Practice Problem Solutios for Lecture 04.1 1) A theory exists that at the time of ovulatio, a woma s body temperature rises by 0.5 to 1.0 degrees Fahreheit ad therefore chages i body temperature ca be used to guess the day of ovulatio. However, i order to use this method, a good estimate of basal body temperature i periods whe ovulatio is defiitely ot occurrig is eeded. Suppose a woma measures her body temperature o awakeig o the first 10 days after mestruatio. She fids that her average basal body temperature for the 10 days is 97.2 F with a stadard deviatio of 0.062 F. a) What is the best estimate of her true basal body temperature? The best estimate of her true basal body temperature is her 10 day average of 97.2 F. b) Costruct a iterval estimate of her true basal body temperature. Use a cofidece level of 80%. To costruct a cofidece iterval usig the methods we kow, we must assume that basal body temperature is at least approximately ormally distributed. 0.062 97.2 1.383 97.2 0.0271 97.1729,97.2271 10 c) Iterpret the cofidece iterval that you costructed i part b). We are 80% cofidet that the iterval (97.17, 97.03) captures the woma s true mea basal body temperature. d) Provide a iterpretatio of 80% cofidece. I repeated samplig of 10 days durig which ovulatio is defiitely ot occurrig, 80 out of 100 cofidece itervals will capture the true mea basal body temperature. e) What is the probability that the iterval calculated i part b) captures the true mea basal body temperature? A cofidece iterval either captures the parameter or it does t therefore the probability is either 0 or 1. However, we caot say whether a specific iterval captures the parameter or ot we do t kow. 2) Researchers would like to estimate the cocetratio (µg/ml) of a specific dose of ampicilli i the urie after a certai period of time. They recruited 25 voluteers that received a specific dose of ampicilli ad they fid that the voluteers have a mea cocetratio of 7.0 µg/ml with a stadard deviatio of 2.0 µg/ml. Assume that the uderlyig distributio of ampicilli cocetratio is ormal. a) Compute a 90% cofidece iterval for ampicilli cocetratio. 2.0 7.0 1.711 7.0 0.6844 25 6.3156,7.6844 1
b) Compute a 95% cofidece iterval for ampicilli cocetratio. 2.0 7.0 2.064 7.0 0.8256 25 6.1744,7.8256 c) Compare the cofidece itervals i part a) ad part b). Why are they differet? The 95% cofidece iterval is wider. There is a trade off betwee cofidece ad precisio. The more cofidece we have, the less precise we will be (holdig all else costat). Also, the t critical values chage with differet cofidece levels. Note that the t critical value for the 95% cofidece iterval is greater tha the t critical value for the 90% cofidece iterval. The larger t critical value will make for a larger margi of error ad thus wider cofidece iterval. d) Compute a 90% cofidece iterval assumig that the researchers were oly able to recruit 9 voluteers. 2.0 7.0 1.86 7.0 1.24 9 5.76,8.24 e) Compare the cofidece itervals i part a) ad part d). Why are they differet? The cofidece iterval i a) with a sample size of 25 is arrower tha the cofidece iterval i d) with a sample size of 9. The sample size,, directly iflueces the width of the CI. As icreases, the SEM decreases ad as the SEM decreases, the margi of error (margi of error = t crit *(SEM)) also decreases thus creatig a arrower iterval. f) How may subjects must the researchers recruit to have a 95% CI with fial width of at most 2? Use the Z table. A fial width of 2 requires that we have to have a margi of error of 1 (oe uit o either side of the sample mea). 2.0 2.0 1 z crit 1.96 1.96 We would eed 16 subjects. 2.0 3.92 15.36 16 g) Why ca t the t table be used for part f)? The t distributio is idexed by degrees of freedom which are based o sample size. Therefore, we ca t get a t crit without kowig the sample size. 2
3) True or False. Justify. a) Assumig all else costat, as the umber of subjects icreases, the width of the cofidece iterval decreases. True. As sample size icreases, SEM decreases ad therefore the cofidece iterval becomes arrower. b) Assumig all else costat, as the cofidece icreases, the width of the cofidece iterval decreases. False. As cofidece icreases, the critical t or z values icrease. This makes the margi of error bigger ad therefore the cofidece iterval becomes wider. c) Assumig all else costat, as the observed sample variace icreases, the width of the cofidece iterval decreases. False. As sample variace icreases, SEM icreases ad therefore the cofidece iterval becomes wider. d) Assumig all else costat, as the observed sample mea icreases, the width of the cofidece iterval icreases. False. The sample mea tells you the locatio of the ceter of the cofidece iterval but gives o iformatio about the width of the cofidece iterval. e) I a study by Deiso et al. (1997), researchers foud that the 95% cofidece iterval for the true mea juice cosumptio i 2 year olds is betwee 4.99 ad 6.95 oz/day. This meas that there is a 95% chace that true mea juice cosumptio lies i the above iterval. False. The true mea juice cosumptio is either i the iterval or ot i the iterval. We do ot kow which. 3
Practice Problem Solutios for Lecture 04.2 1) Cotiie is a metabolite of icotie that ca be used to measure the amout of tobacco smokig. Cotiie i smokers follows a ormal distributio. I a sample of 55 cigarette smokers, the mea cotiie level was 202.3g/ml with a stadard deviatio of 41.9g/ml. a) What is the 95% cofidece iterval for the populatio mea cotiie level of all smokers? Provide a iterpretatio of the cofidece iterval. t of freedom) 54,.025 t50,.025 2.009 (sice 54 degrees of freedom is ot o the table, we roud dow to 50 degrees 41.9 202.3 2.009 202.311.350 (190.95,213.65) 55 We are 95% cofidet that the iterval (190.95, 213.65) captures the true mea cotiie level of all smokers. b) From large populatio studies, it is kow that cigar smokers have a mea cotiie level of 215.7. Is there evidece that the mea cotiie level i cigarette smokers is differet tha that of cigar smokers? Perform a hypothesis test at a level of sigificace of 5%. 1. Assumptios: SRS, Cotiie Levels are ormally distributed. 2. Hypothesis: H : 215.7 0 H1 : 215.7 3. Set α=0.05 4. Test Statistic: x 0 t s ~ t1 uder the ullhypothesis 202.3 215.7 t 2.372 ~ t 41.9 55 5. P Value: p 2P( t 2.372) 2P( t50 54 54 2.372) Sice 2.372 is betwee 2.009 ad 2.403 we kow 0.01 < p/2 <0.025 ad 0.02 < p < 0.05. 6. Coclusio: Sice our p value is less tha α=0.05, we reject the ull hypothesis. 7. Iterpretatio: There is sufficiet evidece to coclude that cotiie levels of cigarette smokers are differet tha 215.17, the mea cotiie levels of cigar smokers, at a 5% level of sigificace. 2) How accurate are rado detectors of a certai type sold to homeowers? To aswer this questio, uiversity researchers placed 12 detectors i a chamber that exposed them to 105 picocuries per liter (pci/l) of rado. The 4
detector readigs are summarized below. (Data provided by Diaa Schelleberg, Purdue Uiversity School of Health Scieces.). summ rado Variable Obs Mea Std. Dev. Mi Max -------------+-------------------------------------------------------- rado 12 104.1333 9.397422 91.9 122.3 a) Is there covicig evidece that the mea readig of all detectors of this type differs from the true value of 105? Perform a appropriate hypothesis test to aswer this questio. 1. Assumptios: SRS, Rado levels are ormally distributed. 2. Hypothesis: H : 105 0 H1 : 105 3. Set α=0.05 4. Test Statistic: x 0 t s ~ t1 uder the ullhypothesis 104.1333105 t 0.3195 ~ t 9.397 12 5. P Value: p 2P( t11 0.3195) Sice 0.3195 is less tha 1.363 we kow p/2>0.10 ad p>0.20. 11 6. Coclusio: Sice our p value is greater tha α=0.05, we fail to reject the ull hypothesis. 7. Iterpretatio: There is isufficiet evidece to coclude that the mea readig of all the detectors is differet tha 105 at a 5% level of sigificace. b) How does the result of the hypothesis test compare to the 95% cofidece iterval for the true populatio mea rado?. ci rado Variable Obs Mea Std. Err. [95% Cof. Iterval] -------------+--------------------------------------------------------------- rado 12 104.1333 2.712802 98.1625 110.1042 Sice the 95% cofidece iterval for the sample mea of rado detectors icludes 105 ( 0 ), we would coclude that the populatio mea rado for could reasoably be 105. This is cosistet with the results of our hypothesis test. 5
c) Is there covicig evidece that rado detectors sold to homeowers cosistetly uder detect amouts of rado? Perform a appropriate hypothesis test to aswer this questio. 1. Assumptios: SRS, Rado levels are ormally distributed. 2. Hypothesis: H : 105 0 H1 : 105 3. Set α=0.05 4. Test Statistic: x 0 t s ~ t1 uder the ullhypothesis 104.1333105 t 0.3195 ~ t 9.397 12 5. P Value: p P( t 0.3195) P( t11 11 11 0.3195) Sice 0.3195 is less tha 1.363 we kow p>0.10. 6. Coclusio: Sice our p value is greater tha α=0.05, we fail to reject the ull hypothesis. 7. Iterpretatio: There is isufficiet evidece to coclude that the mea readig of all the detectors is lower tha 105 at a 5% level of sigificace. d) How do the results of part a) ad part c) differ? I part a we tested to see if the rado levels differed from 105, i part c we tested if they were lower tha 105. The oly thig that differed betwee the two tests was the hypotheses ad the calculatio of the p values (ad iterpretatios). Note that a two sided test is always more coservative. 3) Researchers wat to estimate the mea diastolic blood pressure level of adult females i a small village i South America. From large populatio studies they kow that the variace of the diastolic blood pressure level of adult females is 126 (mmhg) 2. How may adult female villagers must they radomly select ad test if they wat to be 90% cofidet that the true mea diastolic blood pressure level of adult females i the tow is withi 4mmHg of the sample mea? Margi of Error = 4 (ot width =4) Sice MOE Z.05 126 1.645*11.225, we have 4 1.645 4.62 21. 31 4 So i order to have a 90% cofidece iterval with width of 8, we would eed to recruit 22 people. 6
4) A physical therapist estimated the mea maximal stregth of a particular muscle i a group of 15 radomly selected idividuals to be i the rage of [76.3, 92.3]. Assume that stregth scores are approximately ormally distributed with a variace of 144. a) What was the poit estimate of the true mea maximal stregth? Sice cofidece itervals for the sample mea are symmetric, we kow that the sample mea lies exactly betwee 76.3 ad 92.3. 92.3 76.3 x 84.3. 2 Or, otice the width of the cofidece iterval is 16, so the sample mea is 8 below 92.3 or 8 above 76.4. b) What level of cofidece did the physical therapist use to calculate the iterval estimate? We kow the margi of error =8, =15, ad =144 ad that MOE Z / 2 So, 12 8 Z / 2 Z / 2 2.58 15 Sice P(Z>2.58) =.005 we kow: α/2 =.005 ad α=0.01 which correspods to a 99% Cofidece iterval. 2 7