Econ 222: Solution Template Assignment 3, Instructions 1. Data sets for this problem must be downloaded. 7.6 5.4 5.3 7.4 7.8 5.9 6.5 8.7 6.7 6.1 6.9 8.2 9.1 5.9 6.6 Agri-Beef Inc. operates cattle feed lots in several Midwestern states. The company wishes to estimate the average weekly weight gain of cattle on their lots. To do this a simple random sample of cattle is taken and the weekly weight gain, in pounds, is recorded. A mean weekly weight gain of 7.4 pounds is considered ideal for cattle on feed lots. Using your sample data set, test an appropriate hypothesis to determine if the mean weekly weight gain for Agri-Beef differs from 7.4 pounds. Refer to Section 9.3 of the text on pages 368-71. 2. An issue that faces individuals investing for retirement is allocating assets among different investment choices. A study conducted 10 years ago showed that 65% of investors preferred stocks to real estate as an investment. In a recent random sample of 900 investors, 540 preferred stocks to real estate. Is this new data sufficient to allow you to conclude that the proportion of investors preferring stocks to real estate has declined from 10 years ago? Conduct your analysis at the 2% significance level. Refer to Section 9.4 of the text on pages 373-6. Click the link below to submit your responses. After clicking, you may either type in your response or attach a file before clicking Submit. question 1. My data set number is: X = weekly weight gain of cattle on Agri-Beef feed lots. x = mean weekly weight gain of all cattle on Agri-Beef feed lots. Report the following: H0: H1: Using = 0.05, Report the critical values of t (obtained from the t-table) and indicate the rejection region. / 0 2=t . 0/ 22 5 / t2 = /. 20 2 5 Decision Rule: Reject H0 at = 0.05 if the absolute value of the computed t test statistic exceeds the critical values of t (if the computed test statistic is in the rejection region). Compute the test statistic: n= = sx = Conclusion: t x o sx n Reject H0 at = 0.05 or do not reject H0 at = 0.05 and conclude that ... Solution Template Assignment 3, question 2. = proportion of all investors (population proportion) preferring stocks to real estate for retirement investment. Report the following: H0: H1: Compute the test statistic: n= p = sample proportion of investors preferring stock to real estate = Conclusion: p o o(1 o) n 0 z = Decision Rule: Reject H0 at = 0.02 if the computed z test statistic ... z . 0 2 Using = 0.02, Report the critical value of z (obtained from the t-table, bottom row) and indicate the rejection region. ECON 222 Assignment 2 - Hypothesis Testing For each of the following problems, (a) identify the population parameter of interest, (b) define the random variable, and (c) state appropriate null and alternative hypotheses. 1. General Electric has developed a new bulb whose design specifications call for a light output to average 960 lumens. From a sample of 20 new bulbs, the testing committee found a mean light output of 954 lumens. Specify the null and alternative hypotheses General Electric would use to determine whether its new bulb is producing the specified mean light output of 960 lumens? a) Here the population parameter of interest is population mean light output. b) The random variable is the light output of new bulbs. c) The null and alternative hypotheses are, Ho: = 960 and H1: 960 2. Rick Douglas, the new manager of Food Barn, is interested in the percentage of customers who state, when surveyed, that they are satisfied with the store. The previous manager found that 86 percent of the customers at the time were satisfied. Rick claims the proportion of satisfied customers is the same today. Rick sampled 187 customers and found 157 expressed satisfaction. At the 1 percent level of significance, is there sufficient evidence that refute Rick's claim? a) Here the population parameter of interest is population proportion of customers who are satisfied with the store. b) The random variable is the number of customers who are satisfied with the store. c) The null and alternative hypotheses are, Ho: p = 0.86 and H1: p 0.86 3. From a total of 10,200 loans made by a state employee's credit union in the most recent five year period, 350 were sampled to determine what proportion were made to women. This sample showed that 39 percent of the loans were made to women employees. A complete census of loans 5 years ago showed that 41 percent of borrowers at that time were women. At the 2% level, has the proportion of loans made to women changed significantly in the past 5 years? a) Here the population parameter of interest is population of women loan borrowers. b) The random variable is the number of women borrowed loan. c) The null and alternative hypotheses are, Ho: p = 0.41 and H1: p 0.41 4. A television documentary on overeating claimed that Americans are at least 10 pounds overweight on average. To test this claim, eighteen randomly selected individuals were examined; their average excess weight was found to be 12.4 pounds, and the sample standard deviation was 2.7 pounds. At a significance level of 0.01, is there sufficient evidence in the sample data to support the documentary's claim? a) Here the population parameter of interest is population mean excess weight of Americans. b) The random variable is the excess weight of Americans. c) The null and alternative hypotheses are, Ho: = 10 and H1: 10 5. Federal environmental statutes applying to a particular nuclear power plant specify that recycled water must, on average, be no warmer than 84 degrees F (28.9 C) before it can be released into the river beside the plant. From 70 samples, the average temperature of recycled water was found to be 86.3 F. If the population standard deviation is 13.5 F, should the plant be cited for exceeding the limitation of the statute? a) Here the population parameter of interest is population mean temperature of recycled water. b) The random variable is the temperature of recycled water. c) The null and alternative hypotheses are, Ho: = 84 and H1: > 84 6. Two years ago the average 2-week-advance-purchase airfare between RaleighDurham, North Carolina, and New York City was $235. The population standard deviation was $68. A recent sample of 90 randomly chosen travelers between these two cities found that they had paid $218.77, on average, for their tickets. Did the average airfare on this route change significantly in the past two years? a) Here the population parameter of interest is population mean 2-week-advancepurchase airfare between Raleigh-Durham, North Carolina, and New York City. b) The random variable is the 2-week-advance-purchase airfare between Raleigh-Durham, North Carolina, and New York City. c) The null and alternative hypotheses are, Ho: = 235 and H1: 235 Econ 222: Economic and Business Statistics II Lecture 5 Introduction to Hypothesis Testing Hypothesis Testing Example The student union automatic coffee machine has been accused of filling cups with less than the advertised 6 ounces of coffee. A group of ten statistics students decide to test the machine after class. Each student buys a cup of coffee, and this sample results in a mean of 5.93 ounces and a standard deviation of 0.13 ounces. a. Construct a 98 percent confidence interval estimate for the true mean fill amount (x). b. Does this interval support or refute the advertising on the machine? c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 a. Construct a 98 percent confidence interval estimate for the true mean fill amount (x). x t/2,n-1 s/ n Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? The sample result offers evidence against the advertising claim (the null hypothesis). The question to be answered is, \"How strong is this evidence?\" c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? The sample result offers evidence against the advertising claim (the null hypothesis). The question to be answered is, \"How strong is this evidence?\" c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? To find out if the evidence is strong or weak, we ask, \"What is the probability that could be 6 oz. and we find the mean of a sample of size 10 equal to 5.93 or less? How unlikely is this sample result?\" What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter: population mean Example: The mean amount of coffee per cup dropped by the machine is = 6 oz. population proportion Example: A marketing company claims that it receives a response rate of 8%, p = .08, from its mailings. There are ALWAYS two hypotheses specified, a null and alternative. The Null Hypothesis, H0 States the assumption (numerical) to be tested Example: The mean amount of coffee per cup dropped by the machine is x = 6 oz. Rules for the Null Hypothesis, H0, \"H naught\" Is always about an unknown population parameter, never about a sample statistic Must include an equality H0 : 3 H0 : x 3 The Alternative Hypothesis, HA or H1 Is the opposite of the null hypothesis. Challenges the status quo (null is always given the \"benefit of the doubt\"). Is generally the hypothesis that is believed (or needs to be supported) by the researcher, hence it is often called the research hypothesis. In practice, the sample data can only provide evidence to refute the null hypothesis and support the alternative hypothesis and never to support the null. Errors and Error Probabilities Possible Hypothesis Test Outcomes State of Nature Decision Key: Outcome (Probability) H0 True H0 False Accept H0 No error Type II Error () Reject H0 Type I Error () No Error Note: Two types of errors Reject H0 when H0 is true: Type I error is the probability of committing a type I error Accept H0 when H0 is false: Type II error P(type II error) = (called beta) Hypothesis testing analogy Possible Hypothesis Test Outcomes State of Nature Ho: The defendant is innocent H1: The defendant is guilty Decision H0 True (Innocent) H0 False (Guilty) Type II Error ( ) Free the Let guilty person innocent person go free (Acquit) No error Accept Ho (Convict) Type I Error( ) No Error Reject Ho Send innocent person to jail What is the cost of each type of error? Send the guilty to jail Note: Two types of errors Drug testing Example Two hypotheses: This drug is safe and This drug is not safe. Which should be the null hypothesis and which the alternative? Define type I an type II error for each What is the cost of each error? H0: The drug is not safe H1: The drug is safe Hypothesis testing analogy Possible Hypothesis Test Outcomes State of Nature Ho: The drug is not safe H1: The drug is safe Decision H0 True (not safe) H0 False (Safe Drug) (Do not Type II Error ( ) No error sell drug) Unsafe drug not Safe drug kept off Accept Ho marketed market (Sell drug) Reject Ho Type I Error( ) No Error Unsafe drug Safe drug sold to sold to public public What is the cost of each type of error? Type I & II Error Relationship Type I and Type II errors can not happen at the same time Type I error can only occur if H0 is true Type II error can only occur if H0 is false If Type I error probability ( ) Type II error probability ( ) , then Note: Two types of errors If we reject H0, then only a Type I error is possible If we accept H0, then we can commit a Type II error We can reject H0 and report If we accept H0, we would have to report In practice, we will always know , and rarely . Given this, our choices reduce to: Reject H0 and report Do not reject H0 and leave the test inconclusive. Note: Do not reject H0 is not the same as accept H0. Outcomes and Probabilities Possible Hypothesis Test Outcomes Key: Outcome (Probability) State of Nature Decision H0 True Do not Reject No error Inconclusive ( Do not need to report ) Type I Error (Report ) No error H0 Reject H0 H0 False Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Determine the decision rule. Select the significance level and define the rejection region. 3. Obtain sample evidence and compute the test statistic. 4. Reach a decision (Is test statistic in rejection region?) and interpret the result. Steps 2 and 3 can be reversed as in the next slide. Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Obtain sample evidence and compute the test statistic. 3. Determine the decision rule. Compute the observed significance level (p-level) for the test. 4. Reach a decision (Is observed significance level sufficiently low?) and interpret the result. Step 1: The Null Hypothesis, H0 Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until proven guilty Refers to the status quo Always contains the equality or \"=\" , \"\" or \"\" sign. Step 1: The Alternative Hypothesis Is the opposite of the null hypothesis Never contains the \"=\" , \"\" or \"\" sign May be specified using \">\\fEcon 222: Economic and Business Statistics II Lecture 6 Z & t Hypothesis Tests for Mean Using the p-level approach Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Determine the decision rule. Select the significance level and define the rejection region. 