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Suppose that several insurance companies conduct a survey. They randomly surveyed 400 drivers and found that 240 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up a. (.20) n= b. (.20) p, = c. (.20) The standard deviation for p d. (.20) The z value for a 95% confidence interval is e. (.20) Construct a 95% confidence interval for the population proportion that claim to always buckle Fill in the blanks to clarify the following diagram. LL (lower limit) = UL (upper limit) =

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Ans.: A 95% confidence interval for the population Excel formula n that claim to alwavs buckle n- 95% Confidence Interval 0.60 E4E3 Std Error(S.E.0024 SQRT(E5*(1-E5yE3) Confidence = 97% 95% α , l-Co fidence% 1--=1 el-, , l-0.95 0.025 l=1-(1-E992 0.975 196 NORMSINV(E10) 0.5521书-E12E7 0.648 |-65HE12E7 10 l-a/2 = 4ms = 196-> Use the Excel fonnula-NRMSINV(0.975) 14 Lower Limit ofCL= 15 Upper limit ofCL= 17 18Suppose that several insurance companies conduct a survey. They randomly surveyed 400 drivers and found that 240 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up. a. (.20) n= 400 b. (.20) p0.60 c. (20) The standard deviation for p0.0245 d.(20)The z value for a 95% confidence interval is | 1.96 e. (.20) Construct a 95% oonfidence interval for the population proportion that claim to always buckle following diagram. Fill in the blanks to clarify the LL (lower limit) = 0.552 | UL (upper limit) = 0.648

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