Key Ideas Section 81: Overview hypothesis testing Hypothesis Hypothesis Test Section 82: Basics of Hypothesis Testing Null Hypothesis


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1 Chapter 8 Key Ideas Hypothesis (Null ad Alterative), Hypothesis Test, Test Statistic, Pvalue Type I Error, Type II Error, Sigificace Level, Power Sectio 81: Overview Cofidece Itervals (Chapter 7) are great for estimatig the value of a parameter, but i some situatios estimatio is ot the goal. For example, suppose that researchers somehow kow that the average umber of fast food meals eate per week by families i the 1990s was 3.5. Furthermore, they are iterested i seeig if families today eat more fast food tha i the 1990s. I this case, researchers do t eed to estimate how much fast food families eat today. All they eed to kow is whether that average is larger tha 3.5. For the purposes of this example, let µ deote the average umber of fast food meals eate per week by families today. The researchers questio could the be formulated as follows: Is the average umber of meals eate today the same as i the 1990s (µ = 3.5) or is it more (µ > 3.5)? There are two differet possibilities, ad the correct aswer is impossible to kow, sice µ is ukow. However, the researchers could observe the sample mea ad determie which possibility is the most likely. This process is called hypothesis testig. Defiitio Hypothesis A claim or statemet about a property of the populatio (e.g. µ = 3.5 from above) Hypothesis Test A stadard procedure for testig a claim about a property of the populatio. To guide this discussio of hypothesis testig, remember the Rare Evet Rule : If, uder a give assumptio, the probability of a particular observed evet is exceptioally small, we coclude that the assumptio is probably ot correct. Your fried claims to have a fair coi, ad you wat to test this claim. If p = P(Head), the we are testig the claim that p = 0.5. Suppose you flip the coi 100 times. a. If you get 54 heads ad 46 tails, would you coclude p = 0.5? Here, you probably would. We should expect aroud 50 heads ad 50 tails, ad this is ot far off from the mark. This is the kid of outcome oe would expect from a fair coi. b. If you get 89 heads ad 11 tails, would you coclude p = 0.5? You probably would ot say p = 0.5 i this case. Why does the coi seem ufair? If it really was fair, the this kid of outcome (while possible) would have a very low chace of occurrig. It is much more likely that the coi is actually ufair. This is the rare evet rule metioed above. Here, the give assumptio is that the coi is fair. Sice the probability of 89 heads ad 11 tails o a fair coi is exceptioally small, we coclude that the give assumptio (fair coi) is probably ot correct. c. What would you say if you got 40 heads ad 60 tails? 65 heads ad 35 tails? This is gettig ito a kid of gray area where it is hard to decide whether a fair coi would have a good chace of gettig these outcomes. Whe observatios are o the borderlie like this, it is harder to choose betwee the two possible scearios. Makig these decisios is where hypothesis testig comes ito play. Sectio 82: Basics of Hypothesis Testig Defiitios Null Hypothesis The ull hypothesis, deoted H 0, is a statemet that the value of a populatio parameter (e.g. populatio mea, proportio, or stadard deviatio) is equal to a particular value. For the purposes of the test, we assume that the ull hypothesis is true, ad the decide whether there is eough evidece to reject that assumptio. For example, here are some ull hypotheses: H 0 : µ = 3.5 H 0 : p = 0.5 H 0 : σ = 12 (first example) (secod example) Alterative Hypothesis The alterative hypothesis, deoted H 1, is a statemet that the parameter has some value that is differet tha the oe i the ull hypothesis. These statemets are all iequalities ad come i 3 forms: >, <, ad.
2 Here are some examples of alterative hypotheses: H 0 : µ > 3.5 H 0 : p 0.5 H 0 : σ < 12 (first example) (secod example) Test Statistic The test statistic is a value that is used to decide whether to reject the ull hypothesis. It is a quatity based o the sample data ad has a kow distributio whe the ull hypothesis is true. This process will be discussed i more detail soo. The Geeral Idea of a Hypothesis Test To ru a hypothesis test, there are a few geeral steps, which will be elaborated o later. 1. Based o the questio you wat to aswer, formulate the test as a choice betwee two hypotheses (the ull ad alterative). 2. Fid a test statistic whose distributio is kow whe the ull hypothesis H 0 is true. 3. Figure out if the value of the test statistic computed for your sample is a ulikely value from that distributio. 4. If it is ulikely eough, reject H 0 ad coclude that H 1 is more likely. If it is ot ulikely, the coclude that there is ot eough evidece to reject H 0. Note that this is ot sayig H 0 is true rather it is just sayig that there is t eough evidece to coclude that it is false. : I the fast food example from the start of this chapter, researchers wat to test whether the average umber of fast food meals eate per week by moder families is larger tha it was i the 1990s, whe it was 3.5. Suppose that they kow the populatio stadard deviatio is σ = 0.6, the sample size is 100, ad the sample mea was x = There are two possibilities: the ew mea is still 3.5 or the ew mea is larger tha 3.5. Thus, the hypotheses are: H 0 : µ = 3.5 H 1 : µ > If H 0 were true, is there some expressio (a test statistic) whose distributio is kow? Notice that here we also kow that σ = 0.6, = 100, ad x = Recall: x µ x x µ Z = = has a kow distributio. σ x σ This distributio is the stadard ormal distributio. 3. What is the value of Z for this sample? x µ Z = = σ = Is this value of Z ulikely for a stadard ormal distributio? Yes, it is. Remember, the chace of Zscores beig above 3 is very low. I this case, it is much more likely that the ew mea is larger tha 3.5. Therefore, we would reject H 0. Determiig Which Values are Ulikely I the previous example, we kow that was a ulikely value for the stadard ormal distributio. However, the defiitio of ulikely is subjective, ad it is importat to fid a way to make a clear cutoff for too ulikely versus likely eough. To do this, we itroduce some ew termiology: Defiitios Critical Regio The set of all values of the test statistic that lead to rejectio of the ull hypothesis (i.e. the area where values would be cosidered too extreme). Sigificace Level The probability that the test statistic will fall i the critical regio whe H 0 is actually true. I other words, this is the chace that we would mistakely reject H 0 eve though it is actually true. This probability is deoted α, ad it is typically a small value, like α = 0.05 or α = Critical Value The cutoff value betwee the critical regio ad the rage of acceptable values.
3 DecisioMakig i Hypothesis Testig There are 3 geeral approaches to hypothesis testig, ad the differeces betwee the methods are i the decisiomakig process at the ed of the test. These approaches are: 1. The Traditioal Method (ot covered i this course) 2. The PValue Method 3. The Cofidece Iterval Method The PValue Method Oe drawback to the traditioal method is that a ew critical value must be computed for each value of the sigificace level α. Thus, while oe perso thiks a area of α = 0.05 is small eough, someoe else might thik α must be smaller, like The secod perso would have to recompute the critical value for α = 0.01 to ru the test at their sigificace level. To solve this problem, we ca use the pvalue method. 1. Formulate the questio ito a ull hypothesis H 0 ad a alterative hypothesis H Idetify a test statistic which has a kow distributio whe H 0 is true. 3. Compute the test statistic from the sample data. 4. Fid the pvalue for the test statistic, which is the probability of havig a value at least as extreme as the value of the statistic: If H 1 has a > sig, the pvalue is the area above the test statistic ( at least as extreme meas larger ) If H 1 has a < sig, the pvalue is the area below the test statistic ( at least as extreme meas smaller ) If H 1 has a sig, the pvalue is the area above ad below the test statistic ad its opposite. So if the statistic is positive, it is the area above the test statistic ad below its egative. If the statistic is egative, it is the area below the statistic ad above its absolute value. ( at least as extreme meas larger i absolute value )
4 5. If the pvalue is less tha α, reject H 0. Otherwise, there is ot eough evidece to reject H 0. The Cofidece Iterval Method for twosided tests: 1. Compute a cofidece iterval for the populatio parameter as i Chapter Sice the cofidece iterval cotais all likely values of the parameter, reject H 0 if the quatity i the ull hypothesis does ot fall i the cofidece iterval. If it does fall i the iterval, there is ot eough evidece to reject H 0. Types of Error There are two differet ways that the hypothesis test could give the wrog coclusio. 1. H 0 is i reality true, but the test rejects H 0. This is called Type I Error, ad is deoted α (it is also called the sigificace level). 2. H 0 is i reality false, but the test does ot reject H 0. This is called Type II Error, ad is deoted β. Reality H 0 True H 0 False Test Decisio Do ot reject H 0 Reject H 0 Correct Aswer Type I Error (α) Type II Error (β) Correct Aswer I testig situatios, α is chose (most scietists select α = 0.05 or α = 0.01). β caot be selected, but there are techiques to reduce it: For fixed α, icreasig the sample size will reduce β. For fixed sample size, icreasig α will reduce β. To decrease both α ad β, icrease the sample size. Remark: The quatity 1 β is the probability of rejectig H 0 whe it is actually false (see table above). It has a special ame, which is the power of a test. Power is somethig we will ot worry about i this class, but to icrease it, oe ca use the techiques above (reducig β is the same as icreasig power). See p for more iformatio. Sectio 83: Testig a Claim About a Proportio I order to test a claim about a proportio, the hypothesis test requires that a few coditios be met. These coditios satisfy some of the theoretical assumptios made i usig this test. I particular, testig claims about a proportio requires that the Cetral Limit Theorem be used i order to approximate a Biomial distributio with a Normal distributio (we did ot cover that sectio). Coditios 1. The sample must be a simple radom sample. 2. The coditios for a Biomial distributio must be met (i.e. idepedet trials with 2 outcomes, P(success) = p ad is the same for each trial) 3. p 5 ad q 5 (Coditio 2 makes sure it is biomial, ad Coditio 3 is for the approximatio with a ormal distributio) Notatio = sample size p = the true populatio proportio of successes q = 1 p ˆp = the sample proportio of successes Usig the Hypothesis Test for a Proportio To ru the hypothesis test, we use the followig test statistic (otice that the compoets of it are similar to the oes used i the cofidece itervals i the previous chapter):
5 ˆp p Test Statistic: Z = Critical Values ad PValues come from the Stadard Normal Distributio pq (Here, p is the value used i the ull hypothesis) To test a hypothesis about the populatio proportio, follow these steps: 1. Write dow the ull ad alterative hypotheses, as give i the statemet of the problem. 2. Idetify the values of importace: p, q, ˆp,, α 3. Calculate the test statistic Z. 4. Note which sig is used i the alterative hypothesis H Usig either the traditioal or pvalue method, determie whether the test statistic is ulikely. With a > sig, you look at the area at the high ed of the distributio. With a < sig, you look at the area at the low ed of the distributio. With a sig, you divide the area betwee the high ad low eds of the distributio. 