3. Obtain sample evidence and compute the test statistic. 4. Reach a decision (Is test statistic in rejection region?) and interpret the result. Steps 2 and 3 are reversed when using the plevel approach. Step 1: The Null Hypothesis, H0 Begin with the assumption that the null hypothesis is true Always contains the equality or \"=\" , \"\" or \"\" sign. Step 1: The Alternative Hypothesis Never contains the \"=\" , \"\" or \"\" sign The alternative is either one-sided (\">\" or \"<\") or two-sided (\"\"). These are often referred to as one-tail or two-tailed tests. Only use the one-tail test when it can be justified without the sample results. Note: Two types of errors Reject H0 when H0 is true: Type I error is the probability of committing a type I error Accept H0 when H0 is false: Type II error P(type II error) = (called beta) In practice we either: Reject H0 and report - or Do not reject H0 at the level of significance Step 2: Decision Rule Level of Significance, Rejection Region Level of significance is designated by . May either be selected by the researcher in advance based upon the cost of committing a type I error - or Typical values are .01, .05, or .10. The p-level method of testing is used. The selection of will determine what the critical value(s) that define the boundary of rejection region will be Level of Significance and the Rejection Region Level of significance = H0: 6 H1: < 6 H0: 6 H1: > 6 H0: = 6 H1: 6 Represents critical value 0 Lower tail test Z or t Rejection region is shaded 0 Upper tail test /2 Two tailed test Z or t /2 0 Z or t Step 3: The test statistic Convert sample statistic (e.g.: x ) to test statistic ( Z or t statistic ). The test statistic is the distance, in units of the standard error, the point estimate is from the hypothesized value. x z n t n 1 x s n Step 4: Make the decision Compare the critical value(s) for a specified level of significance from the z or t table or from a computer printout to the computed test statistic. Step 4: Make the decision If the test statistic falls in the rejection region, reject H0 at that value of alpha (level of significance); If the test statistic is not in the rejection region, do not reject H0 at that level of significance and leave the result inconclusive. State that there is not sufficient evidence in the sample information to warrant rejection of the null at the specified significance level. Acquittal does not mean the defendant is proven innocent. It means there is not sufficient evidence to obtain a conviction at the significance level chosen. Hypothesis Testing Example Test the student's suspicion that the coffee machine is under filling the cups. 1. Specify appropriate null and alternative hypotheses H : 6 H1: < 6 (This is a one-tail, 0 lower-tail test) 2. Decision Rule: Suppose that = .05 is chosen for this test Hypothesis Testing Example 2. Decision Rule: Determine the rejection region d.f. = n -1 = 9 = .05 t Reject H0 at = 0.05 -t= -1.833 Do not reject H0 0 Sample results: n = 10, x = 5.93 , sx = 0.13 Then the test statistic is: t x o 5.93 6.0 .07 -1.70 sx 0.13 .0411 n 10 Hypothesis Testing Example 4. Reach a decision and interpret the result d.f. = n -1 = 9 = .05 t Reject H0 at = 0.05 Do not reject H0 -1.833 0 -1.70 Since t = -1.7 is not < -1.833, we cannot reject the null hypothesis at = .05. There is not sufficient evidence in the sample to indicate that the mean fill of the machine is less than the advertised 6 oz. Testing Ho at = .10 4. Reach a decision and interpret the result d.f. = n -1 = 9 = .10 t Reject H0 Do not reject H0 -1.383 0 -1.70 Since t = -1.7 is < -1.383, we can reject the null hypothesis (H0) at = .10 that the mean fill of the machine is the advertised 6 oz. and conclude the alternative hypothsis (H1) is correct. The machine is underfilling the cups. p-Value Approach to Testing p-value: Probability of obtaining a test statistic more extreme ( or ) than the observed sample value given H0 is true The p-level is also called observed level of significance The p-level is the smallest value of for which H0 can be rejected (How low can you go?) p-Value Approach to Testing Convert Sample Statistic (e.g. x ) to Test Statistic ( Z or t statistic) Obtain the p-value from a table or computer then: Compare the observed p-value with preselected If p-value < preselected , reject H0 at sig. level If p-value , do not reject H0 at sig. level Report observed sig. level (p-value) and defer the decision to reject to decision maker. p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 t -1.383 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 = .05 t -1.383 -1.833 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 = .05 t -1.383 -1.70 -1.833 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 p-value =.0617 = .05 t p-value = Area beyond the observed test statistic. -1.383 -1.70 -1.833 0 Finding p-values Finding p-values A Two-tailed Test Example Let's test the same hypothesis but from the perspective of the coffee machine repair person. The null hypothesis will be: The coffee machine is operating properly and the alternative is that the machine needs adjustments. Form hypothesis test: H0: = 6 the mean coffee dropped is 6 oz. HA: 6 the mean is either > 6 or < 6 but in either case the machine will have to be adjusted. Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 /2=.05 t 0 Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 Critical Value: t0.05, 9 = 1.833 /2=.05 /2=.05 t 0 Reject H0 at -t/2 -1.833 Do not reject H0 at t/2 1.833 Reject H0 at Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 t 0 Reject H0 at is unknown, so use a t statistic Critical Value: t0.05, 9 = 1.833 /2=.05 -t/2 -1.833 t n 1 Do not reject H0 at t/2 1.833 x o 5.93 6.0 1.70 sx 0.13 n 10 Reject H0 at Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 t 0 Reject H0 at is unknown, so use a t statistic Critical Value: t0.05, 9 = 1.833 /2=.05 -t/2 -1.833 t n 1 Do not reject H0 at t/2 Reject H0 at 1.833 x o 5.93 6.0 1.70 sx 0.13 n 10 Do not reject H0: not sufficient evidence that true mean coffee fill differs from 6 oz. Example: Two-Tail Test The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $174.50 and s = $15.40. Test at the = 0.05 level. (Assume the population distribution is normal) H0: = 168 H1: 168 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 Critical Value: /2=.025 /2=.025 0 -t/2 t/2 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 Do not reject H0 at t/2 2.064 Reject H0 at Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 x 174.50 168 2.11 s 15.40 n 25 Reject H0 at Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 Reject H0 at x 174.50 168 2.11 s 15.40 n 25 Reject H0 at = 0.05. There is sufficient evidence in the sample data to indicate that true mean cost is different than $168. Finding the p -Value Finding the approximate p-value using the t-table p-value is between = .05 and .02 Test Statistic: t = 2.11 in a two-tail test t d.f. = 24 How low can we go? Can we go this low? Y Y Y N N Finding the p-value Using Excel Using Excel to Test Hypotheses Hypothesis Testing - Significance Tests Section 9.3 Do problems 9.27 a. and b., 9.29 a. and b., and 9.33. Section 9.4 Do problems 9.37 all parts, 9.39 all parts. \fEcon 222: Economic and Business Statistics II Lecture 7 Z Test for Population Proportion Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Select the significance level and find critical values of z or t. 3. Compute the test statistic and compare to critical value(s). OR 2. Compute the test statistic and the observed sig. level, the plevel. 3. Compare p-level to acceptable significance level. 4. Reach a decision and interpret the result. Last Class: Tests Concerning a Population Mean The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $174.50 and s = $15.40. Test at the = 0.05 level. (Assume the population distribution is normal) H0: = 168 H1: 168 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = /2=.025 0 Reject H0 at n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 Reject H0 at x 174.50 168 2.11 s 15.40 n 25 Reject H0 at = 0.05. There is sufficient evidence in the sample data to indicate at the 5% significance level that true mean cost is different than $168. Today: Hypothesis Tests for Proportions Involves categorical variables of only two levels. Two possible outcomes \"Success\" or \"Failure.\" Fraction or proportion of population in the success category is denoted by p. Fraction or proportion of success in the sample data is denoted by p. Proportions Sample proportion in the success category is denoted by p p x number of successes in sample n sample size When both np and n(1-p) are at least 5, p can be approximated by a normal distribution with mean and standard deviation p (1 p ) p p p n Hypothesis Tests for Proportions Hypothesis Tests for p when np 5 and n(1-p) 5 The sampling distribution of p is normal, so the test statistic is a z value (never t and never small sample sizes): Test Statistic: p po z po(1 po) n Example: z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses returned. Test the claim at the = .05 significance level. Check: n p = (500)(.08) = 40 > 5 n(1-p) = (500)(.92) = 460 > 5 Sample results: p = 25 / 500 = 0.05 1 - p = 1 - 0.05 = 0.95 n = 500 Z Test for Proportion: Solution H0: p = .08 H1: p .08 Test Statistic: Decision: Conclusion: Z Test for Proportion: Solution (continued) H0: p = .08 H1: p .08 Test Statistic: = .05 = .025 n = 500, p = .05 Decision: Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 = .05 = .025 n = 500, p = .05 Decision: Critical Values: 1.96 .025 .025 -1.96 0 1.96 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 = .05 = .025 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Reject H0 at = .05 Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Reject H0 at = .05 Conclusion: There is sufficient evidence in the sample data to reject the company's claim of 8% response rate at the 5% significance level ( = 0.05). Finding the p -Value Finding the approximate p-value using the t-table p-value is between = .02 and .01 Test Statistic: z = - 2.47 in a two-tail test How low can we go? Can we go this low? z Y Y Y Y N Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 .4932 .4932 z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 .0068 .4932 .4932 z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 .4932 z 0 z = -2.47 2(.5 .4932) 2(.0068) 0.0136 .0068 .4932 P(z 2.47) P(z 2.47) z = 2.47 p-value = .0136 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 2(.5 .4932) 2(.0068) 0.0136 .0068 .4932 .4932 z 0 z = -2.47 P(z 2.47) P(z 2.47) p-value = .0136 z = 2.47 Reject H0 at any as low as .0136. Can reject H0 at = .02 but not at = .01. From a total of 10,200 loans made by a state employee's credit union in the most recent five year period, 350 were sampled to determine what proportion were made to women. This sample showed that 39 percent of the loans were made to women employees. A complete census of loans 5 years ago showed that 41 percent of the borrowers then were women. At the 2% level, has the proportion of loans made to women changed significantly in the past 5 years? Hypothesis Testing - Significance Tests Section 9.4 Do problems 9.37 all parts, 9.39 all parts. Type II Error Probabilities and the power of a test. A Type II error is the error of accepting (failing to reject) H0 when H0 is false. In order to accept H0, we must \"assess the quality of our inference\" by reporting the probability of our error, . and the Power of a Test Suppose we want to accept H0: 50 against H1: < 50 To find the probability of our error, we must then, in fact, calculate for every possible value of under H1. The power of the test = 1 - . and the Power of a Test is the area under the alternative distribution in the acceptance region A Type II error is the error of accepting H0 when H1 is true. Is the area under the null dist. in the rejection region This is the true distribution of x if = 45 Null Distribution: assumes Ho is correct 45 50 Reject H0: 50 Accept H0 : 50 x and the Power of a Test We have n = 100 , s = 25 , and = .10 critical x o z (for H0 : 50) sx 25 50 1.282 46.795 n 100 So = P( x 46.80 ) if = 45 45 Reject H0: 50 46.8 50 Accept H0 : 50 x and the Power of a Test We have n = 100 , s = 25 , and = .10 46.80 45 P( x 6.80 | 50) P z P(z 0.72) .5 .2642 .2358 25 100 Probability of type II error: = .2358 45 Reject H0: 50 46.8 50 Accept H0 : 50 x and the Power of a Test and the Power of a Test Mu = 50 Mu = 48 Mu = 46 Mu = 44 Mu = 42 Mu = 40 30 35 40 45 50 Rejection Region | Acceptance Region 55 60 Mu = 38 Critical value and the Power of a Test Beta and Power Functions 1.2 0.8 Beta 0.6 Power 0.4 0.2 0.0 41.5 42.0 42.5 43.0 43.5 44.0 44.5 45.0 45.5 46.0 46.5 47.0 47.5 48.0 48.5 49.0 49.5 50.0 Values of Mu under H1 Beta and 1-beta 