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QNT561 Night Three I Hypothesis Testing Objectives Confidence Intervals Sample Size One-Sample Hypothesis Tests Two-Sample Hypothesis Tests F-Test, ANOVA Confidence Intervals The Statistics for Business Text, Chapter Five into Six are quite excellent, but the concepts are a bit challenging. Let’s dive right into Excel and explore how to get those calculations done, and what they mean! A Quick YouTube Primer General Steps in Hypothesis Testing 1) Get and idea or a hunch—really! 2) Formulate testable hypothesis 3) Design the experiment 4) Set up a decision rule 5) Collect data 6) Make a decision 7) Revise your idea or hunch 8) Do it again! Five Steps for Testing a Hypothesis 1) State the null and alternate hypothesis 2) Select a level of significance 3) Identify the test statistic 4) Formulate a decision rule 5) Take a sample, do the math, arrive at a decision Null and Alternate Hypothesis The Null Hypothesis is the hypothesis of NO CHANGE. It is the hypothesis you typically wish to reject with your experiment. (The status quo.) The Alternate Hypothesis is the hypothesis of change in some direction—more, less, a difference. This is typically the hypothesis that you are hoping to not reject. Always formed in pairs, they are mutually exclusive and collectively exhaustive statements: H0:Null Hypothesis H1:Alternate Hypothesis Can a Hypothesis be Proved? No---it is not possible to prove a null hypothesis. We can only fail to reject a null hypothesis (which means the alternate hypothesis is not rejected). Statistics and hypothesis testing cannot prove anything! Avoid using the word “prove” when explaining or supporting your findings from testing. What can be said if prove is off limits? Examples of Null and Alternate Hypothesis H0: Defendant is Innocent (status quo as all are presumed innocent until proven guilty…proven?) H1: Defendant is Guilty H0: Olympian did not use illegal steroids (status quo) H1: Olympian did use illegal steroids H0: Green chile containers have 8.0 ounces of product in each (no change—containers are designed for 8.0 oz. and the machine set at 8.0 oz.) H1: Green chile containers do not have 8.0 ounces of product in each; OR has more than 8.0 ounces; OR has less than 8.0 ounces Your Turn Fred wants to measure the time his crew takes to submit their end-of-day reports. He thinks his new crew takes too long. What is the null and alternate hypothesis? Sally needs to see if the changes made to her salon helped increase traffic. What is the null and alternate hypothesis? George works for PLM Industries and fills orders that have between 5-10 items each. He want to see if his fill times have increased. What is the null and alternate hypothesis? Types of Error (Level of Significance) Type I Error (α): Rejecting H0 when, in fact, it is true! α is alpha in the Greek alphabet (also know as the level of significance) A “false positive” Type II Error (β): Failing to reject H0 when, in fact, it is false! β is beta in the Greek alphabet A “false negative” Types of Error (Level of Significance) Example: H0: Olympian did not use illegal steroids (status quo) H1: Olympian did use illegal steroids What is the α impact? (False positive) What is the β impact? (False negative) How to reduce α and β risk? Both can be simultaneously reduced by increasing the sample size, which is not always feasible or cost-effective. Choosing the Level of Significance The smaller, the better, but there are temptations to fiddle with this level to “get the answer you want.” Typical levels are 0.01, 0.05, 0.10, but can assume any value between 0 and 1, but not 0 or 1. The smaller the level of significance, the more difficult it will be to reject the null hypothesis. Statistical Hypothesis Testing A statistical hypothesis is a statement about the value of a population parameter (θ) that we are interested in analyzing. θ is theta in the Greek alphabet and represents a population parameter One-tailed and two-tailed tests H0 : 0 H0 : 0 H0 : 0 H1 : 0 H1 : 0 H1 : 0 Select the Test Statistic This is the statistical tool you will use to “crunch” the data For this chapter: z-tests Student’s t-test x z n z p 1 n x t s n Formulate the Decision Rule Different for a one-tail