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William Christensen, Ph.D. What is a Hypothesis? Hypothesis – (in statistics) a claim or statement about a property of a population. Example: a claim that the mean weight of men is greater than 175 lbs. Example: a claim that less than 50% of households in the U.S. have internet access Example: a claim that annual income levels in St. George have a standard deviation greater than $10,000 Note: Hypothesis is singular and hypotheses is plural Hypothesis Testing In this section you will learn how to test hypotheses related to means, proportions, and standard deviations (or variances) We use sample data to test claims about population parameters (means, proportions, and standard deviations) Statistics is all about chance or probability, so our hypothesis tests are always based on some given chance of being right (or wrong) in our answer Null & Alternative Hypothesis Since we do not know in advance what the results are going to be of a hypothesis test, we ALWAYS state the hypothesis in two forms: 1. H0 (read H-oh or H-zero) is the sign to represent the “Null Hypothesis” and always contains an =, , or sign 2. H1 (read H-one) is the sign to represent the “Alternative Hypothesis” and always contains the opposite sign (≠, , ) Null & Alternative Hypothesis Here are examples of all the kinds of hypothesis testing we learn in this section, shown in their proper format: Hypothesis tests for µ (population Mean) The claim being tested always determines the value that is used, but in this case we use a value of 100 just as an example Null Hypothesis Alternative Hypothesis Case 1 H0: µ = 100 H1: µ ≠ 100 Case 2 H0: µ 100 H1: µ 100 Case 3 H0: µ 100 H1: µ 100 Null & Alternative Hypothesis Here are examples of all the kinds of hypothesis testing for p (population Proportion) The claim being tested always determines the value that is used, but in this case we use a value of 0.50 (50%) just as an example. Note that proportions like probabilities must be between 0.0 and 1.0 Null Hypothesis Alternative Hypothesis Case 1 H0: p = 0.50 H1: p ≠ 0.50 Case 2 H0: p 0.50 H1: p 0.50 Case 3 H0: p 0.50 H1: p 0.50 Null & Alternative Hypothesis Here are examples of all the kinds of hypothesis testing for σ and σ2 (population Standard Deviation and Variance) The claim being tested always determines the value that is used, but in this case we use a value of 10 for standard deviation and 100 for variance just as an example Null Hypothesis Alternative Hypothesis Case 1 H0: σ = 10 H1: σ ≠ 10 Case 2 H0: σ 10 H1: σ 10 Case 3 H0: σ 10 H1: σ 10 Null Hypothesis Alternative Hypothesis Case 1 H0: σ2 = 100 H1: σ2 ≠ 100 Case 2 H0: σ2 100 H1: σ2 100 Case 3 H0: σ2 100 H1: σ2 100 Null & Alternative Hypothesis You may have noticed that every Null hypothesis contains a “condition of equality” ( =, , or sign) And that every Alternative hypothesis has the opposite sign as shown in the following table Case 1 Case 2 Case 3 Null Hypothesis Alternative Hypothesis = ≠ Doing a Hypothesis Test The previous slides show every possible combination of Null and Alternative Hypotheses for population Mean, Proportion, and Standard Deviation / Variance. Now that you know what hypotheses look like, the next step is to learn how to test them. Doing a Hypothesis Test It is always the “Null Hypothesis” that is being tested. If the test “fails” then we reject the Null Hypothesis, or in other words, accept the Alternative Hypothesis If the test does not fail, then we accept the Null Hypothesis Doing a Hypothesis Test Whenever we test hypotheses we are testing a “claim” that has been made Example: The mean weight of men is greater than 175 lbs. Whenever possible, it is a good practice to state the claim in terms of the Alternative Hypothesis (not equal, less than, or greater than) rather than the Null Hypothesis (equal, less than or equal, or greater than or equal) Doing a Hypothesis Test Here are the steps in doing a hypothesis test State the Null and Alternative hypotheses Determine the “Critical Value(s)” and “Critical Region(s)” – this is what we’ll learn next Calculate the “Test Statistic” and compare it against the Critical Value(s) 1. 2. 3. If the Test Statistic lies within the Critical Region then reject the null hypothesis (accept the alternative hypothesis) If the Test Statistic does not lie within the Critical Region then fail to reject the null hypothesis (accept the null hypothesis) Critical Region(s) and Value(s) Critical values are the z-scores or x values associated with the standard normal distribution. These values are established based on the level of confidence or alpha value we choose. Critical regions are bordered by a critical value and include all the area outside (toward the tail). Whenever we do hypothesis testing we do it based on some level of “confidence” or α (alpha) value For example, an α = 0.05 represents a 95% confidence level or 95% chance that our answer will be truly correct and a 5% chance that our answer will not be correct. The total area contained in the critical region(s) equals alpha. E.g., if α = 0.05 then the total combined area in these critical regions equals 5% of the total area (0.05/2 = 0.025 on each side) Critical Values Critical Regions Hypothesis Test: population Mean using large sample (n 30) Assumptions When testing claims about population means from large samples we assume the following conditions exist: 1. The sample is a random sample 2. The sample is large (n 30) Critical Value(s) You actually already know how to find the critical value for hypothesis tests of population Mean using large samples. Remember the Excel function NORMSINV(probability)? It allows us to determine the x value (same as zscore) for a standard normal distribution where we know the area (probability) as measured from the extreme left out to the critical value. For example, if α = 0.05 (total combined area) then the area from the extreme left to the critical value on the left can be found by NORMSINV(0.025) and the critical value of the right can be found by NORMSINV(0.975) Critical Values Area from left = 0.975 Area from left = 0.025 0.025 0.025 µ Critical Values Remember the 3 types of null / alternative hypothesis combinations we have: This is very important in order to properly determine the critical value(s) for nearly ALL HYPOTHESIS TESTS Null Hypothesis Alternative Hypothesis Two-tail test (the area represented by alpha is split equally between the left and right tails) = ≠ Right-tail test (the area represented by alpha is entirely in the right tail with no left tail at all) Left-tail test (the area represented by alpha is entirely in the left tail with no right tail at all) The easy way to remember this is that the sign of the Alternative hypothesis points the direction of the test. A ≠ sign kind of points both ways and represents a twotail test. A sign points right, and a sign points left. µ Critical Value Example α = 0.05 CV = -1.96 Null Hypothesis Alternative Hypothesis = ≠ Two-tail test CV = 1.96 Area = α/2 = 0.025 Area = α/2 = 0.025 Right-tail test Left-tail test CV = -1.645 CV = 1.645 Area = α = 0.05 Area = α = 0.05 Notice how critical values for the left-tail are ALWAYS negative and critical values for the right-tail are ALWAYS positive µ Critical Value Exercise α = 0.01 CV = -2.576 Null Hypothesis Alternative Hypothesis = ≠ Two-tail test CV = 2.576 Area = α/2 = 0.005 Area = α/2 = 0.005 Right-tail test Left-tail test Use Excel function NORMSINV to confirm these critical values are correct CV = -2.326 Area = α = 0.01 CV = 2.326 Area = α = 0.01 Test Statistic A “Test Statistic” is a value we calculate from the sample data We use the Test Statistic to test our hypothesis by comparing the Test Statistic against the critical value(s) The following graph shows how the position of the Test Statistic determines whether we accept the Null hypothesis or the Alternative hypothesis If the Test Statistic falls in the critical region(s) then reject the null hypothesis (accept the alternative hypothesis) If the Test Statistic does not fall in the critical region(s) then “fail to reject H0 (accept the null hypothesis) Test Statistic for µ For large samples, testing claims about population means we calculate the Test Statistic using the following formula (you must remember this one) Test Statistic = x-µ σ n Test Statistic for µ For large samples This is the sample average/mean which we calculate from our sample data Test Statistic = n is simply the size of our sample. E.g., if our sample consists of 50 men, then n=50 x-µ σ n We never really know the population mean (µ) but we do have a value included in the hypothesis (e.g., H0: µ=100) and it is this value that we put here – the value from the hypothesis If we know σ (population std. deviation) then we put it here. However, we usually only have sample data, so we use s (sample std. deviation) here most of the time Hypothesis Test Example I testing a claim about µ from a large sample We generally assume that the average or normal human body temperature is 98.6 degrees F. To test this assumption or claim, data was collected from 106 healthy adults (n = 106). From this sample, the following statistics were obtained. Mean temperature = 98.2o Standard deviation (s) = 0.62o