Step 3: Find the p-value. The p-value is the probability that the test statistic is as large or larger in the direction(s) specified from Ha assuming that H0 is true. This, to get the p-value you need the form of Ha, the value of the test statistic, and the distribution of the test statistic under the H0. Step 4: Decide whether or not the result is statistically significant. The results are statistically significant if the p-value is less than alpha, where alpha is the significance level (usually 0.05). Based on the pvalue we have two possible conclusions If p-value < α we reject the null and we claim the alternative is true. If p-value > α we fail to reject the null, we claim that there isn't enough evidence to report Ha is true. In this case, we do not claim that the null is true! Step 5: Report the conclusion in the context of the situation. Finally, write one or two sentences explaining what is the conclusion in terms of the problem. One sample t-test (one mean or paired data) In this case we consider the following hypotheses: H a : µ µ0 H0 : µ = µ0 vs one of the following Ha : µ > µ0 Ha : µ < µ0 We have the following t-test statistic t Sample estimate- Null value x 0 s Null Std. Error n Assuming that H0 is true, the test statistic has a t-distribution with (n1) degrees of freedom, if one of the following conditions is true: 1. the random variable of interest is bell-shaped (in practice, for small samples the data should show no extreme skewness or outliers). 2. the random variable is not bell-shaped, but a large random sample is measured, n ≥ 30. One sample t-test (one mean or paired data) If we denote with T a random variable with t-distribution with (n-1) d.f, and t is the value of out test statistic, then Ha µ < µ0 µ > µ0 µ µ0 p-value P( T < t ) P( T > t ) 2P( T > |t| ) We can find the p-value using the t-table (table A3), and draw a conclusion about the hypotheses in the usual manner. Two-Sample t-test (difference between two means). There are two populations (Population 1 and 2) of interest having unknown means µ1 and µ2 respectively. In this case we consider the following hypotheses: H a : µ1 - µ2 < 0 H0 : µ1 - µ2 = 0 (i.e. no difference) vs one of the following Ha : µ1 - µ2 > 0 H a : µ1 - µ2 0 We have the following t-test statistic (x x ) 0 t 1 2 2 2 s1 s2 n1 n2 Assuming H0 is true, the test statistic has a t-distribution with approximate d.f. equal to the minimum of n1-1 and n2-1 if the following conditions are true: 1. 2. The two samples are independent. Each sample is either coming from a bell shaped population or the sample size is ≥30. Using the appropriate d.f, we can get the p-value from table A.3 in the same way as in the one-mean case. Difference between two proportions Now we will consider two populations having some unknown proportions of interest, p1 and p2 respectively. In this case we consider the following hypotheses: Ha : p1 - p2 < 0 H0 : p1 - p2 = 0 (i.e. no difference) vs one of the following Ha : p1 - p2 > 0 Ha : p1 - p2 0 ˆ1 pˆ 2 . To get the null standard The sample statistic we will use is p error of the statistic we assume that H0 is true, so let p1 = p2 =p. Thus, ˆ1 and pˆ 2 in the s.e formula, we will substitute them instead of using p with an estimate of the common population proportion. That is, we use the proportion in all available data n1 pˆ 1 n2 pˆ 2 pˆ n1 n2 and we have the null s.e. pˆ (1 pˆ ) pˆ (1 pˆ ) n1 n2 Difference between two proportions Thus, we have the following t-test statistic z pˆ1 pˆ 2 pˆ (1 pˆ ) pˆ (1 pˆ ) n1 n2 pˆ1 pˆ 2 1 1 pˆ (1 pˆ ) n1 n2 Assuming H0 is true, the test statistic has a standard normal distribution if the following conditions are true: 1. The two samples are independent. 2. All the quantities n1 pˆ 1, n1(1 pˆ 1), n2 pˆ 2 and n2(1 pˆ 2) are at least 5 and preferably at least 10. We can obtain the p-value using the standard normal table A1. If we denote with Z a random variable with standard normal distribution, and z is the value of out test statistic, then Ha p-value p 1 - p2 < 0 P( Z < z ) p 1 - p2 > 0 P( Z > z ) p1 - p2 0 2P( Z > |z| ) Table For Hypothesis Testing Type One Mean (or Paired mean) . = Difference Between Means One Proportion Parameter Statistic µ or µd x or d µ1- µ2 x1 x 2 p pˆ Null Std. Error s or n sd n s12 s 2 2 n1 n 2 po(1 po) n pˆ1 pˆ 2 pˆ (1 pˆ ) pˆ (1 pˆ ) n1 n2 Test Statistic t df = n-1 t df=min(n1-1,n2-1) z Difference Between Proportions p1-p2 n1 pˆ 1 n2 pˆ 2 pˆ = n1 n2 This is the overall proportion-like if we had one big sample! pˆ 1 pˆ 2 z Detailed Example 1 A box of corn flakes is advertised as containing 16 oz. of cereal. John works for consumer reports and is interesting in verifying the companies claim. He suspects that the average amount of cereal is less than 16oz. He takes a random sample of 100 boxes of cereal and records the weight of each box. The sample mean is 15.7oz and the sample standard deviation is 1.2oz. Example 1: Overview Before you carry out the hypothesis test it is a good idea to label the information you have. Since there is 1 quantitative variable we are doing a hypothesis test of 1-mean. The parameter of interest is µ, the mean weight of cereal in all corn flakes boxes. The statistic of interest is x = 15.7oz, the sample mean weight. Other information: The sample size n = 100, and