Question 8 10 points A doctor at a walkin clinic is studying

Question 8 [10 points]

A doctor at a walk-in clinic is studying high blood pressure among patients of the ages twenty to forty. According to the American Heart Association, a person has high blood pressure if their Systolic and Diastolic readings are above 139 and 89 respectively. A random sample of 14 patients revealed the following blood pressures, where the readings (x,y) correspond to (Systolic,Diastolic) blood pressures.

Assume that the population standard deviation is not known.
For full marks your answer should be accurate to at least two decimal places.

a) Find the mean diastolic reading of blood pressure for this sample.

Mean = 0

b) Obtain a point estimate of the population mean of diastolic blood pressures.

Estimate of Population Mean = 0

c) Construct a 99-percent confidence diastolic reading estimate for the population mean.

Confidence Interval = (0,0)

(129,80)

(130,89)

(190,110)

(120,84)

(160,109)

(160,109)

(159,99)

(130,89)

(159,90)

(179,100)

(159,90)

(139,85)

(160,100)

(139,85)

A doctor at a walk-in clinic is studying high blood pressure among patients of the ages twenty to forty. According to the American Heart Association, a person has high blood pressure if their Systolic and Diastolic readings are above 139 and 89 respectively. A random sample of 14 patients revealed the following blood pressures, where the readings (x,y) correspond to (Systolic,Diastolic) blood pressures. Assume that the population standard deviation is not known. For full marks your answer should be accurate to at least two decimal places. a) Find the mean diastolic reading of blood pressure for this sample. Mean = 0 b) Obtain a point estimate of the population mean of diastolic blood pressures. Estimate of Population Mean = 0 c) Construct a 99-percent confidence diastolic reading estimate for the population mean. Confidence Interval = (0,0)

Solution

(a) mean = (80+...+85)/14 =94.21429

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(b)Estimate of Population Mean = 94.21429

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(c) The degree of freedom =n-1=14-1=13

Given a=1-0.99= 0.01, t(0.005, df=13) =3.01 (from student t table)

Sample standard deivation =10.16161

So the lower bound is

xbar - t*s/vn= 94.21429 -3.01*10.16161/sqrt(14) =86.03972

So the upper bound is

xbar + t*s/vn =94.21429 +3.01*10.16161/sqrt(14)=102.3889

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(d)From our findings we can not conclude that the patients have a high mean blood pressure.