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## Chapter 9

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**Chapter 9**Hypothesis Testing • Null and Alternative Hypotheses and Errors in Testing • Large Sample Tests about a Mean: Rejection Points • Small Sample Tests about a Population Mean • Hypothesis Tests about a Population Proportion**Introduction**A Hypothesis Test is a statistical procedure that involves formulating a hypothesis and using sample data to decide on the validity of the hypothesis. • In this chapter we will focus on hypothesis tests about population means and proportions (since we are knowledgeable about how sample means and proportions are distributed).**In order to test any hypothesis (even non statistical**hypotheses) you need three elements: 1. The Hypotheses (Both the hypothesis that you are testing and the alternative hypothesis, which is the opposite of the hypothesis) 2. An unbiased test statistic- A measure that you will use to evaluate the hypotheses. 3. A rejection rule- the rule that you will use to ultimately decide if a hypothesis should be rejected.**Non Statistical Examples**• Example 1: • Hypothesis: I should be admitted to Harvard Law School Alternative Hypothesis: I should not be admitted to Harvard Law School • Test Statistic: LSAT • Rejection Rule: Harvard may have a cut off LSAT score for admitting students. • Example 2: • Hypothesis: OJ is innocent Alternative Hypothesis: OJ is guilty • Test Statistic: A jury of 12 of his peers. • Rejection Rule: If 12 of 12 jurors rule that he is guilty beyond a reasonable doubt.**Hypothesis Testing Rules**• In order to test a hypothesis, you must first find the test statistic and rejection rule that is appropriate for evaluating your hypotheses (i.e., we could not evaluate OJ’s innocence based upon his LSAT score). • All hypothesis tests will always end in 1 of 2 ways: 1. You conclude that you must reject the null hypothesis (This is the same as concluding that you have proven the alternative true); or 2. You conclude that you do not have enough evidence to reject the null hypothesis (This is the same as concluding that you do not have enough evidence to prove that the alternative is true).**Developing Null and Alternative Hypotheses**• Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected. • The null hypothesis, denoted H0, is a statement of the basic proposition being tested. The statement is not rejected unless there is convincing sample evidence that it is false. • The alternativeorresearch hypothesis, denoted Ha, is an alternative (to the null hypothesis) statement that will be accepted only if there is convincing sample evidence that it is true.**Developing Null and Alternative Hypotheses**• Testing the Validity of a Claim • If you wish to find evidence to contradict a claim, the claim should be stated as the null hypothesis. • If you wish to prove a claim to be true, then you should state the claim as the alternative hypothesis. • Claims that test whether the mean is equal to a specific value must be stated as the null hypothesis. Anderson, Sweeney, and Williams**A Summary of Forms for Null and Alternative Hypotheses about**a Population Mean • The equality part of the hypotheses always appears in the null hypothesis. • In general, a hypothesis test about the value of a population mean must take one of the following three forms • Where 0 is a specific value H0: >0 Ha: < 0 H0: <0 Ha: > 0 H0: = 0 Ha: 0 One-tailed One-tailed Two-tailed**Example 9.1: Metro EMS**A major west coast city provides one of the most comprehensive emergency medical services in the world. Operating in a multiple hospital system with approximately 20 mobile medical units, the service goal is to respond to medical emergencies with a mean time of 12 minutes or less. The director of medical services wants to formulate a hypothesis test that could use a sample of emergency response times to determine whether there is sufficient evidence to prove that they are not meeting their goal. Anderson, Sweeney, and Williams**Null and Alternative Hypotheses**HypothesesConclusion and Action H0: < The emergency service is meeting the response goal; no appropriate follow-up action is necessary. Ha:> The emergency service is not meeting the response goal; appropriate follow-up action is necessary. Where: = mean response time for the population of medical emergency requests. By defining the claim as the null hypothesis we can set out to try to find sufficient evidence to reject the claim.**Example 9.2: The Potato Chip Manufacturer**Many people eat chips with their soda. Suppose a potato chip manufacturer is concerned that the bagging equipment may not be functioning properly when filling 10-oz bags. You have been asked to set up a hypothesis test that will help determine if there is a problem with the bagging equipment. What null and alternative hypothesis would you use? Pelosi and Sandifer**HypothesesConclusion and Action**H0: =0 The machine is working properly; no appropriate follow-up action is necessary. Ha:0 The machine is not working properly; appropriate follow-up action is necessary. Where: = mean filling weight for the machine.