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7 Hypothesis Testing Elementary Statistics Larson Farber Section 7.1 Introduction to Hypothesis Testing A statistical hypothesis is a claim about a population. Null hypothesis H0 contains a statement of equality such as , = or . Alternative hypothesis Ha contains a statement of inequality such as < , or > Complementary Statements If I am false, you are true If I am false, you are true Writing Hypotheses Write the claim about the population. Then, write its complement. Either hypothesis, the null or the alternative, can represent the claim. A hospital claims its ambulance response time is less than 10 minutes. claim A consumer magazine claims the proportion of cell phone calls made during evenings and weekends is at most 60%. claim Hypothesis Test Strategy Begin by assuming the equality condition in the null hypothesis is true. This is regardless of whether the claim is represented by the null hypothesis or by the alternative hypothesis. Collect data from a random sample taken from the population and calculate the necessary sample statistics. If the sample statistic has a low probability of being drawn from a population in which the null hypothesis is true, you will reject H0. (As a consequence, you will support the alternative hypothesis.) If the probability is not low enough, fail to reject H0. Decision Errors and Level of Significance Actual Truth of H0 Do not reject H0 Reject H0 H0 True Correct Decision Type I Error H0 False Type II Error Correct Decision A type I error: Null hypothesis is actually true but the decision is to reject it. Level of significance, Maximum probability of committing a type I error. Types of Hypothesis Tests Ha is more probable Ha is more probable Ha is more probable Right-tail test Left-tail test Two-tail test P-values The P-value is the probability of obtaining a sample statistic with a value as extreme or more extreme than the one determined by the sample data. P-value = indicated area Area in left tail Area in right tail z z For a right tail test For a left tail test If z is negative, twice the area in the left tail If z is positive, twice the area in the right tail z z For a two-tail test Finding P-values: 1-tail Test The test statistic for a right-tail test is z = 1.56. Find the P-value. Area in right tail z = 1.56 The area to the right of z = 1.56 is 1 – .9406 = 0.0594. The P-value is 0.0594. Finding P-values: 2-tail Test The test statistic for a two-tail test is z = –2.63. Find the corresponding P-value. z = –2.63 The area to the left of z = –2.63 is 0.0043. The P-value is 2(0.0043) = 0.0086. Test Decisions with P-values The decision about whether there is enough evidence to reject the null hypothesis can be made by comparing the P-value to the value of the level of significance of the test. If If , reject the null hypothesis. fail to reject the null hypothesis. Using P-values The P-value of a hypothesis test is 0.0749. Make your decision at the 0.05 level of significance. Compare the P-value to . Since 0.0749 > 0.05, fail to reject H0. If P = 0.0246, what is your decision if 1) Since , reject H0. 2) Since 0.0246 > 0.01, fail to reject H0. Interpreting the Decision Claim Decision Claim is H0 Reject H0 Fail to reject H0 Claim is Ha There is enough evidence to reject the claim. There is enough evidence to support the claim. There is not enough evidence to reject the claim. There is not enough evidence to support the claim. Steps in a Hypothesis Test 1. Write the null and alternative hypothesis. Write H0 and Ha as mathematical statements. Remember H0 always contains the = symbol. 2. State the level of significance. This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.) 3. Identify the sampling distribution. The sampling distribution is the distribution for the test statistic assuming that the equality condition in H0 is true and that the experiment is repeated an infinite number of times. 4. Find the test statistic and standardize it. Perform the calculations to standardize your sample statistic. 5. Calculate the P-value for the test statistic. This is the probability of obtaining your test statistic or one that is more extreme from the sampling distribution. 6. Make your decision. If the P-value is less than (the level of significance) reject H0. If the P value is greater , fail to reject H0. 7. Interpret your decision. If the claim is the null hypothesis, you will either reject the claim or determine there is not enough evidence to reject the claim. If the claim is the alternative hypothesis, you will either support the claim or determine there is not enough evidence to support the claim. Section 7.2 Hypothesis Testing for the Mean (n 30) The z-Test for a Mean The z-test is a statistical test for a population mean. The z-test can be used: (1) if the population is normal and s is known or (2) when the sample size, n, is at least 30. The test statistic is the sample mean and the standardized test statistic is z. When n 30, use s in place of . The z-Test for a Mean (P-value) A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim? 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.05 3. Determine the sampling distribution. Since the sample size is at least 30, the sampling distribution is normal. 4. Find the test statistic and standardize it. n = 52 s = 10 Test statistic 5. Calculate the P-value for the test statistic. Since this is a right-tail test, the P-value is the area found to the right of z = 1.44 in the normal distribution. From the table P = 1 – 0.9251 P = 0.0749. Area in right tail z = 1.44 6. Make your decision. Compare the P-value to . Since 0.0749 > 0.05, fail to reject H0. 7. Interpret your decision. There is not enough evidence to reject the claim that the mean sodium content of one serving of its cereal is no more than 230 mg. Rejection Regions Sampling distribution for Rejection Region z z0 Critical Value z0 The rejection region is the range of values for which the null hypothesis is not probable. It is always in the direction of the alternative hypothesis. Its area is equal to . A critical value separates the rejection region from the non-rejection region. Critical Values The critical value z0 separates the rejection region from the non-rejection region. The area of the rejection region is . Rejection region Rejection region z0 z0 Find z0 for a right-tail test with = .05. Find z0 for a left-tail test with = .01. z0 = –2.33 Rejection region z0 Rejection region z0 Find –z0 and z0 for a two-tail test with z0 = 1.645 –z0 = –2.575 and z0 = 2.575 = .01. Using the Critical Value to Make Test Decisions 1. Write the null and alternative hypothesis. Write H0 and Ha as mathematical statements. Remember H0 always contains the = symbol. 