Using an alpha of 0.05 (a 95% confidence level that our solution will be correct, with a 5% chance of being wrong) test the claim that the mean body temperature of all healthy adults is equal to 98.6o. Hypothesis Test Example I testing a claim about µ from a large sample 1. Write out the claim and the hypotheses • The claim is that the mean body temperature of healthy adults is equal to 98.6o or µ = 98.6o • Notice that we are making a claim about the population of all healthy adults, but we only have a sample of 106 adults upon which to test this claim • Thus, the null hypothesis H0: µ = 98.6o • The alternative hypothesis H1: µ ≠ 98.6o • You must be able to read ‘word problems’ and be able to translate them into null and alternative hypotheses Hypothesis Test Example I testing a claim about µ from a large sample 2. Draw a picture of the normal distribution and determine the critical values • Note that the sign of the alternative hypothesis is ≠ so we have a two-tail test Note: since the standard normal distribution is perfectly symmetrical, you can simply take the negative critical value on the left and change it to a positive value for the right side – or go ahead and use NORMSINV again to solve for the right side α = 0.05, α/2 = 0.025 0.4750 0.025 0.4750 0.025 -1.96 1.96 Critical values are determined using Excel NORMSINV Hypothesis Test Example I testing a claim about µ from a large sample 3. Calculate the Test Statistic and form your final conclusion • • Remember we were given the following information: • n = 106, s = 0.62o, sample mean = 98.2o, α = 0.05, and our null hypothesis is µ = 98.6o We can now calculate the Test Statistic using the formula we learned: x µ zTS = s n = 98.2 - 98.6 0.62 106 = - 6.64 Note: If the test statistic falls inside the critical values then we always accept the null hypotheses (mean body temperature equals 98.6), and if the test statistic falls outside the critical values we always accept the alternative hypothesis (mean body temperature does not equal 98.6). Hypothesis Test Example I testing a claim about µ from a large sample 3. Calculate the Test Statistic and form your final conclusion We calculated the test statistic zTS = -6.64 CONCLUSION: Accept the alternative hypothesis H1: µ ≠ 98.6o The mean body temperature of healthy adults is not 98.6o F Accept the alternative hypothesis Accept the alternative hypothesis Accept the null hypothesis 0.025 Test statistic falls outside the critical value -1.96 0.025 1.96 Critical values determined using Excel NORMSINV Hypothesis Test Example II testing a claim about µ from a large sample Men are often accused of channel-surfing (constantly changing channels using the remote control). With this in mind, a study was done to test the claim that, on average, men change channels every 5.00 seconds or less. To investigate this claim, sample data was collected and the following statistics were obtained: n = 80 (80 men were observed) Sample mean = 5.25 seconds standard deviation = 2.50 seconds Use an alpha = 0.10 to test this claim. Hypothesis Test Example II testing a claim about µ from a large sample 1. Write out the claim and the hypotheses • The claim is that men change channels every 5 seconds or less or µ 5.00 seconds • Notice that we are making a claim about the population of all TV-watching men, but we only have a sample of 80 men upon which to test this claim • Thus, the null hypothesis H0: µ 5.0 seconds • The alternative hypothesis H1: µ 5.0 seconds • REMEMBER that the sign for the null hypothesis MUST ALWAYS be =, , or and the sign for the alternative hypothesis is always ≠, , or (the opposite of the null hypothesis) Hypothesis Test Example II testing a claim about µ from a large sample 2. Draw a picture of the normal distribution and determine the critical values • Note that the sign of the alternative hypothesis is so we have a right-tail test Note: since we have a one-tail test to the right, the entire critical region is on the right side and has an area equal to alpha. We can either use NORMSINV(0.10) and change the answer to a positive value, or use NORMSINV(0.90) which will return the same positive value. α = 0.10 0.500 0.400 0.100 1.28 Critical value is determined using Excel NORMSINV Hypothesis Test Example II testing a claim about µ from a large sample 3. Calculate the Test Statistic and form your final conclusion • • Remember we were given the following information: • n = 80, s = 2.50 seconds, sample mean = 5.25 seconds, α = 0.10, and our null hypothesis is µ 5.0 seconds We can now calculate the Test Statistic using the formula we learned: x µ zTS = s n = 5.25 – 5.00 2.50 80 = 0.89 Note: If the test statistic falls inside the critical value then we always accept the null hypotheses (men switch channels every 5 seconds or less), and if the test statistic falls outside the critical value we always accept the alternative hypothesis (men wait more than 5 seconds before switching channels). Hypothesis Test Example II testing a claim about µ from a large sample 3. Calculate the Test Statistic and form your final conclusion We calculated the test statistic zTS = 0.89 CONCLUSION: Accept the null hypothesis H0: µ 5.0 seconds Men switch channels every 5 seconds or less Accept the alternative hypothesis Accept the null hypothesis 0.100 Test statistic falls inside the critical value 1.28 Critical value determined using Excel =NORMSINV(0.90) Hypothesis Test Example II testing a claim about µ from a large sample • • • • • Discussion: Perhaps you noticed in the last exercise that the sample mean was 5.25 seconds and yet we concluded that men change channels every 5 seconds or less. How is this possible? You must remember that we are talking about probabilities here. We did this problem based on a 90% confidence of being correct (10% chance of being wrong). Therefore, we are not certain about our conclusion, it is just our best estimate based on the info we were given. We do know that a sample of 80 men does NOT give us exactly the same data as we would have if we knew the channel-surfing habits of all men in our population. Certainly if the mean channel-surfing interval for all men was 5.25 seconds then it would be ridiculous to make any other claim. However, in this case we must make our conclusion about the entire population of men based on our sample of only 80 men. Since our sample mean of 5.25 seconds was pretty close to our claimed population mean of 5.00 seconds, and since we had only 80 men in our sample, statistically (at alpha = 0.10) we did not have enough evidence to reject the null hypothesis that the population mean could actually be 5 seconds or less even though the sample mean was more than 5 seconds. On your own, try doing the problem again assuming the sample size was 200 men instead of 80 men. Does it change the answer? Hypothesis Test: population Mean using small sample (n 30) Assumptions When testing claims about population means from small samples we assume the following conditions exist: 1. The sample is a random sample 2. The sample is small (n 30) Critical Value(s) Finding the critical value(s) when doing hypothesis testing of population means using small samples is similar to the method we learned when we have large samples, but with some important differences. For one thing, rather than using the Excel function NORMSINV(probability) as we did with large samples, we must now use the Excel function TINV(probability,deg_freedom) where degrees of freedom equals n-1. Also, and this is really important, this function assumes we are doing a two-tail test, so we enter alpha (not alpha/2) when doing a two-tail test, and we must enter 2 times alpha (alpha x 2) if we are doing a one-tail test. Finally, TINV always returns a positive (right-side) value, so we must know and remember to make the value negative if using it on the left-side For example, if α = 0.05 (total combined area), and n=20, then a left-tailed test critical value would be found by negating the value we get with TINV(0.10,19). For a right-tailed test critical value we can simply use TINV(0.10,19). For a two-tailed test we take the positive and negative value of TINV(0.05,19). Critical Value(s) two-tail hypothesis test For example, if α = 0.05 (total combined area), and n=20, then the right-side critical values is found by TINV(0.05,19) and for the left-side critical value we simply take this same value and make it negative (-2.09). Critical Values 0.025 -2.09 0.025 2.09 Critical Value(s) right-tail hypothesis test For example, if α = 0.05 (total area), and n=20, then, for a right-tail test, the critical value is found by TINV(0.10,19) (note we must double alpha). If this was a left-tail test we would do the same thing except take the value and make it negative. Critical Value 0.05 1.729 Critical Value(s) small samples The smaller the sample size, the larger the critical value. The t-distribution (TINV) we use takes into account smaller samples and requires the sample data to evidence greater differences from the claimed µ (population mean) before the results will reject the null hypothesis (i.e., accept the alternative hypothesis). Test Statistic for µ For small samples, when testing claims about population means we use the same formula we used with large samples to calculate the Test Statistic (you must remember this) Test Statistic = x-µ σ n Test Statistic for µ For small or large samples This is the sample average/mean which we calculate from our sample data Test Statistic = n is simply the size of our sample. E.g., if our sample consists of 50 men, then n=50 x-µ σ n We never really know the population mean (µ) but we do have a value included in the hypothesis (e.g., H0: µ=100) and it is this value that we put here – the value from the hypothesis If we know σ (population std. deviation) then we put it here. However, we usually only have sample data, so we use s (sample std. deviation) here most of the time Hypothesis Test Example testing a claim about µ from a small sample Because of the expense involved, car crash tests often use small samples. When 5 BMW cars are crashed under standard conditions, the repair costs (in dollars) are: $797, $571, $904, $1147, and $418 (mean=$767.40, s=$284.73). Use a 0.05 significance level to test the claim that the mean repair cost for all BMW cars is less than $1,000. Would BMW be justified in advertising that the average repair cost is less than $1,000? Hypothesis Test Example testing a claim about µ from a small sample 1. Write out the claim and the hypotheses • The claim is that the average repair cost for a BMS is less than $1,000 or µ $1,000 • Notice that we are making a claim about the population of all BMWs, but we only have a sample of 5 crashed BMWs upon which to test this claim • Thus, the null hypothesis H0: µ $1,000 • The alternative hypothesis H1: µ $1,000 • REMEMBER that our claim may be the null hypothesis or the alternative hypothesis, but the sign for the null hypothesis MUST ALWAYS be =, , or and the sign for the alternative hypothesis is always ≠, , or (the opposite of the null hypothesis) Hypothesis Test Example testing a claim about µ from a small sample 2. Draw a picture of the normal distribution and determine the critical values • Note that the sign of the alternative hypothesis is so we have a left-tail test α = 0.05 Note: since we have a one-tail test we must enter the probability in TINV as 2 x alpha (0.05 x 2 = 0.10). Degrees of freedom = n-1 = 5 – 1 = 4 0.450 0.500 0.05 -2.132 Change sign to negative since value is on left-side Critical value is determined using Excel TINV Hypothesis Test Example testing a claim about µ from a small sample 3. Calculate the Test Statistic and form your final conclusion • • Remember we were given the following information: • n = 5, sample mean = $767.40 and sample standard deviation (s) = $284.73 • α = 0.05, and our null hypothesis is µ $1,000 We can now calculate the Test Statistic using the formula we learned: 767.4 – 1000 tTS = xs - µ =284.73 = -1.83 n 5 Note: If the test statistic falls inside the critical value then we always accept the null hypotheses, and if the test statistic falls outside the critical value we always accept the alternative hypothesis. Hypothesis Test Example testing a claim about µ from a small sample 3. Calculate the Test Statistic and form your final conclusion We calculated the test statistic tTS = -1.83 CONCLUSION: Accept the null hypothesis H0: µ $1,000 The average BMW repair cost is greater than or equal to $1,000 Accept the alternative hypothesis Accept the null hypothesis 0.05 Test statistic falls inside the critical value -2.132 Critical value determined using Excel =TINV(0.10,19) Hypothesis Test Example testing a claim about µ from a small sample • • • • • Discussion: Perhaps you noticed in the last exercise that the sample mean was $767.40 and yet we concluded that the mean BMW repair cost is greater than or equal to $1,000. How is this possible? You must remember that we are talking about probabilities here. We did this problem based on a 95% confidence of being correct (5% chance of being wrong). Therefore, we are not certain about our conclusion, it is just our best estimate based on the info we were given. We do know that a sample of 5 BMWs does NOT give us a very good understanding of ALL BMWs (the population). Certainly if the mean repair cost for all BMWs involved in crashes was less than $1,000 then we could make that claim without equivocation. However, in this case we must make our conclusion about the entire population of BMWs involved in crashes based on our sample of only 5 crashed BMWs. Since our sample mean of $767.40 is not that far from $1,000 (our claimed population mean), and since we had only 5 BMWs in our sample, statistically (at alpha = 0.05) we did not have enough evidence to reject the null hypothesis that the population mean could actually be $1,000 or more, even though the sample mean was less than $1,000. On your own, try doing the problem again using an alpha of 0.10. Does it change the answer? Can you see how statistics can be used to sell almost anything? Hypothesis Test: population Proportions Assumptions When testing claims about population proportions we assume the following conditions exist: 1. The sample is a random sample 2. The conditions for a binomial experiment are satisfied The are only two possible outcomes (e.g., yes/no, is/isn’t, etc.) 3. Assuming the n*p 5 and n*q 5, we can use the standard normal distribution to determine our critical values with µ = n*p and σ = sqrt(n*p*q) Population Proportions: Notation n = number of trials p̂ (p-hat) = sample proportion p-hat is sometimes given directly example: 10% of observed sports cars are red, thus p-hat = 0.10 remember p-values must always be between 0 and 1 other times p-hat must be calculated (example: 96 households have cable TV and 54 do not, thus the proportion of households with cable TV = 96 / (96 + 54), so p-hat = 0.64 p = population proportion (used in the null hypothesis) q= 1-p Critical Value(s) Finding the critical value(s) when doing hypothesis testing of population proportions is exactly the same as finding critical value(s) for population means using large samples. That is, we use the Excel function NORMSINV(probability) Please review finding critical values for population means using large samples if you need to review this Critical Value(s) two-tail hypothesis test For example, if α = 0.05 (total combined area) then the critical values are NORMSINV(0.025) for the leftside and NORMSINV(0.975) for the right-side (or simply change the left-side critical value to a positive number and use it for the right-side critical value) Critical Values 0.025 -1.96 0.025 1.96 Critical Value(s) right-tail hypothesis test For example, if α = 0.05 (total area), for a right-tail test the critical value is NORMSINV(0.95). Or, we could take the left-tail critical value NORMSINV(0.05) and simply change it to a positive value. Critical Value 0.05 1.645 Test Statistic for p For testing claims about population proportion we use the following formula for the Test Statistic (you must remember this) Test Statistic (zTS) = p̂ p pq n Test Statistic for p This is the sample proportion which we are given or can calculate from the sample data We never really know the population proportion (p) but we do have a value included in the hypothesis (e.g., H0: p=0.50) and it is this value that we put here – the value from the hypothesis Test Statistic (zTS) = This is the same value (p) used in the numerator and found in the hypotheses. The value of q is simply 1 – p (p+q=1) p̂ p pq n n is simply the size of our sample. E.g., if our sample consists of 150 households then n=150 Test Statistic for p CAUTION When calculating the test statistic or p-hat, do not round until the very end. Rounding during calculation can cause large errors in the final answer. Remember, p-values are always between 0 and 1 and so they are often very small numbers. Hypothesis Test Example testing a claim about p Geico Insurance company obtains a simple random sample of 850 drivers and finds that 544 of them use cell phones while driving. Use a 0.05 significance level (α = 0.05) to test the claim that most drivers (that means more than 50% or p 0.50) use cell phones while driving. Hypothesis Test Example testing a claim about p 1. Write out the claim and the hypotheses • The claim is that most drivers use cell phone while driving or p 0.50 • Notice that we are making a claim about the population of all drivers, but we only have a sample of 850 drivers upon which to test this claim • Thus, the null hypothesis H0: µ 0.50 • The alternative hypothesis H1: p 0.50 • REMEMBER that our claim may be the null hypothesis or the alternative hypothesis, but the sign for the null hypothesis MUST ALWAYS be =, , or and the sign for the alternative hypothesis is always ≠, , or (the opposite of the null hypothesis) Hypothesis Test Example testing a claim about p 2. Draw a picture of the normal distribution and determine the critical value(s) • Note that the sign of the alternative hypothesis is so we have a right-tail test α = 0.05 Critical Value 0.05 1.645 Critical value is determined using Excel NORMSINV Hypothesis Test Example testing a claim about p 3. Calculate the Test Statistic and form your final conclusion • • p̂ p pq n Remember we were given the information: n = 850 and 544 of these 850 drivers use cell phones while driving. Thus we can calculate p-hat (drivers in our sample who whose cell phones while driving) as 