**Type I and Type II Errors**• Since hypothesis tests are based on sample data, we must allow for the possibility of errors. • A Type I error is rejecting H0 when it is true. • The person conducting the hypothesis test specifies the maximum allowable probability of making a Type I error, denoted by and called the level of significance.**A Type II error is accepting H0 when it is false.**• Generally, we cannot control for the probability of making a Type II error, denoted by . • Statistician avoids the risk of making a Type II error by using the phrase “do not reject H0” instead of “accept H0”.**Large Sample Tests about Mean (n30) and knownORSmall**Sample Tests about Mean, Normal pop. and known H0: >0 Ha: < 0 Test Statistic: Rejection Rule Reject H0 if z<- z Lower-Tailed test H0: <0 Ha: > 0 Test Statistic: Rejection Rule Reject H0 if z>z Upper-Tailed test H0: = 0 Ha: 0 Test Statistic: Rejection Rule Reject H0 if |z| >z Two-Tailed test**Steps for Computing z (The z value with an upper tail**area of . • In order to determine the z value with an upper tail area of , we need the area beneath the normal curve between the mean and the z value of interest. area= .5- 2.Go to the area section of the standard normal table and find the area closest to the area computed in 2. The corresponding z value is z.**One-Tailed Test about a Population Mean: Large n** = P(Type I Error) Sampling distribution of sample mean (assuming H0 is true) Reject H0 Do Not Reject H0 Anderson, Sweeney, and Williams 0 z (Critical value)**Two-Tailed Test about a Population Mean: Large n** = P(Type I Error) Sampling distribution of sample mean (assuming H0 is true) Reject H0 Reject H0 /2 /2 Anderson, Sweeney, and Williams z z/2 -z/2 0 (Critical values)**Steps of Hypothesis Testing**• Determine the null and alternative hypotheses. • Specify the level of significance . • Select the test statistic that will be used to test the hypothesis. Using the Test Statistic • State the rejection rule for H0 and use to determine the critical value for the test statistic. (Find the rejection rule that is appropriate for your null hypothesis) • Collect the sample data and compute the value of the test statistic. • Use the value of the test statistic and the rejection rule to determine whether to reject H0. 7. State the conclusion in layman’s terms. Anderson, Sweeney, and Williams**Example 9.1 (Revisited)**Recall example 8.1. Suppose we collected a sample of n = 40 EMS calls and computed = 13.25 minutes and σ = 3.2 minutes. Using =.05 conduct a hypothesis test to see if you can find the evidence to refute their claim that the average response time is less than 12 minutes. (The sample standard deviation s can be used to estimate the population standard deviation .) Step 1: H0: <12 Ha: >12 Step 2: =.05 Step 3:**Step 4: Reject H0 if z > zα = 1.645 (α=0.05)**Step 5: Step 6. Is z> 1.645? Since 2.47 > 1.645, we reject H0. Conclusion: We are 95% confident that Metro EMS is not meeting the response goal of 12 minutes; appropriate action should be taken to improve service.**Example 9.3**Consider a company that is trying a new and cheaper package design for its product. The average sales for this product are currently $1500/month. Suppose they wish to prove that sales will decrease as a result of the new method. In order to test this claim they used n=36 test stores and computed an average sale, =$1450 with σ = $250. (Use =.10) Step 1: H0: >$1500 Ha: <$1500 Step 2: =.1 Step 3:**Step 4: Reject H0 if z <- zα = -1.28 (α=0.10)**Step 5: Step 6. Is z<-1.28? Since -1.2 is not less than -1.28, we cannot reject H0. Thus we cannot find sufficient evidence to prove that the new advertising method will decrease sales**Example 9.4: The Chapperel Steel Company**Another recent management approach is to have employees become actual partners of the business. Chapperel Steel Company has done exactly this and the company feels that one of the benefits of this concept is that the average number of sick days will decrease. Prior to implementing this program, Chapperel had an average of 7.2 sick days per employee. Set up the null and alternative hypothesis to test if the average number of sick days per employee is different from 7.2. After implementing this program, a sample of 40 employees provides a sample mean of 6.5 day and a population standard deviation of 2.5 days. Test this hypothesis test using =.01. Pelosi and Sandifer**Step 1: H0: 7.2 **Ha: 7.2 Step 2: =.01 Step 3: Step 4: Reject H0 if |z| >zα/2 =2.575 (=.01)**Step 5:**Step 6: Is |z| > 2.575 ? Since |-1.77|= 1.77 does not exceed 2.575, we cannot reject this claim. Thus we can’t refute the claim that this program does alter the number of employee sick days.