Section 7.1 Introduction to Hypothesis Testing Schrodinger s cat quantum mechanics thought experiment (1935)
Statistical Hypotheses A statistical hypothesis is a claim about a population. Null hypothesis H 0 contains a statement of equality such as, = or. Alternative hypothesis H a contains a statement of inequality such as <, or > If I am false, you are true Complementary Statements If I am false, you are true Accept or reject null, never prove null is true
Writing Hypotheses Write the claim about the population. Write its complement. Either hypothesis, can represent the claim. Example: A hospital claims its ambulance response time is less than 10 minutes. claim Example: A consumer magazine claims the proportion of cell phone calls made during evenings and weekends is at most 60%. claim H 0 always contains the = condition
Hypothesis Test Strategy 1) Assume the equality condition in the null hypothesis is true, regardless of whether the claim is represented by the null or alternative hypothesis. 2) Collect data from a random sample taken from the population and calculate the necessary sample statistics. 3a) 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. (i.e. you will support the alternative hypothesis.) 3b) If the probability is not low enough, fail to reject H 0.
Errors and Level of Significance Decision Do not reject H 0 Reject H 0 Actual Truth of H 0 H 0 True Correct Decision Type I Error H 0 False Type II Error Correct Decision Type I Error: Null hypothesis is true but reject it. Level of significance, (e.g. 0.01, 0.05, 0.10) Maximum probability of committing a Type I Error. Type II Error: Null hypothesis is false but accept it. Probability of Type II Error 1-β (power of test) Innocent until proven guilty Beyond reasonable doubt
Types of Hypothesis Tests H a is more probable One-tail test Right-tail test H a is more probable Left-tail test H a is more probable Two-tail test
The P-value is the probability of obtaining a sample statistic with a value as or more extreme than the one determined by the sample data. P-value = indicated area Area in left tail P-values Area in right tail z For a left tail test If z is negative, twice the area in the left tail z z For a two-tail test z For a right tail test If z is positive, twice the area in the right tail
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 arbitrary level of significance of the test. If reject the null hypothesis. If fail to reject the null hypothesis.
Method I: 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 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.
Interpreting the Decision Claim Claim is H 0 Claim is H a Decision Reject H 0 Fail to reject H 0 There is enough evidence to reject the claim. 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.
Steps in a Hypothesis Test 1. Write the null and alternative hypothesis. Write H 0 and H a as mathematical statements (H 0 always contains the = symbol). 2. State the level of significance. Maximum probability of rejecting the null hypothesis when it is true. (Making a Type I error.) 3. Identify the sampling distribution. Sampling distribution is the distribution for the test statistic assuming that H 0 is true and that the experiment is repeated an infinite number of times.
Steps in a Hypothesis Test 4. Find the test statistic and standardize it. Perform 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 α, reject H 0. If the P value is greater than α, fail to reject H 0. 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 σ is known or (2) when the sample size, n, is at least 30. Test statistic is the sample mean test statistic is z. and the standardized When n ³ 30, use s in place of.
The z-test for a Mean (P-value) Example: 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 α= 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 (Central Limit Theorem, Chptr. 5).
4. Find the test statistic and standardize it. n = 52 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. Test statistic s = 10 (n>30) Area in right tail z = 1.44
6. Make your decision. Compare the P-value to. Since 0.0749 > 0.05, fail to reject H 0. 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.
Method II: Rejection Regions Sampling distribution for Rejection Region z z 0 Critical Value z 0 Rejection region is the range of values for which the null hypothesis is not probable. Always in the direction of the alternative hypothesis. Its area is equal to. A critical value separates rejection region from the nonrejection region.
Critical Values Rejection region Rejection region z 0 Find z 0 for a left-tail test with =.01. z 0 = 2.33 Rejection region z 0 Find z 0 for a right-tail test with =.05. Rejection region z 0 = 1.645 z 0 z 0 Find z 0 and z 0 for a two-tail test with α =.01 z 0 = 2.575 and z 0 = 2.575
Using Critical Values to Make Test Decisions 1. Write the null and alternative hypothesis. Write H 0 and H a as mathematical statements. Remember H 0 always contains the = symbol. 2. State the level of significance. The maximum probability of rejecting the null hypothesis when it is actually true (Type I Error.) 3. Identify the sampling distribution. 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.
4. Find the critical value. Rejection Region z 0 6. Find the test statistic. 5. Find the rejection region. Critical value separates rejection region of the sampling distribution from the non-rejection region. 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 H 0. Otherwise, fail to reject H 0. 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 Example: 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 α = 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 H a contains the > symbol, this is a right-tail test. z 0 Rejection region 4. Find the critical value. 1.645, α = 0.05, cum. area = 0.95 5. Find the rejection region. 6. Find the test statistic and standardize it. n = 52 = 232 s = 10 7. Make your decision. Test statistic does not fall in the rejection region, so fail to reject H 0 8. Interpret your decision. z = 1.44 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 Rejection area 0.05 z 0 z Test statistic = 0.05 z 0 = 1.645 z = 1.23 P = 0.1093 For a P-value decision, compare areas. If reject H 0. If fail to reject H 0. 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. *Decision same, critical values & z-scores, P-values & areas*
Section 7.3 Hypothesis Testing for the Mean (n < 30)
The t Sampling Distribution Find the critical value t 0 for a left-tailed test given α = 0.01 and n = 18. Area in left tail d.f. = 18 1 = 17 t 0 = 2.567 t 0 Find the critical values t 0 and t 0 for a two-tailed test given = 0.05 and n = 11. d.f. = 11 1 = 10 t 0 = 2.228 and t 0 = 2.228 t 0 t 0 *If > 30 d.f. then ~normal*
Testing Small Sample Example: A university says the mean number of classroom hours per week for full-time faculty is 11.0. A random sample 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 α= 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 H a contains the symbol, this is a two-tail test. 4. Find the critical values. t 0 t 0 3.499 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 H 0 at = 0.01 8. Interpret your decision. 5. Find the rejection region. There is not enough evidence to reject the university s claim that faculty spend a mean of 11 classroom hours.
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 is normal. The standardized test statistic is:
Test for Proportions Example: 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 1,036 Americans, 456 said they or a family member owned a cellular phone. Test the spokesperson s claim at α = 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. What is p-value? 0.0043 z = 2.63 falls in the rejection region, so reject H 0 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 s 2 is the test statistic for the population variance. Its sampling distribution is a χ 2 distribution with n 1 d.f. Find a χ 2 0 critical value for a left-tail test when n = 17 and = 0.05. χ 2 0 = 7.962 Find critical values χ 2 0 for a two-tailed test when n = 12, = 0.01. χ 2 L = 2.603 and χ 2 R = 26.757 The standardized test statistic is
Test for Example: 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 α = 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 χ 2 with 10 1 = 9 d.f. 1- or 2-tailed test?
4. Find the critical value. 2.088 6. Find the test statistic. n = 10 s = 28.8 7. Make your decision. 5. Find the rejection region. χ 2 = 8.2944 does not fall in the rejection region, so fail to reject H 0 8. Interpret your decision. There is not enough evidence to support the administrator s claim that the standard deviation is less than 30.