Question

In: Statistics and Probability

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) What is the critical value of z (denoted zα/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
zα/2 =

(c) What is the margin of error (E) for a 90% confidence interval? Round your answer to 3 decimal places.
E =

(d) Construct the 90% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to 3 decimal places.
< p <

(e) Based on your answer to part (d), are you 90% confident that more than 5% of all U.S. adults have pinworm?

Yes, because 0.05 is above the lower limit of the confidence interval.No, because 0.05 is below the lower limit of the confidence interval.    No, because 0.05 is above the lower limit of the confidence interval.Yes, because 0.05 is below the lower limit of the confidence interval.


(f) In Sludge County, the proportion of adults with pinworm is found to be 0.15. Based on your answer to (d), does Sludge County's pinworm infestation rate appear to be greater than the national average?

No, because 0.15 is above the upper limit of the confidence interval.Yes, because 0.15 is below the upper limit of the confidence interval.    No, because 0.15 is below the upper limit of the confidence interval.Yes, because 0.15 is above the upper limit of the confidence interval.

Solutions

Expert Solution

Solution:

a)   

Let denotes the sample proportion. It is point estimate of the population proportion.

     = x/n   = 74/830 = 0.089

= 0.089

b)

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.65

c)

Now , the margin of error is given by

E = /2 *  

= 1.65* [0.089 *(1 - 0.089)/830]

= 0.016

Margin of error = 0.016

d)

Now the confidence interval is given by

( - E)   ( + E)

(0.089 - 0.016)   (0.089 + 0.016)

0.016    0.105

e)

No, because 0.05 is above the lower limit of the confidence interval.

f)

Yes, because 0.15 is above the upper limit of the confidence interval.


Related Solutions

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 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) Construct the 95% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to...
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 from...
Pinworm: In a random sample of 820 adults in the U.S.A., it was found that 76...
Pinworm: In a random sample of 820 adults in the U.S.A., it was found that 76 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) What is the critical value of z (denoted zα/2) for a 90% confidence interval? Use the value from...
In a random sample of 820 adults in the U.S.A., it was found that 86 of...
In a random sample of 820 adults in the U.S.A., it was found that 86 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) What is the critical value of z (denoted zα/2) for a 90% confidence interval? Use the value from the...
In a random sample of 810 adults in the U.S.A., it was found that 76 of...
In a random sample of 810 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? (b) What is the critical value of z for a 99% confidence interval? (c) What is the margin of error (E) for the 99% confidence interval? (d)...
A poll found that 74% of a random sample of 1029 adults said that they believe...
A poll found that 74% of a random sample of 1029 adults said that they believe in ghosts. determine the margin of error
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults...
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults from Germany were asked whether they supported the death penalty. 560 American adults and 590 German adults indicated that they supported the death penalty. Researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the deal penally in the US is different than that in Germany. Use a confidence interval (at confidence interval of 98%) to...
Pinworm: In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that...
Pinworm: In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 10% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 2.5 percentage points at the 90% confidence level. Enter your answers as whole numbers. (a) What is the minimum sample size required to obtain this type of accuracy? Use the prior...
A random sample of 125 adults was given an IQ test. It was found that 60...
A random sample of 125 adults was given an IQ test. It was found that 60 of them scored higher than 100. Based on this, compute a 95% confidence interval for the proportion of all adults whose IQ score is greater than 100. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. What is the lower limit of the 95% confidence interval? What is the upper limit...
In 1995, a random sample of 100 adults that had investments in the stock market found...
In 1995, a random sample of 100 adults that had investments in the stock market found that only 20 said they were investing for the long haul rather than to make quick profits. A simple random sample of 100 adults that had investments in the stock market in 2002 found that 36 were investing for the long haul rather than to make quick profits. Let p1995 and p2002 be the actual proportion of all adults with investments in the stock...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT