Lakshvinkar Bappoo HW03 PDF

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HLTH 204. Section 105 Lakshvinkar Bappoo Student ID: 20761368

HW 04

TA: Judy Castañeda

Page 1 of 5

Page 238-240 No 16 Let X be the pulserates of females Since X is normally distributed , X Normal( 74.0, 12.5 ) P (70 < X < 90 )=NormalCDF ( 70, 90,74, 12.5) =0.5252431367 The probability of a pulse

rate between 70 beats per minute and 90 beats per minute is 0.5252431367. Page 238-240 No 24 Let X be the lengths of pregnancies

a.

Since X is normally distributed , X Normal ( 268, 15 )

9 P ( X ≥ 308 ) =NormalCDF ( 308, 9 ,268, 15 ) =0.0038304251 The probability of a

pregnancy lasting 308 days or longer is 0.0038304251 and there is a stronger chance that the husband is not the father of the child. b.

P ( X 17 ) =NormalCDF (17 , 9 ,14.4 , 1 .0 )=0.0046612218 The probability of 1

randomly selected man having a hip breadth greater than 17 inches is 0.0046612218

b.

n=122

X 1 Normal ( 14.4 ,1 .0)

X 122 Normal ( 14.4 , 1.0 )

Let X be the average of hipbreadth of men

(

Since the individual hip breadths is normally distributed , X Normal 14.4 ,

1.0 √ 122

)

X Normal ( 14.4, 0.90535746) 9 P ( X >17 ) =Normal CDf ( 17 , 9 , 14.4 , 0.90535746) =0 The probability that the 122

randomly selected females have a mean hip breadth greater than 17 is 0.

c. The result from (a) should be taken into consideration because the passengers of the airline cannot be only men. There could be also women as passengers, and they may have also larger hip breadth.

HLTH 204. Section 105 Lakshvinkar Bappoo Student ID: 20761368

HW 04

TA: Judy Castañeda

Page 4 of 5

Page 295-299 No 22 Let p1placebo denotes the true population proportion of the individuals who got allergic reactions when being treated with placebo. Let p1drug denotes the true population proportion of the individuals who got allergic reactions when being treated with drug. We are given: n placebo =270. x placebo=7

ndrug =863 . x drug=8 Since we are using the plus-four method, we

will use: n placebo =270+ 4=274 . x placebo =7+2=9

ndrug =863+4=867 . xdrug =8+ 2=10

C . L=0.95 α=0.05 The function we need to use is 1-PropZInterval. Using calculator; The 95% for p1placebo is (0.01174, 0.05395) or (1.174%, 5.395%) The 95% for p1 drug is (0.00443, 0.01864) or (0.443%, 1.864%) Since the 95% confidence intervals of placebo group (1.174%, 5.395%) and the group treated with the drug (0.443%, 1.864%) overlap, the difference between the groups is not significant. We can say that there are less allergic reactions from the group treated with the drug.

Page 309-314 No 18 This question cannot be answered because we do not have normality of the population and we cannot assume that the population is normally distributed.

Page 309-314 No 24

HLTH 204. Section 105 Lakshvinkar Bappoo Student ID: 20761368

HW 04

TA: Judy Castañeda

Page 5 of 5

a. Let acupuncture denote the true population mean of people getting migraine attacks treated with acupuncture. We are given: X acupuncture =1.8

S acupuncture =1.4

n=142( n ≥ 30 )

C . I =0.95 ⟹ α=0.05 Using

the decision tree for acupuncture, with the help of CLT, we compute 95% CI via the approximate Z-Interval. Using Calculator; 95% CI for acupuncture is (1.5697, 2.0303) b. Let sham denote the true population mean of people getting migraine attacks when given a sham treatment. We are given: X sham=1. 6

S sham =1. 2

n=80 ( n≥ 30 )

C . I =0.95 ⟹ α=0.05 Using the

decision tree for sham, with the help of CLT, we compute 95% CI via the approximate Z-Interval. Using Calculator; 95% CI for sham is (1.337, 1.863)

c. Since the 95% confidence intervals of group treated with acupuncture (1.5697, 2.0303) and the group received a sham treated (1.337, 1.863) overlap, the difference between the groups is not significant. We can say that there are more migraine attacks from the group treated with acupuncture compared to the group which received a sham treatment....