MATHEMATICAL ASSOCIATION OF AMERICA webwork m366ryu set6chap7 l 6 Set6Chap7: Pro

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MATHEMATICAL ASSOCIATION OF AMERICA webwork m366ryu set6chap7 l 6 Set6Chap7: Problem 6 point) How many rounds of gordo those physicians who play got play per year?Asurvey of 12 physicians revealed the following numbers 6, 44, 14, 4, 33, 37, 21, 15, 15, 29, 12, 47 Estimate with 92% contoence the mean number of rounds played assuming that the po is normaty asatuted With a Standara deviation per year by physicians Note: For each conndence interval, enter your answer in the form (LcL ucu You must nclude the parentheses and the comma between You have attempted this peobenn 0 times You have urimited attempts remaning

Explanation / Answer

6)

Sample size, N = 12
Sample mean, x = (6+44+14+4+33+37+21+15+15+29+12+47)/12 = 277/12,
Population Standard deviation, = 6

For 92% CI, zc = 1.75

CI = x +- zc* /sqrt(N)
Lower bound, LCL = 277/12 – [1.75*6/sqrt(12)] = 20.05
Upper bound, UCL = 277/12 + [1.75*6/sqrt(12)] = 26.11

92% CI = (20.05, 26.11)

7) a)

Sample size, N = 12
Sample mean, x = (183+222+303+262+178+232+268+201+244+183+201+140)/12 = 2617/12,
Population Standard deviation, = 50

For 95% CI, zc = 1.96

CI = x +- zc* /sqrt(N)
Lower bound, LCL = 2617/12 – [1.96*50/sqrt(12)] = 189.8
Upper bound, UCL = 2617/12 + [1.96*50/sqrt(12)] = 246.4

95% CI = (189.8, 246.4)

b) Width of confidence interval = 29 hours, Margin of error = 29/2 = 14.5 hours

Margin of error, ME = zc* /sqrt(N), need to find N
14.5 = 1.96*50/sqrt(N) -> N = (1.96*50/14.5)2 = 45.67 = 46 repair times

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