6. If it is ulikely, reject H 0. Otherwise, there is ot eough evidece to reject H 0. Poll workers wat to determie if the atioal percetage of Americas who idetify themselves as supportig a Idepedet political party is larger tha 8%. To do this, they sample 64 people ad fid that 9 of them are Idepedet. Determie whether the poll workers suspicios are correct usig sigificace level α = = 64, α = 0.05, p = 0.08, q = = 0.92, p ˆ = 9 = H 0 : p = 0.08 H 1 : p > 0.08 pˆ p Test Statistic: Z = = = = pq (0.08)(0.92) PValue Method Sice H 1 has a > sig, we wat to fid area above Z = for the stadard ormal distributio. From the ZTable, this area is = Now we compare this area to α ad see that < Therefore, we reject H 0 ad coclude that the atiowide percetage of Idepedets is larger tha 8%. The ewspaper claims that 50% of Americas ow a pet, but Fred thiks that it is differet tha that. To test this claim, he takes a simple radom sample of 100 Americas ad fids that 57 of them ow pets. Test his claim with sigificace level α = = 100, α = 0.05, p = 0.50, q = = 0.50, p ˆ = 57 = H 0 : p = 0.50 H 1 : p 0.50 pˆ p Test Statistic: Z = = = = pq (0.50)(0.50) PValue Method
6 Sice H 1 has a sig, we wat to fid area above Z = 1.40 ad below  Z = for the stadard ormal distributio. From the ZTable, this area is 2(0.0808) = Now we compare this area to α ad see that > Therefore, we do ot reject H 0. I other words, we coclude that there is ot eough evidece to claim that the percetage of Americas owig pets is ot 50%. Sectio 84: Testig a Claim About a Mea: σ Kow To test a claim about a mea, we must agai rely o the Cetral Limit Theorem. As a result, there are a few requiremets that must be met i order for these techiques to work well. Coditios 1. The sample must be a simple radom sample. 2. The populatio stadard deviatio σ is kow. 3. Either the populatio is ormally distributed or > 30. (Coditio 3 is for the approximatio with a ormal distributio usig the CLT) Notatio = sample size µ = the true populatio mea σ = the populatio stadard deviatio x = the sample mea Usig the Hypothesis Test for a Mea To ru the hypothesis test, we use the followig test statistic (otice that the compoets of it are similar to the oes used i the cofidece itervals i the previous chapter): x µ Test Statistic: Z = Critical Values ad PValues come from the Stadard Normal Distributio σ (Here, µ is the value used i the ull hypothesis) To test a hypothesis about the populatio mea, follow these steps: 1. Write dow the ull ad alterative hypotheses, as give i the statemet of the problem. 2. Idetify the values of importace: µ, σ, x,, α 3. Calculate the test statistic Z. 4. Note which sig is used i the alterative hypothesis H Usig either the traditioal or pvalue method, determie whether the test statistic is ulikely. With a > sig, you look at the area at the high ed of the distributio. With a < sig, you look at the area at the low ed of the distributio. With a sig, you divide the area betwee the high ad low eds of the distributio. 6. If it is ulikely, reject H 0. Otherwise, there is ot eough evidece to reject H 0. The ewspaper claims that the average umber of pets owed by Americas is 1.5, but Fred thiks that it is differet tha that. To test this claim, he takes a simple radom sample of 100 Americas ad fids a sample average of 1.4. Test his claim with sigificace level α = 0.05, assumig that the populatio stadard deviatio is 0.4. = 100, α = 0.05, µ = 1.5, σ = 0.4, x = 1. 4 H 0 : µ = 1.5 H 1 : µ 1.5 x µ Test Statistic: Z = = = = 2. 5 σ
7 PValue Method Sice H 1 has a sig, we wat to fid area above Z = 2.5 ad below  Z = 2.5 for the stadard ormal distributio. From the ZTable, this area is 2(0.0062) = Now we compare this area to α ad see that < Therefore, we reject H 0. The evidece suggests that the average umber of pets owed by Americas is less tha 1.5. Egieers i a bottle maufacturig facility wat to see if the average width of the bottle wall is less tha 0.1 cm. To test this, they radomly sample 500 bottles ad measure the wall width. The sample average width is cm. They also kow from previous experiece that the stadard deviatio i wall widths for the populatio is 0.01 cm. Ru a hypothesis test with α = 0.05 to see if the populatio average wall width is less tha 0.1 cm. = 500, α = 0.05, µ = 0.1, σ = 0.01, x = H 0 : µ = 0.1 H 1 : µ < 0.1 x µ Test Statistic: Z = = = = σ PValue Method Sice H 1 has a < sig, we wat to fid area below Z = for the stadard ormal distributio. From the ZTable, this area is Now we compare this area to α ad see that < Therefore, we reject H 0. The evidece suggests that the average wall width is less tha 0.1 cm. Sectio 85: Testig a Claim About a Mea: σ Ukow Whe σ is ukow, we must rely o a differet techique for the hypothesis test. However, the Cetral Limit Theorem is still used, ad these coditios verify that the theoretical assumptios are met. Coditios 1. The sample must be a simple radom sample. 2. The populatio stadard deviatio σ is ukow. 3. Either the populatio is ormally distributed or > 30. (Coditio 3 is for the CLT) Notatio = sample size µ = the true populatio mea x = the sample mea s = the sample stadard deviatio df = 1 = degrees of freedom Usig the Hypothesis Test for a Mea To ru the hypothesis test, we use the followig test statistic (otice that the compoets of it are similar to the oes used i the cofidece itervals i the previous chapter):
8 x µ Test Statistic: t = Critical Values ad PValues come from the Studett Distributio with 1 degrees of freedom s (Here, µ is the value used i the ull hypothesis) To test a hypothesis about the populatio mea, follow these steps: 1. Write dow the ull ad alterative hypotheses, as give i the statemet of the problem. 2. Idetify the values of importace: µ, x, s,, α 3. Calculate the test statistic t. 4. Note which sig is used i the alterative hypothesis H Usig either the traditioal or pvalue method, determie whether the test statistic is ulikely. With a > sig, you look at the area at the high ed of the distributio. With a < sig, you look at the area at the low ed of the distributio. With a sig, you divide the area betwee the high ad low eds of the distributio. 6. If it is ulikely, reject H 0. Otherwise, there is ot eough evidece to reject H 0. A power compay wats to see if the average amout of curret passig through a series of coectios is larger tha 30 milliamperes (ma). They radomly select 35 of the coectios ad fid a sample average curret of 32 ma, with a sample stadard deviatio of 5 ma. Ru a hypothesis test with α = 0.05 to see if the populatio average curret is more tha 30 ma. = 35, α = 0.05, µ = 30, x = 32, s = 5, df = 1 = 34 H 0 : µ = 30 H 1 : µ > 30 x µ Test Statistic: t = = = = s PValue Method Sice H 1 has a > sig, we wat to fid area above t = for the t distributio. Now, otice that the ttable does ot allow oe to directly fid this area. However, for df = 34 we see that has a area above of 0.01 ad has a area above of Sice < t < 2.441, the pvalue will fall betwee 0.01 ad (see picture) Now we compare this area to α ad see that 0.01 < p < < 0.05 = α. Sice p < 0.05, we reject H 0. The evidece suggests that the average curret is more tha 30 ma. 10 years ago, the cesus bureau estimated the average umber of U.S. citizes residig i rural couties to be 75,000. Now, they wish to see if the average has chaged. They take a sample of 31 couties across the U.S. From the couties sampled, the average umber of residets was 77,000, with a stadard deviatio of Usig a sigificace level of α = 0.05, ru a test to see if the average has chaged over the past decade. = 31, α = 0.05, µ = 75,000, x = 77, 000, s = 5700, df = 1 = 30 H 0 : µ = 75,000 H 1 : µ 75,000 x µ 77,000 75, Test Statistic: t = = = = s
9 PValue Method Sice H 1 has a sig, the pvalue will be the area above t = ad below  t = for the t distributio. Agai, otice that the ttable does ot allow oe to directly fid this area. However, for df = 30 we see that has a twotailed area of 0.10 ad has a twotailed area of Sice < t < 2.042, the pvalue will fall betwee 0.05 ad (see picture) Now we compare this area to α ad see that p > 0.05 = α. Sice p > 0.05, we do ot reject H 0. There is ot sufficiet evidece to coclude that the average umber of residets has chaged.
1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
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