Question

Pinworm: In Sludge County, a sample of 60 randomly selected citizens were tested for pinworm. Of...

Pinworm: In Sludge County, a sample of 60 randomly selected citizens were tested for pinworm. Of these, 14 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.01 significance level.

(a) What is the sample proportion of Sludge County residents with pinworm? Round your answer to 3 decimal places. p̂ =

(b) What is the test statistic? Round your answer to 2 decimal places. zp hat =

(c) What is the P-value of the test statistic? Round your answer to 4 decimal places. P-value =

(d) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0

(e) Choose the appropriate concluding statement.

The data supports the claim that the infestation rate in Sludge County is greater than the national average.

There is not enough data to support the claim that that the infestation rate in Sludge County is greater than the national average.

We reject the claim that the infestation rate in Sludge County is greater than the national average.

We have proven that the infestation rate in Sludge County is greater than the national average.

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In Sludge County, a sample of 40 randomly selected citizens were tested for pinworm. Of these,...
In Sludge County, a sample of 40 randomly selected citizens were tested for pinworm. Of these, 9 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.01 significance level. a) What is the sample proportion of Sludge County residents with pinworm? Round your answer to 3 decimal places. p̂ = (b) What is the test...
Pinworm: In Sludge County, a sample of 60 randomly selected citizens were tested for pinworm. Of...
Pinworm: In Sludge County, a sample of 60 randomly selected citizens were tested for pinworm. Of these, 11 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.05 significance level. (a) What is the sample proportion of Sludge County residents with pinworm? Round your answer to 3 decimal places. p̂ = (b) What is the...
Pinworm: In a random sample of 790 adults in the U.S.A., it was found that 76...
Pinworm: In a random sample of 790 adults in the U.S.A., it was found that 76 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places.    (b) What is the critical value of z (denoted zα/2) for a 99% confidence interval? Use the value...
Pinworm: In a random sample of 790 adults in the U.S.A., it was found that 80...
Pinworm: In a random sample of 790 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 95% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) What is the critical value of z (denoted zα/2) for a 95% confidence interval? Use the value from...
Pinworm: In a random sample of 780 adults in the U.S.A., it was found that 76...
Pinworm: In a random sample of 780 adults in the U.S.A., it was found that 76 of those had a pinworm infestation. You want to find the 95% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) What is the critical value of z (denoted zα/2) for a 95% confidence interval? Use the value from...
Pinworm: In a random sample of 810 adults in the U.S.A., it was found that 84...
Pinworm: In a random sample of 810 adults in the U.S.A., it was found that 84 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) Construct the 99% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 74...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 74 of those had a pinworm infestation. You want to find the 90% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) Construct the 90% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to...
In a random sample of 900 teenagers, 148 had used tobacco of some form in the...
In a random sample of 900 teenagers, 148 had used tobacco of some form in the last year. The managers of an anti-tobacco campaign want to claim that less than 20% of all teenagers use tobacco. Test their claim at the 0.01 significance level. (a) What is the sample proportion of teenagers who use tobacco? Round your answer to 3 decimal places. p̂ = (b) What is the test statistic? Round your answer to 2 decimal places. zp hat =...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 70...
Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 70 of those had a pinworm infestation. You want to find the 90% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? (b) What is the critical value of z (denoted zα/2) for a 90% confidence interval? zα/2 =   (c) What is the margin of error (E) for...
Spam: Larry claims that more than a quarter of all his email is spam. In a...
Spam: Larry claims that more than a quarter of all his email is spam. In a random sample of 60 of his emails, 20 of them are spam. Test his claim at the 0.10 significance level. (a) What is the sample proportion of spam in his email? Round your answer to 3 decimal places. p̂ = (b) What is the test statistic? Round your answer to 2 decimal places. zp hat = (c) What is the P-value of the test...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT