Sampling Distributions Solutions

Sampling Distributions Solutions

<p>Sampling Distributions – Solutions</p><p>1 Suppose that medical researchers want to estimate the true proportion of all teenagers with high blood pressure whose blood pressure would decrease if they took calcium supplements. To test this, they plan a clinical trial in which n = 200 teenagers with high blood pressure will be given calcium supplements. a. Suppose the clinical trial is completed, and 120 of the 200 participants experienced a decrease in blood pressure. i. What is the value of the sample proportion (also called the point estimate) pˆ for this sample? pˆ = 120/200 = 0.60</p><p> ii. Using the result found in part i, determine the value of the standard error of pˆ . Note that in many situations we will not know the true parameter, p, and thus will substitute our sample proportion for p in the calculation of the SE. This formula then for the SE( pˆ ) when p is pˆ(1 pˆ) .6(0.4) unknown is: SE( pˆ ) = = 0.035 n 200</p><p> iii. Verify that normal approximation methods would be appropriate for this study. Both n* pˆ and n*(1- pˆ ) are greater than 10 – (120 and 80) respectively.</p><p>2 Vehicle speeds at a certain highway location are believed to have approximately a normal distribution with mean =60 mph and standard deviation =6 mph. a. Speeds for a randomly selected sample of n = 16 vehicles will be recorded. Determine the values that fill in the blanks in the following sentence. For samples of n = 16 vehicles, there is about a 68% chance that the mean vehicle speed will be between ___ and ___. 6 60 1* , which is 58.5 to 61.5. 16 b. Speeds for a randomly selected sample of n = 36 vehicles will be recorded. Determine the values that fill in the blanks in the following sentence. For samples of n = 36 vehicles, there is about a 95% chance that the mean vehicle speed will be between ___ and ___. 6 60  2 , which is 58 to 62. 36 c. Using the information in part b, find the following:</p><p> i. The standard error of the sample mean, X</p><p> 6 SE( X ) =   1.00 n 36</p><p> ii. If the sample mean speed of these 36 cars was 58 mph, use the Standard Normal Table to find P( X < 58).         X   58  60 P( < 58) = P Z    P Z    P(Z  2.00) = 0.0228 X     6       n   36  iii. Write an interpretation for this outcome in part ii.</p><p>There is a approximately a 2.3% chance that a sample of 36 cars would have a mean of less than 58 mph.</p>

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