Transcript Chapter 9
Chapter 9 Part C III. One-Tailed Tests B. P-values • Using p-values is another approach to conducting a hypothesis test, yielding the same result. • In general: a p-value is the probability of obtaining a sample result that is at least as unlikely as what is observed. • For example, suppose you calculate a z-score of negative 2.80. A diagram The standard normal probability table tells us that there is a .0026 probability of observing a sample mean less than or equal to what was observed. P-value is P=.0026 .4974 Z=-2.80 0 Z How do you use it? •If you’re testing the hypothesis at the 99% level, then =.01. •If p-value< , reject Ho. =.01 P-value is P=.0026 0 Z=-2.80 Z=-2.33 Z Result of the test. • So in this example, .0026<.01, so we reject Ho. • Had we been using our previous methodology of comparing the test statistic to the critical value, we would get the same decision. • Since Z=-2.80 is greater (absolute value) than Z=.01=-2.33, we would reject Ho. An Example Ho: 10 Ha: < 10 n=36, x =8, s=5, =.01, Z=.01=-2.33 Test statistic, Z=(-2)/(5/6)=-2.4 p-value for a Z=-2.4 is .0082 and since this is less than =.01, reject the null Ho. MAKE SURE YOU CAN DO THIS! IV. Two-Tailed Tests about : (large sample) • The big important gold star difference when you do a two-tailed test is that the rejection range is split equally in each tail of the sampling distribution. • If you get a test statistic that is either too low, or too high, you will reject Ho. A. An Example • A lathe is set to cut bars of steel into perfect lengths of 6 centimeters. If the bars are anything but 6 cm, our customers will incur substantial costs in resizing the bars and will find other suppliers. Ho: = 6 cm Ha: 6 cm The rejection range If you are testing this hypothesis at the =.05 level of significance, then .025 (/2) is in each tail. .025 .025 .4750 .4750 =6 x Critical Z and Test Statistic • Find the z-score that is associated with .4750, and you’ll find Z=.025=±1.96. • In a sample of 121 bars, you find a sample mean of 6.08 cm with a standard deviation of .44 cm. Z 6.08 6 .44 / 121 2.0 • With a test statistic of 2.0, you would reject Ho and conclude, with 95% confidence, that the bars are not exactly 6 cm. B. P-values for 2-tailed tests • Because the significance level () is split between each tail of the sampling distribution, so is the pvalue. • The p-value is reported as 2 times the area in one tail. • If 2*(area in tail) < , reject Ho. Example • In the previous example, Z=2.0. The area in the tail, beyond Z=2.0 is (.5-4772)=.0228 • The p-value = 2(.0228) = .0456, and since this is less than =.05, reject Ho. /2=.025 p=.0228 /2=.025 0 Z C. Confidence Intervals Recall, that with a (1- ) confidence coefficient, a confidence interval is constructed as: x z /2 If you have a 2-tailed hypothesis test like: Ho: = 0 Ha: 0 n If you take a sample, and your confidence interval includes the hypothesized value of 0, then you cannot reject the null. If your 0 is outside of the confidence interval, reject the null. Back to our example Our sample mean was 6.08 cm, with a standard error of (.44/11)=.04. So our 95% confidence interval would look like: 6.08 1.96(.04) or a range between 6.0016 and 6.1584. Since 6 cm is NOT in the interval, we reject Ho.