the sample standard deviation s = 1.2oz Example 1: Hypotheses and Conditions 1. Define the null and alternative hypotheses: Ho: µ = 16oz Ha: μ < 16oz 2. Check conditions: Either A or B A) The distribution of weights is normal. B) The sample size is ≥ 30. Example 1: Test Statistic and P-Value 3. Calculate the test statistic. Test Statistic Sample Statistic Null Value Null Standard Error Null Value = 16, Sample Statistic = 15.7 s Null Std. Error = = 1.2/10 = .12 n Test Statistic = 15.7 16 2.5 .12 P-Value = P(T<-2.5) Use Table A.3! degrees of freedom = df = n-1 = 100-1 = 99 Example 1: P-value & Conclusion T-Distribution, df = 99 Density 0.2 0.1 P-Value = P(T<-2.5) 0.0 Look up the absolute value of the test statistic and df in the table. If they do not have the exact test statistic, then take your best guess OR use the next smallest test statistic. Using the next larger and next smallest value for tstatistic on the table you can get an interval for the p-value. 0.3 Table A.3 gives the one-sided p-values for t-tests. For 2-sided hypothesis tests make sure you multiply the p-value by 2! -4 -2 0 2 4 T In our case, df = 99 and t-statistic = -2.5. Since 2.5 is not in the table we can use 2.33, and 90 df (since 99 is not in the table). We get that the pvalue is less than .011. The p-value is < .05 so we REJECT the null hypothesis and we conclude that the average weight IS less than 16oz. Detailed Example 2 A potential presidential candidate in 2002 election wished to know if men and women had equal preferences for her candidacy versus this not being the case (she suspected that men were more likely to favor her). Her pollster polled a random sample of 400 men and 400 women, with the following results. Preference: Would vote for her? No Yes Total Men 164 236 400 Women 180 220 400 Total 344 556 800 Example 2: Overview Here there are 2 categorical variables, both with 2 levels so we are interested in a test of 2-proportions! Is the proportion of men who favor the candidate greater than the proportion of women who favor the candidate? The parameter of interest is pm-pw. The sample statistic of interest is pˆ m pˆ w =.59-.55 = .04 Example 2: 1. Calculate the null and alternative hypotheses: Ho: pm = pf or pm-pf = 0 Ha: pm > pf or pm-pf > 0 Let’s denote with pm with p1 and pf with p2. 2. Check the conditions: All quantities, n1 pˆ 1, n1(1 pˆ 1) and are greater than or equal to 10. n2 pˆ 2, n2(1 pˆ 2) Example 2: Test Statistic and P-Value 3. Calculate the test statistic Test Statistic z Sample Statistic Null Value Null Standard Error Null Value 0 (since theparameteis p1 p2 ) Statistic pˆ 1 pˆ 2 236/400– 220/400 0.04 n1 pˆ 1 n2 pˆ 2 236 220 pˆ 0.57 n1 n2 80 Null Std. Error pˆ (1 pˆ )1 n1 1 n2 .57(1 .57)1 400 1 400 0.035 T est Statistic ( pˆ 1 pˆ 2 ) 0 .04 1.14 pˆ (1 pˆ ) pˆ (1 pˆ ) .035 n1 n2 Example 2: P-Value and Conclusion P-Value = P(Z>1.14) = 1-P(Z<1.14) = 1-0.8729 = 0.127 P-Value is >.05 so we can NOT reject the null hypothesis. Therefore, we do NOT have enough evidence to conclude that men are more likely to favor the candidate. MINITAB Output for Example 2 Test and CI for Two Proportions Sample Men Women X 236 220 N Sample p 400 0.590000 400 0.550000 Estimate for p(1) - p(2): 0.04 Test for p(1) - p(2) = 0 (vs > 0): Z = 1.14 P-Value = 0.126 CI’s and Two-sided Alternatives When testing the hypotheses H0: parameter = null value vs. Ha: parameter null value If the null value is covered by a (1 - ) CI, the null hypothesis is not rejected and the test is not statistically significant at level . If the null value is not covered by a (1 - ) CI, the null hypothesis is rejected and the test is statistically significant at level . For instance, for a 95% Confidence Interval, (1- ) = 0.95 = 95%. So for 95% confidence, the significance level is = 0.05, which is the significance level and confidence interval used most frequently. Problem 13.38 in the Text Each of the following presents a 95% CI and the alternative hypothesis of a corresponding hypothesis test. In each case, state a conclusion for the test, including the level of significance you are using. A) CI for µ is (101 to 105), Ha: µ ≠1 00 B) CI for p is (.12 to .28), Ha: p<.10 C) CI for µ1-µ2 is (3 to 15), Ha: µ1-µ2 > 0 D) CI for p1-p2 is (-.15 to .07), Ha: p1-p2 ≠ 0 Possible Errors, Power and Sample Size Considerations When we are testing a hypothesis two out four possible decisions lead to an error. Type I error - We reject H0 when it is true Type II error- We fail to reject H0 when Ha is true. There is an inverse relationship between the probabilities of the two types of errors. Increase in the probability of type I error leads to decrease in the probability of type II error and vice versa. When the alternative hypothesis is true, the probability of making the correct decision is called power of the test. The power increases when the sample size increased. Can you see why? The power increases when the difference between the true population parameter value and the null value increases. Can you see why? The hypothesis test may have very low power because of small data set. We should also be careful in cases our conclusions are based on extremely large samples, since in cases like that even a weak relationship (or a small difference) can be statistically significant.