The Department of Transportion (DOT) is attempting to determine the proportion o

Question

The Department of Transportion (DOT) is attempting to determine the proportion of drivers who require all passengers in the car to wear their seatbelt before putting the vehicle in drive. A survey of 114 drivers is performed and 62 people say they will not drive until all passengers in the vehicle are buckled up. To report their finding they want to create a 95% confidence interval. What would be the margin error for this confidence interval?

Question 6 options:

1) 0.0043 2) 0.0914 3) 0.0086 4) 0.0765 5) 0.0466

Explanation / Answer

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.543859649          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.046648775          
              
Now, for the critical z,              

alpha/2 =   0.025          

Thus, z(alpha/2) =    1.959963985          

Thus,              

Margin of error = z(alpha/2)*sp =    0.09142991 = 0.0914 [ANSWER, OPTION 2]      

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