1. 7 Elementary Statistics Larson Farber Hypothesis Testing

2. Introduction to Hypothesis Testing Section 7.1

3. A statistical hypothesis is a claim about a population. Alternative hypothesis H a contains a statement of inequality such as Null hypothesis H 0 contains a statement of equality such as, = or. Complementary Statements If I am false, you are true If I am false, you are true

4. A consumer magazine claims the proportion of cell phone calls made during evenings and weekends is at most 60%. 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. Writing Hypotheses claim

5. 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. Hypothesis Test Strategy 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 H 0. (As a consequence, you will support the alternative hypothesis.) If the probability is not low enough, fail to reject H 0.

6. 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. Actual Truth of H 0 Errors and Level of Significance H 0 TrueH 0 False Do not reject H 0 Reject H 0 Correct Decision Correct Decision Type II Error Type I Error Decision

7. Right-tail test Two-tail test Left-tail test Types of Hypothesis Tests H a is more probable

8. 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. If z is negative, twice the area in the left tail If z is positive, twice the area in the right tail P-values P-value = indicated area zz zz Area in left tail Area in right tail For a left tail test For a right tail test For a two-tail test

9. Finding P-values: 1-tail Test The test statistic for a right-tail test is z = 1.56. Find the P-value. The area to the right of z = 1.56 is 1 –.9406 = 0.0594. The P-value is 0.0594. z = 1.56 Area in right tail

10. The test statistic for a two-tail test is z = –2.63. Find the corresponding P-value. The area to the left of z = –2.63 is 0.0043. The P-value is 2(0.0043) = 0.0086. Finding P-values: 2-tail Test z = –2.63

11. 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 fail to reject the null hypothesis. If reject the null hypothesis.

12. 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 H 0. If P = 0.0246, what is your decision if 1) Since, reject H 0. 2) Since 0.0246 > 0.01, fail to reject H 0. Using P-values

13. There is enough evidence to reject the claim. Claim Interpreting the Decision Claim is H 0 Claim is H a Reject H 0 Fail to reject H 0 Decision There is not enough evidence to reject the claim. There is enough evidence to support the claim. There is not enough evidence to support the claim.

14. 1. Write the null and alternative hypothesis. 2. State the level of significance. 3. Identify the sampling distribution. Write H 0 and H a as mathematical statements. Remember H 0 always contains the = symbol. This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.) The sampling distribution is the distribution for the test statistic assuming that the equality condition in H 0 is true and that the experiment is repeated an infinite number of times. Steps in a Hypothesis Test

15. 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.

16. If the P-value is less than (the level of significance) reject H 0. If the P value is greater, fail to reject H 0. 6. Make your decision. 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.

17. Hypothesis Testing for the Mean (n 30) Section 7.2

18. 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.

19. 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 companys 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. The z-Test for a Mean (P-value)

20. 4. Find the test statistic and standardize it. 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 n = 52 s = 10 Test statistic z = 1.44 Area in right tail P = 0.0749.

21. 6. Make your decision. 7. Interpret your decision. Compare the P-value to. Since 0.0749 > 0.05, fail to reject H 0. 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.

22. Sampling distribution for 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. Rejection Regions Rejection Region Critical Value z 0 zz0z0

23. The critical value z 0 separates the rejection region from the non-rejection region. The area of the rejection region is. Find z 0 for a left-tail test with =.01. Find z 0 for a right-tail test with =.05. Find –z 0 and z 0 for a two-tail test with =.01. z 0 = –2.33 –z 0 = –2.575 and z 0 = 2.575 z 0 = 1.645 Critical Values z0z0 z0z0 Rejection region Rejection region z0z0 z0z0 Rejection region Rejection region

24. 1. Write the null and alternative hypothesis. 2. State the level of significance. 3. Identify the sampling distribution. Write H 0 and H a as mathematical statements. Remember H 0 always contains the = symbol. This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.) The sampling distribution is the distribution for the test statistic assuming that the equality condition in H 0 is true and that the experiment is repeated an infinite number of times. Using the Critical Value to Make Test Decisions

25. 6. Find the test statistic. 5. Find the rejection region. 4. Find the critical value. 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. z0z0 Rejection Region

26. 7. Make your decision. 8. Interpret your decision. If the test statistic falls in the critical region, reject H 0. Otherwise, fail to reject H 0. 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.