544 / 850 = 0.64. Remember, p-values MUST ALWAYS and CAN ONLY be between 0 and 1. We can now calculate the Test Statistic using the formula: We know that p-hat=0.64, and p=0.50 (from the hypotheses), and since q=1-p then q=1-0.5=0.5, and from the problem we know n=850 So, we know all the parts and can calculate the Test Statistic = (0.64-0.50) / sqrt ((0.50*0.50) / 850) = 8.16 Hypothesis Test Example testing a claim about p 3. Calculate the Test Statistic and form your final conclusion We calculated the test statistic = 8.16 CONCLUSION: Accept the alternative hypothesis H1: p 0.50 Most (more than 50%) of all drivers use cell phone while driving Accept the alternative hypothesis Test statistic falls outside the critical value Accept the null hypothesis 0.05 CV =1.645 Critical value determined using Excel =NORMSINV(0.95) Hypothesis Test: population Standard Deviation or Variance Assumptions When testing claims about population standard deviation or variance we assume the following conditions exist: 1. The sample is a random sample 2. The population has values that are normally distributed (this is a strict requirement but confirming this assumption is beyond the scope of this class so see me if you need to know how to do this) 2 Critical Value(s) Finding the critical value(s) when doing hypothesis testing of population standard deviation or variance is a little different than anything we have done so far in this section. Here we use the Chi-squared (2) distribution and the Excel function CHIINV where degrees of freedom (df) = n-1 2 Critical Value(s) The 2 (chi-squared) distribution looks like this. Notice how it starts at 0. That means there are NO NEGATIVE values in a chi-squared distribution. That also means we have NO NEGATIVE critical values when using a chi-squared distribution. In addition, since the distribution is NOT SYMMETRICAL, we must determine left-side values separately from right-side values 0 2 Critical Value(s) two-tail hypothesis test For example, if α = 0.05 (total combined area), and n=11 (df=n-1=10) then the critical values are CHIINV(0.975,10) for the left-side and CHIINV(0.025,10) for the right-side No, I did not get that backwards, in the Chi-squared distribution the “probability” you input is the area to-theright of the critical value. If you get confused about this, just remember that the smaller number goes on the left and the larger number goes on the right. Left-side 2 critical value 0.025 CV=3.247 Right-side 2 critical value 0.025 CV=20.483 2 Critical Value(s) left-tail hypothesis test If α = 0.05 and n=11 for a left-tail test (i.e., the alternative hypothesis has the sign) then the critical value is CHIINV(0.95,10) since the areato-the-right of 0.05 in the left-tail is 0.95 Left-side 2 critical value Area-to-the-right = 0.95 0.05 CV = 3.940 2 Critical Value(s) right-tail hypothesis test If α = 0.05 and n=11 for a right-tail test (i.e., the alternative hypothesis has the sign) then the critical value is CHIINV(0.05,10) since the areato-the-right of 0.05 in the right-tail is simply 0.05 Right-side 2 critical value Area-to-the-right = 0.05 0.05 CV = 18.307 Test Statistic for 2 For testing claims about population standard deviation or variance we use the following formula for the Test Statistic (you must remember this) Test Statistic 2 ( ) = (n 1) s 2 2 Test Statistic for 2 n is simply the size of our sample. E.g., if our sample consists of 81 altimeters then n=81 s is the sample standard deviation, which we must square. Note, if given sample variance (s2) then do not square it again. Test Statistic (2) = σ is population standard deviation, which we don’t really know, but a value for σ is always included in the hypotheses. Therefore, simply take this value from the hypotheses. Note, if the hypotheses relate to variance (σ2) then put that value directly in the equation without squaring it. However, if the hypotheses relate to standard deviation (σ) then make sure to square that value in the equation. (n 1) s 2 2 Test Statistic for 2 CAUTION You cannot have a negative chi-squared value. In a two-tail chi-square hypothesis test, the smaller value always goes on the left. square chi-squared (2). Chisquared is just a name or label, don’t think there is a chi-value that you then must square NEVER Hypothesis Test Example testing a claim about 2 Aircraft altimeters (the instrument that tells a pilot how high the plane is) made by XYZ Company in the past have a standard deviation of 43.7 ft. (that means that on average, they are off by about 43.7 ft.). XYZ recently purchased new new production equipment and a sample of 81 new altimeters have