test and a two-tail test The decision rule is determined from your level of significance. With this information, you determine the critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected. Decision Rule, One Tail Test Select your level of significance. In this case, we will select 0.05. Subtract that from 0.50, you get 0.4500 (0.50 is the half of the normal distribution curve.) Get the chart titled “Areas under the Normal Curve” and in the body of the table, find a value that is very close to 0.4500. You see that 0.4505 is pretty close. Now find the numbers in the margins of the table and you get 1.65—I’ll show you this. One-Tailed Example Decision Rule, Two Tailed Test Select your level of significance, in this slide, we will continue to use 0.05. Divide that by 2, or take half of 0.05, which is 0.025. Subtract 0.025 from 0.50 and you get 0.4750. Get the chart titled “Areas under the Normal Curve” and in the body of the table, find a value that is very close to 0.4750. You see that 0.4750 is listed! Go to the margins, and you get 1.96 and -1.96---I’ll explain how to do this. Two-Tailed Example Combo Slide! Decision Rule Statement “At the 0.05 level of significance, we will reject the null hypothesis if: the computed z value is greater than 1.65.” (one tail test, right) the computed z value is less than -1.65.” (one tail test, left) the computed z value is greater than 1.96 or less than -1.96.” (two tail test) Let’s Try One! PLM Industries manufactures and assembles desks and other office equipment. The weekly production of the Model A325 desk at the Fredonia Plant follows a normal probability distribution with other plants with a population mean of 200 and a population standard deviation of 16. Recently, new production methods have been introduced and new employees hired at the Fredonia plant. The VP of manufacturing would like to investigate whether there has been a change in the weekly production of the A325. Is the mean number of desks produced at the FredoniaPlant different from 200 at the 0.01 significance level? This plant averages 203.5 desks per week over 50 weeks. Step One: State the Null and Alternate Hypothesis In English: “There is a significant difference at the Fredonia Plant in the manufacturing output of the A325 desk.” In Statistics: H0: μ=200 H1: μ ≠200 Two-tailed Test! Step Two: Select the Level of Significance The problem provided the significance level: 0.01. What if the problem didn’t provide one? Pick one! The typical levels are 0.01, 0.05, 0.10, but you get to decide the level—how small of a Type I Error risk do you wish to have? Step Three: Select the Test Statistic The test statistic for a mean when the population standard deviation is known is the z-test. We know the population mean and standard deviation from the problem. [If the population standard deviation (σ) is unknown, the Student’s t-test is appropriate.] Step Four: Formulate the Decision Rule The decision rule is formulated by finding the critical value of z from the table. Since this is a two-tailed test, take 0.01 and divide it by 2, and you get 0.005. Take 0.5000-0.005 and you get 0.4950. Find this value in the body of the table and you find 0.4951, close enough! Take it to the margins, and you have a critical value of 2.58 and -2.58. Step Five: Do the Math, Interpret Your Results 203.5 200 x Plug and chug! z z 16 z========== n z=1.55 50 Is 1.55 greater than 2.58? It is not, therefore: Fail to reject the null hypothesis. Interpret the Results: There is no significant difference in the manufacturing output of the A325 desk at the Fredonia Plant. Lather, rinse, repeat! Those are the five steps in a nutshell. Now, let’s do more of those! But first…the p-value! p Value---What is it and Why Do I Care? The p-value is the actual significance level. We had 0.01 in the previous problem and we did NOT reject the null hypothesis. The p-value tells us the actual risk involved in having a Type I Error. Take the calculated z value, in this last problem it was 1.55. Find the area under the curve that matches 1.55…in this case 0.4394. 