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8.2 Testing a Claim About a Proportion What to read and understand in this section? This S~Qfl desaibe~ i! ~Qntpl~te bypQt;h~ll? te~oog pt~~dur~ fot tesoog i cit!im n,acJe about a population proportion p. We illustrate hypothesis testing with these three meth- ods: (1) the P-value method, (2) the critical value method, (3) the use of confidence intervals. We need to follow the 8 step testing process from the previous section. Equivalent Methods When testing claims about proportions, the P-value method and the critical value method are equivalent to each other in the sense that they both lead to the same conclusion. However, the confidence interval method is not equiv- alent to them. (Both the P-value method and the critical value method use the same standard deviation based on the claimed proportion p, but the confidence interval method uses an estimated standard deviation based on the sample proportion.) So the confidence interval method could result in a different conclusion. Key mements: Testing a Qaim About a Population Proportion (Normal Approxima- tion Method) Objective Conduct a formal hypothesis test of a claim about a population proportion p. Notation n = sample size of trials p = proportion (pis the value used in the statement of the null hypothesis) p =:. (sample proportion) n q=1-p Requirements • The sample observations are a simple random sample. • The conditions for a binomial distribution are satisfied: - There is a fixed number of trials. - The trials are independent. - Each trial has two categories of "success" and "failure." - The probability of a success remains the same in all trials. • The conditions np ~ 5 and nq ~ 5 and are both satisfied., so the binomial dis- tribution of sample proportions can be approximated by a normal distribution with 1' = np and u = npq. Note that p used here is the assumed proportion used in the claim, not the sample proportion p. 14

Test Statistic .I\ 2. :: ? - Q ~ V\ W e. 1,J 1lf l"\u+ be, C:o "'r\A, 1 1 n5 +1,,c t ~ .(. s-t,:\'tsl k M Cd \ l,\,, ) 1 / < ,._ t~h s tt d i 01 " . --rL, c. -z_ tc~ +- 3- /.e.,- I i -;1 le.. w; I I be e,,. ? ~v~ 0~ 14-e r ✓ vc-lv.c Cu,,- t"" -\-c- \ 1 O!A \J ; {,\_ .+ a ck Y'\Olo 5'- 8.2.1 Calculator Command In the previous sections, pertaining to proportions, we used the calculator function lPropZint to find the confidence interval. In this section, for proportions we use the test on the calculator called 1-PropZTest, You can find this function under the keys 'Stat' then 'Tests'. The calculator will calculate the P-value for a particular data set. Your TI84 output for this test is shown in an example below. The first line is prop< .1, this is called a left tail test(LTI), and says that we are testing the claim that a certain proportion is less than 0.1. Second line z = -4.46 (rounded), this is our z-score or z test statistic (formula was provided in section 8.1). In this case z = -4.56 (rounded) is a "unusual" value, not "usual". The 3rd line, P = 4.1151£ - 6, which is almost 7.ero, this is the P-value and the number that we really want in this P-value method. Then we will have to form the conclusion. The P-value is the area under the normal distnbution. We have been calculating these areas in both chapters 6 and 7. Adverse Reactions to Drug The drug Lipitor (atorvastatin) is used to treat high cholesterol. In a clinical trial of Tl-83/84 Plus NORHAL rLOAT AUTO REAL RADIAN HP ~ 1-ProPZTest Prop(.1 z=-4.45929186 P=4.1151493E·6 p=.0544611819 n=863 Lipitor, 47 of 863 treated subjects experienced headaches (based on data from Pfizer). 15

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week after Thanksgiving (based on data from ''Holidays, Birthdays, and Postponement of Cancer Death," by Young and Hade, Journal of the American Medical Association, Vol. 292, No. 24). If people can postpone death until after Thanksgiving, then the proportion of deaths in the week before should be less than 0.5. Use a 0.05 significance level to test the claim that the proportion of deaths in the week before Thanksgiving is less than 0.5. Based on the result, does there appear to be any indication that people can temporarily postpone death to survive the Thanksgiving holiday? Requirement Check: I. sfl~ 'L F~ col 1t1u.VV1(e. ir- o+- i"'kpc.1rJ l!.. <A+ tv;v (s Ci,.,J;viJu.c-l dl co- fl_s j li·d- lA · lt,.> 6, c c.·t-a50 1t ;c-S" ":> . Mp1 S u: 1.,cl "'tr<;- s\.,,to ""' = CD[[ .} sq~ 6 ~ J "' -= \louo {l't,0CJG) C6.S) ') / S-- ?: ~~ ( l LoOG J Co . s=- ) t/S- \l, c... l{ \,.. , /r'c.M C-"'-+ t ~ 1. Claim: p < .50 (before Thanksgiving) 2. Opposite claim p ~ .50 ~'. Ho ; 1! = :~ Hi ; 1! < '.~Q ~ ~ ~ LTI 4. « = .05 5. Normal distribution stat [Tests] [lpropZTest] (enter the stat data) n = 6062 + 5938 = 12000, x = (before Thanksgiving) Ans P-value= 0. 8712. 6. The P-value is greater than 0.05. "If P-value is high then the null will fly'' Fail to reject Ho, 7. From final conclusion flowchart, "There is not sufficient evidence to support the claim that less than 0.5 of the deaths occur the week before Thanksgiv- ing." Based on these results, there is no indication that people can temporarily postpone their death. Cellphones and cancer In a study of 420,095 Danish cell phone users, 135 subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today). Test the claim of a somewhat common belief that such cancers are affected by cell phone use. That is, test the claim 20