1.0 Conclusions about and the Power of a Test Calculating and the power of a test is difficult work. It's even harder for a two-tail test than the example given for a one-tail test. Fortunately we can avoid all this work by not accepting the null hypothesis. A type II error can only be made if Ho is accepted. We therefore never accept Ho. Conclusions about and the Power of a Test Decision choices become: Reject Ho and report or the p-level. Never reject Ho at any larger than 0.10. Do not reject Ho and leave the result inconclusive. State that there is not sufficient evidence in the sample information to warrant rejection of the null at the specified significance level. Acquittal does not mean the defendant is proven innocent. It means there is not sufficient evidence to obtain a conviction at the significance level chosen. ECON 222 Assignment 2 - Hypothesis Testing For each of the following problems, (a) identify the population parameter of interest, (b) define the random variable, and (c) state appropriate null and alternative hypotheses. 1. General Electric has developed a new bulb whose design specifications call for a light output to average 960 lumens. From a sample of 20 new bulbs, the testing committee found a mean light output of 954 lumens. Specify the null and alternative hypotheses General Electric would use to determine whether its new bulb is producing the specified mean light output of 960 lumens? a) Here the population parameter of interest is population mean light output. b) The random variable is the light output of new bulbs. c) The null and alternative hypotheses are, Ho: = 960 and H1: 960 2. Rick Douglas, the new manager of Food Barn, is interested in the percentage of customers who state, when surveyed, that they are satisfied with the store. The previous manager found that 86 percent of the customers at the time were satisfied. Rick claims the proportion of satisfied customers is the same today. Rick sampled 187 customers and found 157 expressed satisfaction. At the 1 percent level of significance, is there sufficient evidence that refute Rick's claim? a) Here the population parameter of interest is population proportion of customers who are satisfied with the store. b) The random variable is the number of customers who are satisfied with the store. c) The null and alternative hypotheses are, Ho: p = 0.86 and H1: p 0.86 3. From a total of 10,200 loans made by a state employee's credit union in the most recent five year period, 350 were sampled to determine what proportion were made to women. This sample showed that 39 percent of the loans were made to women employees. A complete census of loans 5 years ago showed that 41 percent of borrowers at that time were women. At the 2% level, has the proportion of loans made to women changed significantly in the past 5 years? a) Here the population parameter of interest is population of women loan borrowers. b) The random variable is the number of women borrowed loan. c) The null and alternative hypotheses are, Ho: p = 0.41 and H1: p 0.41 4. A television documentary on overeating claimed that Americans are at least 10 pounds overweight on average. To test this claim, eighteen randomly selected individuals were examined; their average excess weight was found to be 12.4 pounds, and the sample standard deviation was 2.7 pounds. At a significance level of 0.01, is there sufficient evidence in the sample data to support the documentary's claim? a) Here the population parameter of interest is population mean excess weight of Americans. b) The random variable is the excess weight of Americans. c) The null and alternative hypotheses are, Ho: = 10 and H1: 10 5. Federal environmental statutes applying to a particular nuclear power plant specify that recycled water must, on average, be no warmer than 84 degrees F (28.9 C) before it can be released into the river beside the plant. From 70 samples, the average temperature of recycled water was found to be 86.3 F. If the population standard deviation is 13.5 F, should the plant be cited for exceeding the limitation of the statute? a) Here the population parameter of interest is population mean temperature of recycled water. b) The random variable is the temperature of recycled water. c) The null and alternative hypotheses are, Ho: = 84 and H1: > 84 6. Two years ago the average 2-week-advance-purchase airfare between RaleighDurham, North Carolina, and New York City was $235. The population standard deviation was $68. A recent sample of 90 randomly chosen travelers between these two cities found that they had paid $218.77, on average, for their tickets. Did the average airfare on this route change significantly in the past two years? a) Here the population parameter of interest is population mean 2-week-advancepurchase airfare between Raleigh-Durham, North Carolina, and New York City. b) The random variable is the 2-week-advance-purchase airfare between Raleigh-Durham, North Carolina, and New York City. c) The null and alternative hypotheses are, Ho: = 235 and H1: 235 Econ 222: Economic and Business Statistics II Lecture 5 Introduction to Hypothesis Testing Hypothesis Testing Example The student union automatic coffee machine has been accused of filling cups with less than the advertised 6 ounces of coffee. A group of ten statistics students decide to test the machine after class. Each student buys a cup of coffee, and this sample results in a mean of 5.93 ounces and a standard deviation of 0.13 ounces. a. Construct a 98 percent confidence interval estimate for the true mean fill amount (x). b. Does this interval support or refute the advertising on the machine? c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 a. Construct a 98 percent confidence interval estimate for the true mean fill amount (x). x t/2,n-1 s/ n Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? The sample result offers evidence against the advertising claim (the null hypothesis). The question to be answered is, \"How strong is this evidence?\" c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? Example 14. x = 5.93, sx = 0.13 ounces, n = 10 b. Does this interval sample result support or refute the advertising on the machine? The sample result offers evidence against the advertising claim (the null hypothesis). The question to be answered is, \"How strong is this evidence?\" c. Could the machine possibly be accurate and still give a sample average of 5.93 ounces? To find out if the evidence is strong or weak, we ask, \"What is the probability that could be 6 oz. and we find the mean of a sample of size 10 equal to 5.93 or less? How unlikely is this sample result?\" What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter: population mean Example: The mean amount of coffee per cup dropped by the machine is = 6 oz. population proportion Example: A marketing company claims that it receives a response rate of 8%, p = .08, from its mailings. There are ALWAYS two hypotheses specified, a null and alternative. The Null Hypothesis, H0 States the assumption (numerical) to be tested Example: The mean amount of coffee per cup dropped by the machine is x = 6 oz. Rules for the Null Hypothesis, H0, \"H naught\" Is always about an unknown population parameter, never about a sample statistic Must include an equality H0 : 3 H0 : x 3 The Alternative Hypothesis, HA or H1 Is the opposite of the null hypothesis. Challenges the status quo (null is always given the \"benefit of the doubt\"). Is generally the hypothesis that is believed (or needs to be supported) by the researcher, hence it is often called the research hypothesis. In practice, the sample data can only provide evidence to refute the null hypothesis and support the alternative hypothesis and never to support the null. Errors and Error Probabilities Possible Hypothesis Test Outcomes State of Nature Decision Key: Outcome (Probability) H0 True H0 False Accept H0 No error Type II Error () Reject H0 Type I Error () No Error Note: Two types of errors Reject H0 when H0 is true: Type I error is the probability of committing a type I error Accept H0 when H0 is false: Type II error P(type II error) = (called beta) Hypothesis testing analogy Possible Hypothesis Test Outcomes State of Nature Ho: The defendant is innocent H1: The defendant is guilty Decision H0 True (Innocent) H0 False (Guilty) Type II Error ( ) Free the Let guilty person innocent person go free (Acquit) No error Accept Ho (Convict) Type I Error( ) No Error Reject Ho Send innocent person to jail What is the cost of each type of error? Send the guilty to jail Note: Two types of errors Drug testing Example Two hypotheses: This drug is safe and This drug is not safe. Which should be the null hypothesis and which the alternative? Define type I an type II error for each What is the cost of each error? H0: The drug is not safe H1: The drug is safe Hypothesis testing analogy Possible Hypothesis Test Outcomes State of Nature Ho: The drug is not safe H1: The drug is safe Decision H0 True (not safe) H0 False (Safe Drug) (Do not Type II Error ( ) No error sell drug) Unsafe drug not Safe drug kept off Accept Ho marketed market (Sell drug) Reject Ho Type I Error( ) No Error Unsafe drug Safe drug sold to sold to public public What is the cost of each type of error? Type I & II Error Relationship Type I and Type II errors can not happen at the same time Type I error can only occur if H0 is true Type II error can only occur if H0 is false If Type I error probability ( ) Type II error probability ( ) , then Note: Two types of errors If we reject H0, then only a Type I error is possible If we accept H0, then we can commit a Type II error We can reject H0 and report If we accept H0, we would have to report In practice, we will always know , and rarely . Given this, our choices reduce to: Reject H0 and report Do not reject H0 and leave the test inconclusive. Note: Do not reject H0 is not the same as accept H0. Outcomes and Probabilities Possible Hypothesis Test Outcomes Key: Outcome (Probability) State of Nature Decision H0 True Do not Reject No error Inconclusive ( Do not need to report ) Type I Error (Report ) No error H0 Reject H0 H0 False Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Determine the decision rule. Select the significance level and define the rejection region. 3. Obtain sample evidence and compute the test statistic. 4. Reach a decision (Is test statistic in rejection region?) and interpret the result. Steps 2 and 3 can be reversed as in the next slide. Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Obtain sample evidence and compute the test statistic. 3. Determine the decision rule. Compute the observed significance level (p-level) for the test. 4. Reach a decision (Is observed significance level sufficiently low?) and interpret the result. Step 1: The Null Hypothesis, H0 Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until proven guilty Refers to the status quo Always contains the equality or \"=\" , \"\" or \"\" sign. Step 1: The Alternative Hypothesis Is the opposite of the null hypothesis Never contains the \"=\" , \"\" or \"\" sign May be specified using \">\\fEcon 222: Economic and Business Statistics II Lecture 6 Z & t Hypothesis Tests for Mean Using the p-level approach Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Determine the decision rule. Select the significance level and define the rejection region. 3. Obtain sample evidence and compute the test statistic. 4. Reach a decision (Is test statistic in rejection region?) and interpret the result. Steps 2 and 3 are reversed when using the plevel approach. Step 1: The Null Hypothesis, H0 Begin with the assumption that the null hypothesis is true Always contains the equality or \"=\" , \"\" or \"\" sign. Step 1: The Alternative Hypothesis Never contains the \"=\" , \"\" or \"\" sign The alternative is either one-sided (\">\" or \"<\") or two-sided (\"\"). These are often referred to as one-tail or two-tailed tests. Only use the one-tail test when it can be justified without the sample results. Note: Two types of errors Reject H0 when H0 is true: Type I error is the probability of committing a type I error Accept H0 when H0 is false: Type II error P(type II error) = (called beta) In practice we either: Reject H0 and report - or Do not reject H0 at the level of significance Step 2: Decision Rule Level of Significance, Rejection Region Level of significance is designated by . May either be selected by the researcher in advance based upon the cost of committing a type I error - or Typical values are .01, .05, or .10. The p-level method of testing is used. The selection of will determine what the critical value(s) that define the boundary of rejection region will be Level of Significance and the Rejection Region Level of significance = H0: 6 H1: < 6 H0: 6 H1: > 6 H0: = 6 H1: 6 Represents critical value 0 Lower tail test Z or t Rejection region is shaded 0 Upper tail test /2 Two tailed test Z or t /2 0 Z or t Step 3: The test statistic Convert sample statistic (e.g.: x ) to test statistic ( Z or t statistic ). The test statistic is the distance, in units of the standard error, the point estimate is from the hypothesized value. x z n t n 1 x s n Step 4: Make the decision Compare the critical value(s) for a specified level of significance from the z or t table or from a computer printout to the computed test statistic. Step 4: Make the decision If the test statistic falls in the rejection region, reject H0 at that value of alpha (level of significance); If the test statistic is not in the rejection region, do not reject H0 at that level of significance and leave the result inconclusive. State that there is not sufficient evidence in the sample information to warrant rejection of the null at the specified significance level. Acquittal does not mean the defendant is proven innocent. It means there is not sufficient evidence to obtain a conviction at the significance level chosen. Hypothesis Testing Example Test the student's suspicion that the coffee machine is under filling the cups. 1. Specify appropriate null and alternative hypotheses H : 6 H1: < 6 (This is a one-tail, 0 lower-tail test) 2. Decision Rule: Suppose that = .05 is chosen for this test Hypothesis Testing Example 2. Decision Rule: Determine the rejection region d.f. = n -1 = 9 = .05 t Reject H0 at = 0.05 -t= -1.833 Do not reject H0 0 Sample results: n = 10, x = 5.93 , sx = 0.13 Then the test statistic is: t x o 5.93 6.0 .07 -1.70 sx 0.13 .0411 n 10 Hypothesis Testing Example 4. Reach a decision and interpret the result d.f. = n -1 = 9 = .05 t Reject H0 at = 0.05 Do not reject H0 -1.833 0 -1.70 Since t = -1.7 is not < -1.833, we cannot reject the null hypothesis at = .05. There is not sufficient evidence in the sample to indicate that the mean fill of the machine is less than the advertised 6 oz. Testing Ho at = .10 4. Reach a decision and interpret the result d.f. = n -1 = 9 = .10 t Reject H0 Do not reject H0 -1.383 0 -1.70 Since t = -1.7 is < -1.383, we can reject the null hypothesis (H0) at = .10 that the mean fill of the machine is the advertised 6 oz. and conclude the alternative hypothsis (H1) is correct. The machine is underfilling the cups. p-Value Approach to Testing p-value: Probability of obtaining a test statistic more extreme ( or ) than the observed sample value given H0 is true The p-level is also called observed level of significance The p-level is the smallest value of for which H0 can be rejected (How low can you go?) p-Value Approach to Testing Convert Sample Statistic (e.g. x ) to Test Statistic ( Z or t statistic) Obtain the p-value from a table or computer then: Compare the observed p-value with preselected If p-value < preselected , reject H0 at sig. level If p-value , do not reject H0 at sig. level Report observed sig. level (p-value) and defer the decision to reject to decision maker. p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 t -1.383 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 = .05 t -1.383 -1.833 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 = .05 t -1.383 -1.70 -1.833 0 p-value example Example: How likely is it to see a sample mean of 5.93 (or something farther below the true mean) if the true mean is = 6.0? = .10 d.f. = n -1 = 9 p-value =.0617 = .05 t p-value = Area beyond the observed test statistic. -1.383 -1.70 -1.833 0 Finding p-values Finding p-values A Two-tailed Test Example Let's test the same hypothesis but from the perspective of the coffee machine repair person. The null hypothesis will be: The coffee machine is operating properly and the alternative is that the machine needs adjustments. Form hypothesis test: H0: = 6 the mean coffee dropped is 6 oz. HA: 6 the mean is either > 6 or < 6 but in either case the machine will have to be adjusted. Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 /2=.05 t 0 Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 Critical Value: t0.05, 9 = 1.833 /2=.05 /2=.05 t 0 Reject H0 at -t/2 -1.833 Do not reject H0 at t/2 1.833 Reject H0 at Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 t 0 Reject H0 at is unknown, so use a t statistic Critical Value: t0.05, 9 = 1.833 /2=.05 -t/2 -1.833 t n 1 Do not reject H0 at t/2 1.833 x o 5.93 6.0 1.70 sx 0.13 n 10 Reject H0 at Example Solution: Two-Tail Test H0: = 6 HA: 6 = 0.10 n = 10, d.f. = 9 /2=.05 t 0 Reject H0 at is unknown, so use a t statistic Critical Value: t0.05, 9 = 1.833 /2=.05 -t/2 -1.833 t n 1 Do not reject H0 at t/2 Reject H0 at 1.833 x o 5.93 6.0 1.70 sx 0.13 n 10 Do not reject H0: not sufficient evidence that true mean coffee fill differs from 6 oz. Example: Two-Tail Test The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $174.50 and s = $15.40. Test at the = 0.05 level. (Assume the population distribution is normal) H0: = 168 H1: 168 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 Critical Value: /2=.025 /2=.025 0 -t/2 t/2 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 Do not reject H0 at t/2 2.064 Reject H0 at Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 x 174.50 168 2.11 s 15.40 n 25 Reject H0 at Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 /2=.025 0 Reject H0 at -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 Reject H0 at x 174.50 168 2.11 s 15.40 n 25 Reject H0 at = 0.05. There is sufficient evidence in the sample data to indicate that true mean cost is different than $168. Finding the p -Value Finding the approximate p-value using the t-table p-value is between = .05 and .02 Test Statistic: t = 2.11 in a two-tail test t d.f. = 24 How low can we go? Can we go this low? Y Y Y N N Finding the p-value Using Excel Using Excel to Test Hypotheses Hypothesis Testing - Significance Tests Section 9.3 Do problems 9.27 a. and b., 9.29 a. and b., and 9.33. Section 9.4 Do problems 9.37 all parts, 9.39 all parts. \fEcon 222: Economic and Business Statistics II Lecture 7 Z Test for Population Proportion Steps in Hypothesis Testing 1. Specify the null and alternative hypotheses. 2. Select the significance level and find critical values of z or t. 3. Compute the test statistic and compare to critical value(s). OR 2. Compute the test statistic and the observed sig. level, the plevel. 3. Compare p-level to acceptable significance level. 