Please help! Insurance companies are interested in knowing the population percen

Question

Please help! Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car They randomly survey 387 drivers and find that 311 claim to always buckle up. Construct a 86% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1.5] Box 1: Enter your answer using interval notation. Example: [2.1,5.6172) Use U for union to combine intervals. Example: (oo,2] U [4,00 Enter DNE for an empty set, oo for Infinity Points nossihle: 1

Explanation / Answer

Sol:

p^=sample proportion=x/n=success/total

=311/387

=0.803618

Z crit for 86%=1.4757

86% confidence interval for population proportion is

p^-Zsqrt(p^(1-p^/n),p^+Zsqrt(p^(1-p^/n)

0.803618-1.4757sqrt(0.803618(1-0.803618)/387),0.803618+1.4757sqrt(0.803618(1-0.803618)/387)

=0.7738,0.8334

lower limit=0.7738

upperlimit=0.8334

ANSWER:0.7738,0.8334

0.7738<p<0.8334

we are 86% confident that the true population proportion that claim to always buckle up lies in between

0.7738 and 0.8334

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