that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of 0.0340% for people who do not use cell phones. Because this issue has such great importance, use a 0. 005 significance level. Based on these results, should cell phone users be concerned about cancer of the brain or nervous system? Requirement Check: . \. SQ. S L fi x.. J. l') l,\.r- b c. v ~¾ ;~ dC'..~ c.. 0<la.v-J-- h ,, JJ l..,) ·1 -1- l-- -lw -o C l. -\- c. SO v°1 c. S ,- ~ ~ "21~ c;:1,-cA.- rp ,,...s f ,.,,_c~ "'- c. 41- GoC\, c.. V\cA. ~: 0 - 00 o 1 ll ('.\ hd q_--; (> , ti 'l 'l {, G 1. Given: n = 420095 and x = 135 2. claim, p -=f .-00034-orp-= · 0.-034% 3. Opposite claim p = . 00034 4. Ho : p = .00034and H1 : p f= .00034 This is a TI"f 5. rt= .005 6. normal distribution Z (TS) 7. Calculator: [stat] [Tests] [lpropZTest] (enter the stat data). Ans P-value = 0.5122 8. The P-value is greater than 0. 005. "If P-value is high then the null will fly". Fail to reject Ho. 9. From final conclusion flowchart. "There is not sufficient evidence to support the claim that the percentage of cancer with cell phone users is not equal to 0. 034% . The conclusion is that cell phone users should not be worried about cancer being caused by their cell phones. 21

P-values: Use technology or use the Student t distribution (Table A-3) with degrees of freedom given by d f = n - 1. (Figure in Section 8.1 summarizes the procedure for finding P-values.) Critical values, Use the Student t distribution (Table A-3) with degrees of freedom given by df = n -1. (When Table A-3 doesn't include the number of degrees of freedom, you could be conservative by using the next lower number of degrees of freedom found in the table, you could use the closest number of degrees of freedom in the table, or you could interpolate. There are two cases just like chapter 7, when u is known and not known. We focus on case 1 when u is not known . For the t-test described in this section, the P-value method, the critical value method, and the confidence interval method are all equiv- alent in the sense that they all lead to the same conclusion. The output for T-Test on the TI 84 is shown below . · Tl-83/84 Plus NORMAL fLOAT AUTO REAL RADIAN MP l!I 101¥11 μ(7 t=·.2897748534 P=.3886888459 x=6.aaaaaaaaa Sx=1.99240984 n=12 Adult Sleep: The author obtained times of sleep for randomly selected adult subjects included in the National Health and Nutrition Examination Study, and those times (hours) are listed below. Here are the unrounded statistics for this sample: n = 12, f = 6. 83333333 hours, s = 1.99240984 hours. A common recommendation is that adults should sleep between 7 hours and 9 hours each night. Use the P-value method with a 0. 05 significance level to test the claim that the mean amount of sleep for adults is less than 7 hours. 4 8 4 4 8 6 9 7 7 10 7 8 23

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Garlic for reducing cholesterol: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/ dL) had a mean of 0.4 and a standard deviation of 21.0 (based on data from "Effect of Raw Garlic vs Com- mercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia," by Gardner et al., Archives of Internal Medicine, Vol. 167). Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Requirement Ol.eck: (J) _s; (L .s @ h -; 1 0 ~ e-c.G<- lA.. ~ o n :; ~q S, 0 ve4u; 1v c.. .,,- .... c CAh C.. Vc vvi c -r 1. Given: n = 49, mean difference between(before and after)= 0.4 and standard deviation s = 21.0 2. Claim: μ > 0 3. Opposite claim μ ~ 0 Are requirements met? SRS and n = 49 > 30 -4. Ho : μ = -0 H1 : μ > -0 This is a RIT 5. « = .05 6. t-distribution stat {Tests} ['ITest} , {Stats} {enter the stat data) Ans P-value = 0.4472 7. The P-value is greater than 0.05. "If P-value is high then the null will fly" Fail to reject Ho. 8. Final conclusion flowchart: There is not sufficient evidence to support the claim that garlic will reduce cholesterol. Lead in Medicine: Listed below are the lead concentrations (in μg I g) measured in dif- ferent Ayurveda medicines. Ayurveda is a traditional medical system commonly used 26

in India. The lead concentrations listed here are from medicines manufactured in the United States (based on data from "Lead, Mercury, and Arsenic in US and Indian Man- ufactured Ayurvedic Medicines Sold via the Internet," by Saper et al., Journal of the American Medical Association,Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14 1'K I g. Assume that a simple random sample has been selected and test the given claim. 3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5 Requirement Check: Are requirements met? SRS but n < 30. Histogram. Normal Quantile plot? 1. Oaim: μ < 14 μ5/ry 2. Opposite clainq, ~ 14 u-5 I j 3. Ho : }l = 14 H1 : }l < 14 This is a LTI 4! g = !Q5 5. t-distribution [stat] [Tests] [TI'est] , [Stats] (enter the 10 data points in the list) Ans P-value = 0.0913 6. The P-value is greater than 0.05. "If P-value is high then the null will fly" Fail to reject Ho. 7. Final conclusion flowchart: There is not sufficient evidence to support the claim that the mean lead levels is less than 14 microgram per gram. 27