4. Reach a decision and interpret the result. Last Class: Tests Concerning a Population Mean The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $174.50 and s = $15.40. Test at the = 0.05 level. (Assume the population distribution is normal) H0: = 168 H1: 168 Example Solution: Two-Tail Test H0: = 168 H1: 1680.05 = /2=.025 0 Reject H0 at n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.064 /2=.025 -t/2 -2.064 t n 1 Do not reject H0 at t/2 2.064 Reject H0 at x 174.50 168 2.11 s 15.40 n 25 Reject H0 at = 0.05. There is sufficient evidence in the sample data to indicate at the 5% significance level that true mean cost is different than $168. Today: Hypothesis Tests for Proportions Involves categorical variables of only two levels. Two possible outcomes \"Success\" or \"Failure.\" Fraction or proportion of population in the success category is denoted by p. Fraction or proportion of success in the sample data is denoted by p. Proportions Sample proportion in the success category is denoted by p p x number of successes in sample n sample size When both np and n(1-p) are at least 5, p can be approximated by a normal distribution with mean and standard deviation p (1 p ) p p p n Hypothesis Tests for Proportions Hypothesis Tests for p when np 5 and n(1-p) 5 The sampling distribution of p is normal, so the test statistic is a z value (never t and never small sample sizes): Test Statistic: p po z po(1 po) n Example: z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses returned. Test the claim at the = .05 significance level. Check: n p = (500)(.08) = 40 > 5 n(1-p) = (500)(.92) = 460 > 5 Sample results: p = 25 / 500 = 0.05 1 - p = 1 - 0.05 = 0.95 n = 500 Z Test for Proportion: Solution H0: p = .08 H1: p .08 Test Statistic: Decision: Conclusion: Z Test for Proportion: Solution (continued) H0: p = .08 H1: p .08 Test Statistic: = .05 = .025 n = 500, p = .05 Decision: Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 = .05 = .025 n = 500, p = .05 Decision: Critical Values: 1.96 .025 .025 -1.96 0 1.96 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 = .05 = .025 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Reject H0 at = .05 Conclusion: Z Test for Proportion: Solution (continued) Test Statistic: H0: p = .08 H1: p .08 p po .05 .08 z 2.4726 po(1 po) .08(1 .08) = .05 = .025 n 500 n = 500, p = .05 Decision: Critical Values: 1.96 Reject Ho at = .05 Reject Ho at = .05 .025 .025 -1.96 -2.47 0 1.96 Cannot reject Ho at = .05 z Reject H0 at = .05 Conclusion: There is sufficient evidence in the sample data to reject the company's claim of 8% response rate at the 5% significance level ( = 0.05). Finding the p -Value Finding the approximate p-value using the t-table p-value is between = .02 and .01 Test Statistic: z = - 2.47 in a two-tail test How low can we go? Can we go this low? z Y Y Y Y N Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 .4932 .4932 z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 .0068 .4932 .4932 z 0 z = -2.47 z = 2.47 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 .4932 z 0 z = -2.47 2(.5 .4932) 2(.0068) 0.0136 .0068 .4932 P(z 2.47) P(z 2.47) z = 2.47 p-value = .0136 Finding the p -value on the ztable Test Statistic: z = - 2.47 in a two-tail test. p-value = area more extreme than the test statistic. The green shaded area below. Taken from the z-table for z = 2.47 = 0.5 - .4932 .0068 2(.5 .4932) 2(.0068) 0.0136 .0068 .4932 .4932 z 0 z = -2.47 P(z 2.47) P(z 2.47) p-value = .0136 z = 2.47 Reject H0 at any as low as .0136. Can reject H0 at = .02 but not at = .01. From a total of 10,200 loans made by a state employee's credit union in the most recent five year period, 350 were sampled to determine what proportion were made to women. This sample showed that 39 percent of the loans were made to women employees. A complete census of loans 5 years ago showed that 41 percent of the borrowers then were women. At the 2% level, has the proportion of loans made to women changed significantly in the past 5 years? Hypothesis Testing - Significance Tests Section 9.4 Do problems 9.37 all parts, 9.39 all parts. Type II Error Probabilities and the power of a test. A Type II error is the error of accepting (failing to reject) H0 when H0 is false. In order to accept H0, we must \"assess the quality of our inference\" by reporting the probability of our error, . and the Power of a Test Suppose we want to accept H0: 50 against H1: < 50 To find the probability of our error, we must then, in fact, calculate for every possible value of under H1. The power of the test = 1 - . and the Power of a Test is the area under the alternative distribution in the acceptance region A Type II error is the error of accepting H0 when H1 is true. Is the area under the null dist. in the rejection region This is the true distribution of x if = 45 Null Distribution: assumes Ho is correct 45 50 Reject H0: 50 Accept H0 : 50 x and the Power of a Test We have n = 100 , s = 25 , and = .10 critical x o z (for H0 : 50) sx 25 50 1.282 46.795 n 100 So = P( x 46.80 ) if = 45 45 Reject H0: 50 46.8 50 Accept H0 : 50 x and the Power of a Test We have n = 100 , s = 25 , and = .10 46.80 45 P( x 6.80 | 50) P z P(z 0.72) .5 .2642 .2358 25 100 Probability of type II error: = .2358 45 Reject H0: 50 46.8 50 Accept H0 : 50 x and the Power of a Test and the Power of a Test Mu = 50 Mu = 48 Mu = 46 Mu = 44 Mu = 42 Mu = 40 30 35 40 45 50 Rejection Region | Acceptance Region 55 60 Mu = 38 Critical value and the Power of a Test Beta and Power Functions 1.2 0.8 Beta 0.6 Power 0.4 0.2 0.0 41.5 42.0 42.5 43.0 43.5 44.0 44.5 45.0 45.5 46.0 46.5 47.0 47.5 48.0 48.5 49.0 49.5 50.0 Values of Mu under H1 Beta and 1-beta 1.0 Conclusions about and the Power of a Test Calculating and the power of a test is difficult work. It's even harder for a two-tail test than the example given for a one-tail test. Fortunately we can avoid all this work by not accepting the null hypothesis. A type II error can only be made if Ho is accepted. We therefore never accept Ho. Conclusions about and the Power of a Test Decision choices become: Reject Ho and report or the p-level. Never reject Ho at any larger than 0.10. Do not reject Ho and leave the result inconclusive. State that there is not sufficient evidence in the sample information to warrant rejection of the null at the specified significance level. Acquittal does not mean the defendant is proven innocent. It means there is not sufficient evidence to obtain a conviction at the significance level chosen