27. 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 companys claim? 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. The z-Test for a Mean 1. Write the null and alternative hypothesis.

28. n = 52 = 232 s = 10 7. Make your decision. 6. Find the test statistic and standardize it. 8. Interpret your decision. 5. Find the rejection region. Rejection region Since H a contains the > symbol, this is a right-tail test. z = 1.44 does not fall in the rejection region, so fail to reject H 0 There is not enough evidence to reject the companys claim that there is at most 230 mg of sodium in one serving of its cereal. 1.645 4. Find the critical value. z0z0

29. Using the P-value of a Test to Compare Areas z0z0 Rejection area 0.05 z 0 = –1.645 z Area to the left of z 0.1093 z = –1.23 For a critical value decision, decide if z is in the rejection region If z is in the rejection region, reject H 0. If z is not in the rejection region, fail to reject H 0. = 0.05 For a P-value decision, compare areas. If reject H 0. If fail to reject H 0. P = 0.1093

30. Hypothesis Testing for the Mean (n < 30) Section 7.3

31. Find the critical value t 0 for a left-tailed test given = 0.01 and n = 18. Find the critical values –t 0 and t 0 for a two-tailed test given d.f. = 18 – 1 = 17 t 0 t 0 = –2.567 d.f. = 11 – 1 = 10 –t 0 = –2.228 and t 0 = 2.228 The t Sampling Distribution = 0.05 and n = 11. Area in left tail t 0

32. 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 universitys 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. Testing –Small Sample

33. t = –1.08 does not fall in the rejection region, so fail to reject H 0 at = 0.01 n = 8 = 10.050 s = 2.485 7. Make your decision. 6. Find the test statistic and standardize it 8. Interpret your decision. There is not enough evidence to reject the universitys claim that faculty spend a mean of 11 classroom hours. 5. Find the rejection region. Since H a contains the symbol, this is a two-tail test. 4. Find the critical values. –3.4993.499 t0t0 –t0–t0

34. T-Test of the Mean Test of = 11.000 vs not = 11.000 Variable N Mean StDev SE Mean T P Hours 8 0.050 2.485 0.879 –1.08 0.32 Enter the data in C1, Hours. Choose t-test in the STAT menu. 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. Minitab Solution

35. Hypothesis Testing for Proportions Section 7.4

36. p is the population proportion of successes. The test statistic is. If and the sampling distribution for is normal. Test for Proportions The standardized test statistic is: (the proportion of sample successes)

37. 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 spokespersons claim at = 0.05. What can you conclude? 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.05

38. 3. Determine the sampling distribution. 7. Make your decision. 6. Find the test statistic and standardize it. 8. Interpret your decision. z = 2.63 falls in the rejection region, so reject H 0 There is enough evidence to support the claim that over 40% of Americans own a cell phone or have a family member who does. 1036(.40) > 5 and 1036(.60) > 5. The sampling distribution is normal. n = 1036 x = 456 4. Find the critical value. 1.645 5. Find the rejection region. Rejection region

39. Hypothesis Testing for Variance and Standard Deviation Section 7.5

40. Find a 2 0 critical value for a left-tail test when n = 17 and = 0.05. s 2 is the test statistic for the population variance. Its sampling distribution is a 2 distribution with n – 1 d.f. Find critical values 2 0 for a two-tailed test when n = 12, = 0.01. The standardized test statistic is 2 0 = 7.962 2 L = 2.603 and 2 R = 26.757 Critical Values for

41. 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 administrators 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 2 with 10 – 1 = 9 d.f. Test for

42. 7. Make your decision. 6. Find the test statistic. 8. Interpret your decision. n = 10 s = 28.8 2 = 8.2944 does not fall in the rejection region, so fail to reject H 0 There is not enough evidence to support the administrators claim that the standard deviation is less than 30. 2.088 4. Find the critical value. 5. Find the rejection region.