a standard deviation of 52.3 ft. Using an alpha of 0.05 test the claim that the new altimeters have a standard deviation different from the old altimeters. Note: different means “not equal to”. Hypothesis Test Example testing a claim about 2 1. Write out the claim and the hypotheses • The claim is that the population of altimeters made on the new equipment have a standard deviation different from the old altimeters (43.7 feet) or σ ≠ 43.7 ft. • Notice that we are making a claim about the population of new altimeters, but we only have a sample of 81 new altimeters upon which to test this claim • Thus, the null hypothesis H0: σ = 43.7 ft. • The alternative hypothesis H1: σ ≠ 43.7 ft. Hypothesis Test Example testing a claim about 2 2. Draw a picture of the chi-squared distribution and determine the critical value(s) • Note that the sign of the alternative hypothesis is ≠ so we have a two-tail test α = 0.05, df = n-1 = 81-1 = 80 0.025 57.153 0.025 106.629 Critical values are determined using Excel CHIINV Hypothesis Test Example testing a claim about 2 3. Calculate the Test Statistic and form your final conclusion • • From the problem we know that the sample standard deviation (s) = 52.3 ft. and the population standard deviation from the hypotheses (σ) = 43.7 ft. Now we can use our Test Statistic formula: Test Statistic (2) = 2 = (81-1)*52.32 / 43.72 = 114.586 (n 1) s 2 2 Hypothesis Test Example testing a claim about 2 3. Calculate the Test Statistic and form your conclusion We calculated the test statistic = 114.586 CONCLUSION: Accept the alternative hypothesis H1: σ ≠ 43.7 ft. It appears the new production equipment does a worse job at controlling standard deviation than the old equipment Test statistic falls outside the critical value – accept the alternative hypothesis Accept the alternative hypothesis Accept the alternative hypothesis Accept the null hypothesis 0.025 57.153 0.025 106.629 Critical values determined using Excel = CHIINV Hypothesis Tests: End Notes Type I Error A Type I Error is made when we accept the alternative hypothesis when, in fact, we should have accepted the null hypothesis Type I Error is represented by α (alpha) Type II Error A Type II Error is made when we accept the null hypothesis when, in fact, we should have accepted the alternative hypothesis Type II Error is represented by (beta) This is the first and last time that I will mention beta in this class – there is nothing you need to remember about beta Controlling Type I and Type II Errors The bottom-line when it comes to controlling for Type I and Type II errors is to have as large a sample size as possible. Larger sample size means less Type I and less Type II error Hint for Test How to correctly state the conclusion of a hypothesis test! (MUST KNOW FOR TEST – see practice exam for more examples) A. B. Remember we are testing a claim, so the correct wording of a conclusion requires us to correctly state the claim One of two outcomes results from any hypothesis test – you must pick the conclusion that matches the test results 1) 2) We find sufficient evidence to support the claim We do not find sufficient evidence to support the claim Hint for Test How to correctly state the conclusion of a hypothesis test! (Example) Question: A researcher claims that more than 44% of voters favor a certain initiative. A test of the hypothesis is conducted and the null hypothesis cannot be rejected. Which of the following sentences most correctly states the conclusion? Hint: first note that the claim (p > 0.44) is H1, so H0 must be p ≤ 0.44. Since we cannot reject the null hypothesis (per the test results), we are sort of accepting the null, which means there is NOT sufficient evidence to support the alternative hypothesis (our claim). Thus, we can analyze each of the following sentences to see which is most correct. First, see if the claim is correct, then see if the conclusion is correct. A. There is sufficient evidence to support the claim that the true proportion is less than 44% (this sentence incorrectly states both the claim and the conclusion) B. There is sufficient evidence to support the claim that the true proportion is greater than 44% (this sentence correctly states the claim, but incorrectly states the conclusion) C. There is not sufficient evidence to support the claim that the true proportion is less than 44% (this sentence incorrectly states the claim, although the conclusion is correct) D. There is not sufficient evidence to support the claim that the true proportion is greater than 44% (this sentence correctly states both the claim and the conclusion) William Christensen, Ph.D.