0.5000.4394=0.0606 or 6.06%. p-Value… The p-value must be smaller than the level of significance if you rejected the null hypothesis! Always! In this case, our significance level was 0.01 and our p-value was 0.0606; clearly, if we did reject the null hypothesis, we would have been quite incorrect! That is the p-value. (It is calculated differently for z, t, and F tests, too.) What if we changed… …the significance level from 0.01 to 0.15? Well, that would not be ethical. We select the significance level BEFORE we conduct the study…always! However, I said I would teach you how you can lie with statistics, and how you are lied to with statistics, so here we go. Desk A325 Problem, new level of significance μ=200, σ=16 Our plant mean 203.5, sample size 50. New significance level 0.15 (instead of 0.01) We will reject the null hypothesis if the computed z-value is greater than 1.44 or less than -1.44. (Where did that come from?!?!?!) Math! 203.5 200 z 1.55 16 50 “OK…what just happened?” Our calculated z-value is 1.55, but we need a number greater than 1.44…and 1.55 is greater than 1.44, therefore, we reject the null hypothesis! We can now say that there was a significant change in the production of the A325 desk at the Fredonia Plant…we’ve INCREASED weekly production! Let’s celebrate!!! (Why increased? How do we know that?) Really? Well, the data says that the production increased…right? [Oh---that is soooo wrong! Yup, but who else knows? Thus, one of the lies I’ve been telling you about!] Statistical Hypothesis Testing Is A VERY Powerful Tool! Just by tweaking the level of significance, what was originally not a “fact” is now a “fact”. Knowing how stats works will make you a better manager, a better consumer of information, and when you apply sound and ethical statistics to help you and your company solve problems, you will demonstrate your abilities as an effective, ethical, and educated manager. Statistical Hypothesis Testing Abuses Changing the significance level after-the-fact is not condoned and the demonstration you just saw was to illustrate how it works---it is not an endorsement! Does it happen? Sure it does, but now you know what to ask and how to spot it. You also know that changing the level of significance after-the-fact is unethical. Another! {Generic problem} A sample of 36 observations is selected from a normal population. The sample mean is 21 and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.05 significance level. H 0 : 20 H1 : 20 Well, some steps are done! Step one is…. H1 : 20 Step two is….. State the null and alternate hypothesis—done. H 0 : 20 Select a level of significance (0.05)---done. Step three is…. Select the test statistic—you have to select this: In this case, a z-test because we know the population mean and standard deviation. (20 and 5) Formulate a Decision Rule One or two tailed test? Why? Because there is direction! See the greater than sign? H : One! That’s direction…a right side! 0.05 significance level .5000-.05=0.4500 Critical value==1.65 0 20 H1 : 20 Formulate a Decision Rule We will reject the null hypothesis IF the calculated z-value is greater than 1.65. Take a Sample… Already done for you in this problem. Math time! 21 20 z 1.2 5 36 Did we reject the null hypothesis? 1.2 is not greater than 1.65, therefore do not reject the null hypothesis. There is no significant increase noted in the sample. Another! {Generic problem} A sample of 64 observations is selected from a normal population. The sample mean is 215 and the population standard deviation is 15. Conduct the following test of hypothesis using the 0.04 significance level! H 0 : 220 H1 : 220 Do It Using the Five Steps! Step One: State the null and alternate hypothesis Step Two: Select a level of significance Step Three: Identify the test statistic Step Four: Formulate a decision rule Step Five: Take a sample, arrive at a decision, INTERPRET THE RESULT! Answer One tail test Critical value: 0.5000-.04=0.4600; therefore, a critical value of 1.76 or -1.76, depending on the direction. Test statistic: z-test—we know σ We will reject the null hypothesis if the computer z-value is less than -1.76. Math! 215 220 z 2.667 15 64 Interpret the Result Because -2.667 is less than -1.76, we reject the null hypothesis and “accept” the alternate hypothesis. There is a significant reduction in the sample…what exactly, we don’t know, but we know it is significant. p-value! Well, calculated z