**Example 9.5 Glow Toothpaste**The production line for Glow toothpaste is designed to fill tubes of toothpaste with a mean weight of 6 ounces. Periodically, a sample of 30 tubes will be selected in order to check the filling process. Quality assurance procedures call for the continuation of the filling process if the sample results are consistent with the assumption that the mean filling weight for the population of toothpaste tubes is 6 ounces; otherwise the filling process will be stopped and adjusted. Anderson, Sweeney, and Williams**Step 1: H0: **Ha: Step 2: Assume a .05 level of significance. • Step 3: • Step 4: Assuming a .05 level of significance, Reject H0 if |z| > zα/2 =1.96 (=.05) Anderson, Sweeney, and Williams**Step 5:**Assume that a sample of 30 toothpaste tubes provides a sample mean of 6.1 ounces and population standard deviation of 0.2 ounces. Let n = 30, = 6.1 ounces, =.2 ounces Step 6: Is |z| > 1.96 ? Since 2.74 > 1.96, we reject H0. Thus, the mean filling weight for the population of toothpaste tubes is not 6 ounces. Anderson, Sweeney, and Williams**Confidence Interval Approach to aTwo-Tailed Test about a**Population Mean • Select a simple random sample from the population and use the value of the sample mean to develop the confidence interval for the population mean . • If the confidence interval contains the hypothesized value 0, do not reject H0. Otherwise, reject H0. Anderson, Sweeney, and Williams**Confidence Interval Approach to a Two-Tailed Hypothesis Test**The 95% confidence interval for is or 6.0284 to 6.1716 Since the hypothesized value for the population mean, 0 = 6, is not in this interval, the hypothesis-testing conclusion is that the null hypothesis, H0: = 6, can be rejected. (As shown in the previous slide, following traditional hypothesis testing steps.)**Large Sample Tests about Mean (n30) and unknownOR**Small Sample Tests about Mean, Normal pop. and unknown If the sampled population is normal, n<30, and is unknown H0: > 0 Ha: < 0 Test Statistic: Rejection Rule Reject H0 if t<- t Lower-Tailed test H0: < 0 Ha: > 0 Test Statistic: Rejection Rule Reject H0 if t> t Upper-Tailed test H0: = 0 Ha: 0 Test Statistic: Rejection Rule Reject H0 if |t| > t Lower-Tailed test**Steps for Computing t(The t value with an upper tail area**of . • Compute =1 – (Confidence Coefficient). • Compute upper tail area= . • Degrees of freedom= n-1. • Go to t chart and find the t value with the upper tail area (computed in step 2), and the degrees of freedom (computed in step 3). This is t .**Summary of Hypothesis Testing Procedures**for a Population Mean Yes No n > 30 ? No Popul. approx. normal ? known ? Yes Yes Use s to estimate No known ? No Use s to estimate Yes Slide 15 Slide 32 Increase n to> 30 Slide 32 Slide 15 Anderson, Sweeney, and Williams**Example 9.6: Highway Patrol**A State Highway Patrol periodically samples vehicle speeds at various locations on a particular roadway. The sample of vehicle speeds is used to test the hypothesis H0: m< 65. The locations where H0 is rejected are deemed the best locations for radar traps. At Location F, a sample of 16 vehicles shows a mean speed of 68.2 mph with a standard deviation of 3.8 mph. Use an a = .05 to test the hypothesis. Anderson, Sweeney, and Williams**Let n = 16, = 68.2 mph, s = 3.8 mph**Step 1: H0: m< 65 Ha: m> 65 Step 2: a = .05 Step 3:**Step 4: Assuming a .05 level of**significance, Reject H0 if t > tα =1.753 (=.05, df=15) Step 5: Compute the value of our test statistic =68.2 – 65 = 3.36 3.8/4 Step 6: Is 3.36>1.753? Yes. So reject Ho. Conclude that vehicle speeds in that area exceed 65.**A Summary of Forms for Null and Alternative Hypotheses about**a Population Proportion • The equality part of the hypotheses always appears in the null hypothesis. • In general, a hypothesis test about the value of a population proportion pmust take one of the following three forms (where p0 is the hypothesized value of the population proportion). H0: p>p0H0: p<p0 Ha: p < p0Ha: p > p0 H0: p= p0 Ha: pp0 One-tailed One-tailed Two-tailed**Tests about a Population Proportion:Large-Sample Case {(np>**5) and n(1 - p) > 5)} • H0: p> p0 • Ha: p<p0 • Test Statistic: • Rejection Rule • Reject H0 if z<- z H0: p<p0 Ha: p> p0 Test Statistic: Rejection Rule Reject H0 if z>z H0: p = p0 Ha: p p0 Test Statistic: Rejection Rule Reject H0 if |z| >z where**Example 9.7: NSC**For a Christmas and New Year’s week, the National Safety Council estimated that 500 people would be killed and 25,000 injured on the nation’s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving. A sample of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC’s claim with a = 0.05.**Example: NSC**Step 1: H0: p = .5 Ha: p .5 Step 2: a = 0.05 Step 3:**Step 4: Reject H0 if |z| > zα/2 =1.96 (α=.05)**Step 5: Step 6: Is |z| > 1.96 ? No, so do not reject H0. Thus, we do not have enough evidence to reject the claim that 50% of the accidents would be caused by drunk driving.