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CHAPTER 10 MORE ON HYPOTHESIS TESTING Outline • The p-value of a test of hypothesis • The probability of a Type II error 1 THE p-VALUE • For some given sample data the null hypothesis may be rejected or not rejected depending on the significance level. • If the significance level is large, the rejection region is large. So, the null hypothesis is more likely to be rejected. • If the significance level is small, the rejection region is small. So, the null hypothesis is less likely to be rejected. • So, a natural question is for how large a level of significance, the null hypothesis will be rejected and for how small a level of significance the null hypothesis will not be rejected. • The p-value answers the above question. 2 THE p-VALUE • If we know the p-value, then for any level of significance larger than the p-value, the null hypothesis will be rejected and for any level of significance smaller than the p-value, the null hypothesis will not be rejected. • The p-value of a test of hypothesis is the smallest value of that would lead to rejection of the null hypothesis. • The p-value depends on the alternative hypothesis. For the same sample data, the p-values are different for different alternative hypothesis. 3 • The p-value for a two-tail ztest Step 1: Using the sample mean, standard deviation and size of the sample, compute the test statistic: x z / n Step 2: The p-value for a twotail test is the sum of two areas - one on the right of |z| and one on the left of -|z| f(x) THE p-VALUE TWO-TAIL p-value of a two-tail z-test is the x sum of two n areas z z 4 Step 2: The p-value for a right-tail test is the area on the right of z. p-value of a right-tail z-test x f(x) • The p-value for a right-tail z-test Step 1: Using the sample mean, standard deviation and size of the sample, compute the test statistic: x z / n f(x) THE p-VALUE RIGHT-TAIL p-value of a right-tail z-test n z x z n 5 Step 2: The p-value for a lefttail test is the area on the left of z. p-value of a left-tail z-test z f(x) • The p-value for a left-tail ztest Step 1: Using the sample mean, standard deviation and size of the sample, compute the test statistic: x z / n f(x) THE p-VALUE LEFT-TAIL p-value of a left-tail z-test x n x z n 6 THE p-VALUE TWO-TAIL Example 7: A machine that produces ball bearings is set so that the average diameter is 0.60 inch. In a sample of 49 ball bearings, the mean diameter was found to be 0.61 inch. Assume that the standard deviation is 0.035. Can we conclude that the mean diameter is not 0.60 inch? State the null hypothesis and alternative hypothesis. Compute and interpret the p-value of the test of hypothesis. 7 THE p-VALUE TWO-TAIL 1. State H0 and HA HO : HA : x 2. Find test statistic z / n 3. Compute the p-value f(x) x n 4. Interpret the p-value 8 THE p-VALUE TWO-TAIL Example 8: A machine that produces ball bearings is set so that the average diameter is 0.60 inch. In a sample of 49 ball bearings, the mean diameter was found to be 0.61 inch. Assume that the standard deviation is 0.035. Can we conclude that the mean diameter is not 0.60 inch? Consider following values of level of significance: 1. = 1% 2. = 3% n 3. = 5% 4. = 10% Use the p-value from Example 7. f(x) x 9 THE p-VALUE RIGHT-TAIL Example 9: A random sample of 100 observations from a normal population whose standard deviation is 50 produced a mean of 145. Does this statistic provide sufficient evidence at the 5% significance level to infer that population mean is more than 140? State the null hypothesis and alternative hypothesis. Compute and interpret the p-value of the test of hypothesis. 10 THE p-VALUE RIGHT-TAIL 1. State H0 and HA HO : HA : x 2. Find test statistic z / n 3. Compute the p-value f(x) x n 4. Interpret the p-value 11 THE p-VALUE LEFT-TAIL Example 10: A random sample of 100 observations from a normal population whose standard deviation is 50 produced a mean of 145. Does this statistic provide sufficient evidence at the 5% significance level to infer that population mean is less than 150? State the null hypothesis and alternative hypothesis. Compute and interpret the p-value of the test of hypothesis. 