is -2.667, table says 0.4962, 0.5000-0.4962=0.0038, which is less than the significance level of 0.04. Student’s t-test Just like the z-test, but we use this tool when we don’t know the population standard deviation σ. Used if the sample is less than 30, too. x t sd n Everything Else…Same (almost)! The five step process is the same. What does change is the table we consult for the critical value and the application of the degrees of freedom to determine the c.v. (critical value) Here we go! Problem The McPatrick Insurance Company Claims Department reports that the mean cost to process a claim is $60. An industry comparison showed this amount to be larger than most other insurance companies, so the company instituted cost-cutting measures. To evaluate the effect of the cost-cutting measures, the supervisor of the department selected a random sample of 26 claims processed last month. Problem, continued n=26 (26 random samples of claims selected) Sample mean 56.42 Sample standard deviation: 10.04 Significance level 0.01 Null and alternate hypothesis? (The manager wants to see if the cost-cutting is helping the cost!) Go! Let’s work this together! H0; H1 State these in “English” and in “Geek” Significance: 0.01 Test statistic: One-tail t-test because we have no σ, therefore we substitute the sample standard deviation. Critical value: One tail, degree of freedom (n-1), 26-1=25. Formulate the decision rule: “We will reject the null hypothesis if the computed/calculated t-statistic is less than -2.485.” Math! x 56.42 60 t s n t 10.04 26 1.818 Interpret the Results -1.818 is NOT less than -2.485, therefore we do NOT reject the null hypothesis The cost-cutting did not yield significant costper-claim savings. Back to the drawing board, Claims Department Supervisor…something else may be the cost culprit. Again! The mean length of a counterbalance bar is 43 millimeters. The production supervisor is concerned that the adjustments of the machine producing the bars has changed…or he suspects they have changed. He asks engineering to investigate. They randomly pull 12 bars from the population and measures each. Raw data to follow… Data Problem: Is the machine out of tolerance? (Has there been a change in the settings?) Raw Data: 42, 39, 42, 45, 43, 40, 39, 41, 40, 42, 43, 42. Calculate the sample mean and standard deviation (using Excel® or your calculator). State the null and alternate hypothesis---one or two tailed test? Use the 0.02 significance level! Five Step Process….GO! What do we have? H 0 : 43 H1 : 43 Significance level 0.02, two-tail test. Critical value: df=12-1=11, c.v.=+2.718 and 2.718…two-tailed test, right? Test Statistic: t-test..we have no sigma, small n. Decision rule: “We will reject the null hypothesis if the computed/calculated t-value is greater than 2.718 or less than -2.718.” Nitty Gritty Math! Sample mean: 41.5 Sample standard deviation: 1.784 Computed t-value: -2.913 x t s n 41.5 43 t 2.913 1.784 12 Bottom Line? Last step: “We reject the null hypothesis— there is a significance change in the machine’s adjustments causing the parts produced to be smaller than specifications allow.” Time to get the machine fixed…possibly alert customers of the issue….recall? Business decisions are often made based on…STATISTICAL HYPOTHESIS TESTING! Test of Hypothesis, One Proportion z p 1 n Aside from the math, this is handled just like a regular z-test! Where: p=the sample proportion π=the population proportion n=sample size One Proportion Example A sample of 100 observations revealed that p=0.75. At the 0.05 significance level can the null hypothesis be rejected? {Generic Problem} H 0 : 0.70 H1 : 0.70 One Proportion Example Step One: Null and alternate given to us. Step Two: Level of Significance: given to us as 0.05, one-tail test. Step Three: Identify the test statistic---one proportion z-test. Step Four: “We will reject the null hypothesis if the computed z-value is greater than 1.65.” One Proportion Example Math Time! z 0.75 0.70 0.70 1 0.70 100 1.09 Interpret the results: The computed z-value of 1.09 is NOT greater than the 1.65 critical value, therefore, we cannot reject the null hypothesis. There is no significant increase in the sample proportion when compared to the population proportion.