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(B) In a hospital, an orthopedic doctor collects a random sample of 52 previous patients and checks to see if pre-surgical physical therapy strength training helps to speed their recovery after a total hip replacement. She finds that in 86.54% of the 52 patients that indeed it does. Since the doctor didn't sample every possible surgical patient she has to use a confidence interval to estimate the proportion of total hip surgical patients that will benefit from pre-surgical PT. (1) p = .8654 (ii) n = 52 (ii) np = (iv) n (1 – p) 45.0008 (this is the number showed improved recovery time.) 6.9992 (this is the number that didn't show improved recovery time.) (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need get the critical z-score. We will use 97% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .985 (vi) This gives a critical z-score of z* = 2.17 (vii) The standard error is: SE (viii) The margin of error is: МОЕ — (ix) Hence, we are 92% confident that the true proportion of surgical total hip patients that will have improved recovery time with pre-surgical PT is between and

(B) In a hospital, an orthopedic doctor collects a random sample of 52 previous patients and checks to see if pre-surgical physical therapy strength training helps to speed their recovery after a total hip replacement. She finds that in 86.54% of the 52 patients that indeed it does. Since the doctor didn't sample every possible surgical patient she has to use a confidence interval to estimate the proportion of total hip surgical patients that will benefit from pre-surgical PT. (1) p = .8654 (ii) n = 52 (ii) np = (iv) n (1 – p) 45.0008 (this is the number showed improved recovery time.) 6.9992 (this is the number that didn't show improved recovery time.) (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need get the critical z-score. We will use 97% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .985 (vi) This gives a critical z-score of z* = 2.17 (vii) The standard error is: SE (viii) The margin of error is: МОЕ — (ix) Hence, we are 92% confident that the true proportion of surgical total hip patients that will have improved recovery time with pre-surgical PT is between and

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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(B) In a hospital, an orthopedic doctor collects a random sample of 52 previous patients and checks to see if pre-surgical physical therapy strength training helps to speed their recovery after a total hip replacement. She finds that in 86.54% of the 52 patients that indeed it does. Since the doctor didn't sample every possible surgical patient she has to use a confidence interval to estimate the proportion of total hip surgical patients that will benefit from pre-surgical PT. (i) p = .8654 (ii) n = 52 (iii) np = 45.0008 (this is the number showed improved recovery time.) (iv) n(1 – p) (this is the number that didn't show improved recovery time.) 6.9992 - (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 97% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .985 (vi) This gives a critical z-score of z* = 2.17 (vii) The standard error is: SE || (viii) The margin of error is: МОЕ — (ix) Hence, we are 92% confident that the true proportion of surgical total hip patients that will have improved recovery time with pre-surgical PT is between and
Transcribed Image Text:(B) In a hospital, an orthopedic doctor collects a random sample of 52 previous patients and checks to see if pre-surgical physical therapy strength training helps to speed their recovery after a total hip replacement. She finds that in 86.54% of the 52 patients that indeed it does. Since the doctor didn't sample every possible surgical patient she has to use a confidence interval to estimate the proportion of total hip surgical patients that will benefit from pre-surgical PT. (i) p = .8654 (ii) n = 52 (iii) np = 45.0008 (this is the number showed improved recovery time.) (iv) n(1 – p) (this is the number that didn't show improved recovery time.) 6.9992 - (v) Since both of the previous answers are greater than or equal to 10 we can proceed with a normal model for the confidence interval. We now need to get the critical z-score. We will use a 97% confidence level. In MS Excel we need to use the "=norm.s.inv()" command to determine the appropriate critical z-score. =norm.s.inv( .985 (vi) This gives a critical z-score of z* = 2.17 (vii) The standard error is: SE || (viii) The margin of error is: МОЕ — (ix) Hence, we are 92% confident that the true proportion of surgical total hip patients that will have improved recovery time with pre-surgical PT is between and
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