12 THE p-VALUE LEFT-TAIL 1. State H0 and HA HO : HA : x 2. Find test statistic z / n 3. Compute the p-value f(x) x n 4. Interpret the p-value 13 TYPE II ERROR • There are two types of errors in a test of hypothesis • The level of significance, sets a limit on the Type I error. • A Type I error is committed when the null hypothesis is actually true, but the sample data leads to the rejection of the null hypothesis. • The Type II error is the opposite of Type I error. • A Type II error is committed when the null hypothesis is actually false, but the sample data does not lead to the rejection of the null hypothesis. 14 TYPE II ERROR • The probability of committing a Type II error is denoted by . • Example: consider the manufacturer of the packaged cereal again. Each cereal box is expected to have a net weight of 100 gm. But, due to some problems in the production system, the average weight is shifted to 98 gm. A Type II error is committed if a sample is collected with average weight nearly 100 gm. Notice that in such a case, the problem with the production system will not be detected by the sample! 15 TYPE II ERROR • Given • The Type II error, is obtained as follows – Null and alternative hypothesis, H0 and – Find the rejection region and HA state it in terms of x-values explained in the next 3 slides – Level of significance, – Type II error, is the probability that the sample – Actual population mean will not lie in the mean, rejection region when the – Standard deviation, actual population mean is , – Sample size n standard deviation is and a random sample of size n is 16 drawn TYPE II ERROR TWO-TAIL Rejection Region x H O z / 2 and n x H O z / 2 n Type II Error z x HO /2 n P x H O z / 2 n 17 TYPE II ERROR RIGHT-TAIL Rejection Region x H O z n Type II Error P x H O z n 18 TYPE II ERROR LEFT-TAIL Rejection Region x H O z n Type II Error P x H O z n 19 TYPE II ERROR PROBABILITY TWO-TAIL Example 11: A machine that produces ball bearings is set so that the average diameter is 0.60 inch. In a sample of 49 ball bearings, the mean diameter was found to be 0.61 inch. Assume that the standard deviation is 0.035. Can we conclude that the mean diameter is not 0.60 inch? If the probability of the of a Type I error is chosen to be 0.05 (in other words, the level of significance, of the test of hypothesis is 0.05), calculate the probability of a Type II error, assuming that the true population mean is 0.605 inch. (Notice that the sample mean of 0.61 inch is irrelevant) 20 TYPE II ERROR PROBABILITY TWO-TAIL 1. State H O : HA : 2. Find rejection region: 3. Convert rejection region in terms of x-values x H O z / 2 , x H O z / 2 n n f(x) x n 4. Find 21 TYPE II ERROR PROBABILITY RIGHT-TAIL Example 12: A random sample of 100 observations from a normal population whose standard deviation is 50 produced a mean of 145. A hypothesis test is conducted to see if this statistic provides sufficient evidence at the 5% significance level to infer that population mean is more than 140. Calculate the probability of a Type II error of the test of hypothesis, assuming that the true population mean is 142. (Notice that the sample mean of 145 is irrelevant) 22 TYPE II ERROR PROBABILITY RIGHT-TAIL 1. State H O : HA : 2. Find rejection region: 3. Convert rejection region in terms of x-values x H O z n f(x) x n 4. Find 23 TYPE II ERROR PROBABILITY LEFT-TAIL Example 13: A random sample of 100 observations from a normal population whose standard deviation is 50 produced a mean of 145. A hypothesis test is conducted to see if this statistic provide sufficient evidence at the 5% significance level to infer that population mean is less than 150. Calculate the probability of a Type II error of the test of hypothesis, assuming that the true population mean is 147. (Notice that the sample mean of 145 is irrelevant) 24 TYPE II ERROR PROBABILITY LEFT-TAIL 1. State H O : HA : 2. Find rejection region: 3. Convert rejection region in terms of x-values x H O z n f(x) x n 4. Find 25 READING AND EXERCISES • Sections 10.4-10.5: – Reading: pp. 346-359 – Exercises: 10.14,10.16,10.20, 10.22, 10.24, 10.32 26

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