STA3032 Quiz 5 solutions - (60 0.025 4.8 2.00 1.116 75 S S...

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STA3032 QUIZ 5 Solutions Spring 2011 Problem 4 on page 335 (Section 5.1) [MODIFIED] Interpolation methods are used to estimate heights ABOVE SEA LEVEL FOR LOCATIONS WHERE DIRECT MEASUREMENTS ARE UNAVAILABLE. In the article “Transformation of Ellipsoid Heights to local leveling heights” (M. Yamalak and O. Baykal, Journal of Surveying Engineering, 2001: 90 – 103), a second-order polynomial method of interpolation for estimating heights from GPS measurements is evaluated. In a random sample of 74 locations, the errors made by the method averaged 3.8 cm, with a standard deviation of 4.8 cm. a. [ NEW] What is the random variable (say X) in this problem? X = the errors made by the method b. [NEW] What are given in the problem? n = 74, x = 3.8 and S X = 4.8 c. Find a 95% confidence interval for the mean error made by this method. [This is part a in the text.] Since n is not very very large you should have used the t-distribution and got (74 1, 0.025)
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Unformatted text preview: (60, 0.025) 4.8 2.00 1.116 75 S S ME t t n n-= = 9 = . Hence A 95% confidence interval for the population mean, μ, is ( 29 ( 29 (3.8 1.116) 2.684 , 4.916 Estimate ME e = = However, your book accepts using the standard normal distribution, considering n = 74 as large. Then, /2 4.8 1.96 1.094 74 S ME z n α = = = and hence, A 95% confidence interval for the population mean, μ, is ( 29 ( 29 (3.8 1.094) 2.706 , 4.894 Estimate ME e = = d. A surveyor claims that the mean error is between 3.2 and 4.4 cm. At what level of confidence can this statement be made? [This is part c in the text.] We need to follow the above steps in reverse order and write (3.2,4.4) CI = hence /2 /2 4.8 (4.4 3.2) / 2 0.6 0.558 74 ME z z =-= = = . Then /2 0.6 1.07 0.558 z = = and P(Z > 1.07) = P(Z < - 1.07) = 0.1423 = α/2. Therefore α = 0.2846 and 1 – α = 0.7154. So the CI is correct at 71.54% level of significance....
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