2. State the level of significance. This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.) 3. Identify the sampling distribution. The sampling distribution is the distribution for the test statistic assuming that the equality condition in H0 is true and that the experiment is repeated an infinite number of times. 4. Find the critical value. Rejection Region z0 6. Find the test statistic. 5. Find the rejection region. The critical value separates the rejection region of the sampling distribution from the non-rejection region. The area of the critical region is equal to the level of significance of the test. Perform the calculations to standardize your sample statistic. 7. Make your decision. If the test statistic falls in the critical region, reject H0. Otherwise, fail to reject H0. 8. Interpret your decision. If the claim is the null hypothesis, you will either reject the claim or determine there is not enough evidence to reject the claim. If the claim is the alternative hypothesis, you will either support the claim or determine there is not enough evidence to support the claim. The z-Test for a Mean A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim? 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.05 3. Determine the sampling distribution. Since the sample size is at least 30, the sampling distribution is normal. Since Ha contains the > symbol, this is a right-tail test. Rejection region z0 1.645 4. Find the critical value. 5. Find the rejection region. 6. Find the test statistic and standardize it. n = 52 = 232 s = 10 7. Make your decision. z = 1.44 does not fall in the rejection region, so fail to reject H 0 8. Interpret your decision. There is not enough evidence to reject the company’s claim that there is at most 230 mg of sodium in one serving of its cereal. Using the P-value of a Test to Compare Areas = 0.05 z0 = –1.645 Rejection area 0.05 z0 z = –1.23 P = 0.1093 z For a P-value decision, compare areas. If reject H0. If fail to reject H0. For a critical value decision, decide if z is in the rejection region If z is in the rejection region, reject H0. If z is not in the rejection region, fail to reject H0. Section 7.3 Hypothesis Testing for the Mean (n < 30) The t Sampling Distribution Find the critical value t0 for a left-tailed test given and n = 18. = 0.01 d.f. = 18 – 1 = 17 t0 = –2.567 Area in left tail t0 Find the critical values –t0 and t0 for a two-tailed test given = 0.05 and n = 11. –t0 = –2.228 and t0 = 2.228 d.f. = 11 – 1 = 10 t0 t0 Testing –Small Sample A university says the mean number of classroom hours per week for full-time faculty is 11.0. A random sample of the number of classroom hours for full-time faculty for one week is listed below. You work for a student organization and are asked to test this claim. At = 0.01, do you have enough evidence to reject the university’s claim? 11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1 1. Write the null and alternative hypothesis 2. State the level of significance = 0.01 3. Determine the sampling distribution Since the sample size is 8, the sampling distribution is a t-distribution with 8 – 1 = 7 d.f. Since Ha contains the ≠ symbol, this is a two-tail test. 4. Find the critical values. 5. Find the rejection region. –t0 –3.499 t0 3.499 6. Find the test statistic and standardize it n=8 = 10.050 s = 2.485 7. Make your decision. t = –1.08 does not fall in the rejection region, so fail to reject H0 at 8. Interpret your decision. There is not enough evidence to reject the university’s claim that faculty spend a mean of 11 classroom hours. = 0.01 Minitab Solution Enter the data in C1, ‘Hours’. Choose t-test in the STAT menu. T-Test of the Mean Test of = 11.000 vs Variable N Hours 8 Mean 0.050 not = 11.000 StDev 2.485 SE Mean T 0.879 –1.08 Minitab reports the t-statistic and the P-value. Since the P-value is greater than the level of significance (0.32 > 0.01), fail to reject the null hypothesis at the 0.01 level of significance. P 0.32 Section 7.4 Hypothesis Testing for Proportions Test for Proportions p is the population proportion of successes. The test statistic is . (the proportion of sample successes) If and the sampling distribution for The standardized test statistic is: is normal. Test for Proportions A communications industry spokesperson claims that over 40% of Americans either own a cellular phone or have a family member who does. In a random survey of 1036 Americans, 456 said they or a family member owned a cellular phone. Test the spokesperson’s claim at = 0.05. What can you conclude? 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.05 3. Determine the sampling distribution. 1036(.40) > 5 and 1036(.60) > 5. The sampling distribution is normal. Rejection region 4. Find the critical value. 5. Find the rejection region. 1.645 6. Find the test statistic and standardize it. n = 1036 x = 456 7. Make your decision. z = 2.63 falls in the rejection region, so reject H0 8. Interpret your decision. There is enough evidence to support the claim that over 40% of Americans own a cell phone or have a family member who does. Section 7.5 Hypothesis Testing for Variance and Standard Deviation Critical Values for s2 is the test statistic for the population variance. Its sampling distribution is a c2 distribution with n – 1 d.f. Find a c20 critical value for a left-tail test when n = 17 and = 0.05. c20 = 7.962 Find critical values c20 for a two-tailed test when n = 12, = 0.01. c2L = 2.603 and c2R = 26.757 The standardized test statistic is Test for A state school administrator says that the standard deviation of test scores for 8th grade students who took a life-science assessment test is less than 30. You work for the administrator and are asked to test this claim. You find that a random sample of 10 tests has a standard deviation of 28.8. At = 0.01, do you have enough evidence to support the administrator’s claim? Assume test scores are normally distributed. 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.01 3. Determine the sampling distribution. The sampling distribution is c2 with 10 – 1 = 9 d.f. 4. Find the critical value. 5. Find the rejection region. 2.088 6. Find the test statistic. n = 10 s = 28.8 7. Make your decision. c2 = 8.2944 does not fall in the rejection region, so fail to reject H0 8. Interpret your decision. There is not enough evidence to support the administrator’s claim that the standard deviation is less than 30.