Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 380 drivers and find that 288 claim to always buckle up. Construct a 97% confidence interval for the population proportion that claim to always buckle up Box 1: Enter your answer using interval notation. Example: [2.1,5.6172) Use U for union to combine intervals. Example: (-o0,2] U [4.00) Enter DNE for an empty set. Use oo to enter Infinity License Points possible: 1 Unlimited attempts.

Please show steps in IT84 calculator please.

Answer 1

A (1-$\alpha$)% for population proportion is given by

p $\in$ ( $\widehat{p} - Z_{\alpha /2}*\sqrt{\widehat{p}*(1-\widehat{p})/n}$ , $\widehat{p} + Z_{\alpha /2}*\sqrt{\widehat{p}*(1-\widehat{p})/n}$ )

where $\widehat{p}$ = X/n , X = number of drivers who claim to buckle up and n = sample size , $\alpha =0.03$

Hence X= 288 , n = 380 and $\widehat{p}$ = 288/380 = 0.758 and $Z_{\alpha /2}$ = Z _0.015 = 2.1701.

Hence p $\in$ ( 0.758 - 2.1701 * $\sqrt{0.758*0.242/380}$ , 0.758 +2.1701 * $\sqrt{0.758*0.242/380}$ )

or p $\in$ ( 0.710 , 0.806 )

We have used simple arithmetic functions and square root function along with values from normal probability tables to find out that a 97% confidence interval for the population proportion that claim to always buckle up is (0.710, 0.806)