The DMV reports that 5.6% of all drivers will get into an accident in the next y

Question

The DMV reports that 5.6% of all drivers will get into an accident in the next year. A researcher wants to know if the proportion of teenage drivers who get into an accident is different. She surveys 10,000 teenage drivers and finds that 614 got into an accident in the last year. Find the 99% confidence interval for the population proportion of teenage drivers that will get into an accident. Find the 90% confidence interval for the population proportion of teenage drivers that will get into an accident. Which confidence interval has a larger margin of error? Explain why it may have a larger margin of error.

Explanation / Answer

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.0614          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.002400626          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.006183602          
lower bound = p^ - z(alpha/2) * sp =   0.055216398          
upper bound = p^ + z(alpha/2) * sp =    0.067583602          
              
Thus, the confidence interval is              
              
(   0.055216398   ,   0.067583602   ) [ANSWER]

***********************

b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.0614          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.002400626          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.003948678          
lower bound = p^ - z(alpha/2) * sp =   0.057451322          
upper bound = p^ + z(alpha/2) * sp =    0.065348678          
              
Thus, the confidence interval is              
              
(   0.057451322   ,   0.065348678   ) [ANSWER]

******************************

c)

The 99% confidence interval has a larger margin of error. [ANSWER]

Larger confidence levels yield greater margin of errors because you have to take a larger interval to be "more confident" that you have the true proportion.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.