A government sample survey plans to measure the LDL (bad) cholesterol level of a

Question

A government sample survey plans to measure the LDL (bad) cholesterol level of an SRS of men aged 20 to 34. Suppose that in fact the LDL cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean mu = 117 milligrams per deciliter (mg/dL) and standard deviation sigma = 25 mg/dL. Use Table A for the following questions, where necessary. (a) Choose an SRS of 100 men from this population. What is the sampling distribution of x? (Use the units of mg/dL.) N(117, 25) N(0.117, 0.025) N(117, 0.25) N(117, 2.5) N(1.17, 0.25) What is the probability that x takes a value between 114 and 120 mg/dL? This is the probability that x estimates mu within plusminus 3 mg/dL. () (b) Choose an SRS of 1000 men from this population. Now what is the probability that x falls within plusminus 3 mg/dL of mu? The larger sample is much more likely to give an accurate estimate of mu. (Use Table A and round your answer to 3 decimal places.)

Explanation / Answer

(a) Population mean = 117 mg/dL and = 25 mg/dL

so for a sample size = 100 men

so sampling distribtuion

sample xbar =   = 117 mg/dL

and standard error sx = /sqrt(n) = 25/sqrt(n) = 25/ sqrt(100) = 25/10 = 2.5 mg/dL

so option d is correct N(117,2.5)

P( 114<= xbar <= 120) = P(xbar <= 120) - P(114<= xbar )

so z values for X = 120 mg/dL => z = (120 - 117)/ 2.5 = 1.2 so respeective P - value = 0.8849

so z - values for X = 114mg/dL => z = (114 - 117)/ 2.5 = -1.2 so respective P - value = 0.1151

P( 114<= xbar <= 120) = P(xbar <= 120) -  P(114<= xbar ) = 0.8849 - 0.1151 = 0.7698

(b) N = 1000 mean so

now standard error of sample mean = /sqrt(n) = 25/sqrt(1000) = 0.79

P( 114<= xbar <= 120) = P(xbar <= 120) - P(114<= xbar )

so z values for X = 120 mg/dL => z = (120 - 117)/ 0.79 = 3.79 so respeective P - value = 0.00007

so z - values for X = 114mg/dL => z = (114 - 117)/ 2.5 = -1.2 so respective P - value = 0.99993

P( 114<= xbar <= 120) = P(xbar <= 120) -  P(114<= xbar ) = 0.99993 - 0.00007= 0.9998

so its. 99.98 % probability that xbar falls within +- 3 mg/dL

Yes, the larger sample size is much much more likely to give an accurate estimate of  

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