Question

The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured...

The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured in the state of Mississippi. A pilot sample of 60 drivers from Mississippi found that 18 were uninsured. Determine the total sample size needed to construct a 99% confidence interval for the proportion of uninsured drivers in Mississippi with a margin of error equal to 5%. (Use the z-score with three decimal places, Remember the rounding policy for sample size))

Homework Answers

Answer #1

Solution :

Given that,

n = 60

x = 18

Point estimate = sample proportion = = x / n =18 / 60 = 0.3

1 - = 1 - 0.3 = 0.7

margin of error = E = 0.05

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 / 0.05)2 * 0.3 * 0.7

= 557.41

sample size = n = 558  

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured...
The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured in the state of Mississippi. A pilot sample of 60 drivers from Mississippi found that 18 were uninsured. Determine the total sample size needed to construct a 99% confidence interval for the proportion of uninsured drivers in Mississippi with a margin of error equal to 5%. Use the z-score with three decimal places, Remember the rounding policy for sample size
A certain region would like to estimate the proportion of voters who intend to participate in...
A certain region would like to estimate the proportion of voters who intend to participate in upcoming elections. A pilot sample of 25 voters found that 18 of them intended to vote in the election. Determine the additional number of voters that need to be sampled to construct a 96​% interval with a margin of error equal to 0.07 to estimate the proportion.
Suppose a hospital would like to estimate the proportion of patients who feel that physicians who...
Suppose a hospital would like to estimate the proportion of patients who feel that physicians who care for them always communicated effectively when discussing their medical care. A pilot sample of 40 patients found that 18 reported that their physician communicated effectively. Determine the additional number of patients that need to be sampled to construct a 95​% confidence interval with a margin of error equal to 3​% to estimate this proportion.
A certain region would like to estimate the proportion of voters who intend to participate in...
A certain region would like to estimate the proportion of voters who intend to participate in upcoming elections. A pilot sample of 25 voters found that 17 of them intended to vote in the election. Determine the additional number of voters that need to be sampled to construct a 96?% interval with a margin of error equal to 0.05 to estimate the proportion.
A certain region would like to estimate the proportion of voters who intend to participate in...
A certain region would like to estimate the proportion of voters who intend to participate in upcoming elections. A pilot sample of 25 voters found that 21 of them intended to vote in the election. Determine the additional number of voters that need to be sampled to construct a 98?% interval with a margin of error equal to 0.04 to estimate the proportion.
In examining the credit accounts of a department store, an auditor would like to estimate the...
In examining the credit accounts of a department store, an auditor would like to estimate the true mean account error (book value – audited value). To do so, the auditor selected a random sample of 50 accounts and found the mean account error to be $60 with a standard deviation of $30. a. Construct a 95% confidence interval for the population mean account error. What conclusion can be made from this confidence interval? b. How large a sample is actually...
What sample size of U.S. adults do you need, if you would like to estimate the...
What sample size of U.S. adults do you need, if you would like to estimate the proportion of U.S. adults who are "pro-choice" with a 2.5% margin of error (at the 95% level)? 40 16 1,000 1,600 Your answer to the above question indicates that if you take a sample of that size, the sample proportion of adults who are pro-choice is: More than 2.5% away from the proportion who are pro-choice among all U.S. adults. Within 2.5% of the...
A consumer agency wants to estimate the proportion of all drivers who wear seat belts while...
A consumer agency wants to estimate the proportion of all drivers who wear seat belts while driving. What is the most conservative estimate of the sample size that would limit the margin of error to within 038 of the population proportion for a 99% confidence interval? Round your answer up to the nearest whole number. n=
A website reported that 29% of drivers 18 and older admitted to texting while driving in...
A website reported that 29% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 800 drivers 18 years and older drawn in​ 2010, 176 of the drivers said they texted while driving. Complete parts a through c below. a. Construct a 95% confidence interval to estimate the actual proportion of people who texted while driving in 2010. A 95% confidence interval to estimate the actual proportion has a lower limit of nothing​​​​​​​...
A website reported that 31% of drivers 18 and older admitted to texting while driving in...
A website reported that 31% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 400 drivers 18 years and older drawn in​ 2010, 92 of the drivers said they texted while driving. Complete parts a through c below. a. Construct a 90​% confidence interval to estimate the actual proportion of people who texted while driving in 2010.A 90​% confidence interval to estimate the actual proportion has a lower limit of nothing and...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT