A sample selected from a population gave a sample proportion equal to 0.59 .

Sinead Mcgee 2021-09-30 Answered
A sample selected from a population gave a sample proportion equal to 0.59 .
Round your answers to three decimal places.
a) Make a \(\displaystyle{95}\%\) confidence interval for p assuming \(\displaystyle{n}={150}\)
( Enter your answer; confidence interval assuming \(\displaystyle{n}={150}\), lower bound ,Enter your answer; confidence interval assuming \(\displaystyle{n}={150}\), upper bound )
b) Construct a \(\displaystyle{95}\%\) confidence interval for p assuming \(\displaystyle{n}={600}\). ( Enter your answer; confidence interval assuming \(\displaystyle{n}={600}\), lower bound ,Enter your answer; confidence interval assuming \(\displaystyle{n}={600}\), upper bound )
c) Make a \(\displaystyle{95}\%\) confidence interval for p assuming \(\displaystyle{n}={1275}\). ( Enter your answer; confidence interval assuming \(\displaystyle{n}={1275}\), lower bound ,Enter your answer; confidence interval assuming \(\displaystyle{n}={1275}\), upper bound )

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

unett
Answered 2021-10-01 Author has 11211 answers
Step 1
a) Given,
\(\displaystyle\hat{{{p}}}={0.59}\)
\(\displaystyle{n}={150}\)
t-table at \(\displaystyle{95}\%\) confidence is
\(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}\)
Confidence interval \(\displaystyle=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.54}\pm{1.96}\sqrt{{{\frac{{{0.59}{\left({1}-{0.59}\right)}}}{{{150}}}}}}\)
\(\displaystyle={0.59}\pm{1.96}\sqrt{{{0}.\infty{16126666}}}\)
\(\displaystyle={0.59}\pm{0.0402}\)
\(\displaystyle={\left({0.550},\ {0.630}\right)}\)
\(\displaystyle\therefore{L}{o}{w}{e}{r}\ {B}{o}{u}{n}{d}={0.550}\)
Upper Bound \(\displaystyle={0.630}\)
Step 2
b) \(\displaystyle{n}={600}\)
\(\displaystyle\hat{{{p}}}={0.59}\)
\(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}\)
\(\displaystyle{C}{I}=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{\left\lbrace{n}\right\rbrace}\rbrace\)
\(\displaystyle={0.59}\pm{1.96}\sqrt{{{0.59}{\left({1}-{0.59}\right)}}}{\left\lbrace{600}\right\rbrace}\rbrace\)
\(\displaystyle={0.59}\pm{1.96}\sqrt{{{0.00040316666}}}\)
\(\displaystyle={0.59}\pm{0.0201}\)
\(\displaystyle={\left({0.570},\ {0.610}\right)}\)
Lower Bound \(\displaystyle={0.570}\)
Upper Bound \(\displaystyle={0.610}\)
Step 3
c) \(\displaystyle{n}={1275}\)
\(\displaystyle\hat{{{p}}}={0.59}\)
\(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}\)
\(\displaystyle{C}{I}=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.59}\pm{1.96}\sqrt{{{\frac{{{0.59}{\left({1}-{0.59}\right)}}}{{{1275}}}}}}\)
\(\displaystyle={0.59}\pm{1.96}\sqrt{{{0.00018972549}}}\)
\(\displaystyle={0.59}\pm{0.0138}\)
\(\displaystyle={\left({0.576},\ {0.604}\right)}\)
Lower bound \(\displaystyle={0.576}\)
Upper bound \(\displaystyle={0.604}\)
Have a similar question?
Ask An Expert
41
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-06
Out of a 54 person sample from our fall Math-150 classes, \(\displaystyle{12.4}\%\) of the students are left-handed. Using a TI 84 graphing calculator:
Construct an \(\displaystyle{80}\%\) confidence interval for the population proportion, p. Round to tenths of a percent and write a sentence answer.
The U.S. has \(\displaystyle{13.1}\%\) of it’s population that are left-handed. Did this fall in your confidence interval from part a?
Be sure to show all calculator commans used.
asked 2021-08-01
Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a \(\displaystyle{95}\%\) two-sided confidence interval on the death rate from lung cancer.
a) Construct a \(\displaystyle{95}\%\) two-sides confidence interval on the death rate from lung cancer. Round your answer 3 decimal places.
\(\displaystyle?\leq{p}\leq?\)
b) Using the point estimate of p obtained from the preliminary sample what sample size is needed to be \(\displaystyle{95}\%\) confident that the error in estimatimating the true value of p is less than 0.00?
c) How large must the sample if we wish to be at least \(\displaystyle{95}\%\) confident that the error in estimating p is less than 0.03 regardless of the value of p?
asked 2021-08-04
On a certain day, a large number of fuses were manufactured, each rated at 15 A. A sample of 75 fuses is drawn from the day’s production, and 17 of them were found to have burnout amperages greater than 15 A.
a) Find a \(\displaystyle{95}\%\) confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A.
b) Find a \(\displaystyle{98}\%\) confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A.
c) Find the sample size needed for a \(\displaystyle{95}\%\) confidence interval to specify the proportion to within \(\displaystyle\pm{0.05}.\)
asked 2021-05-29
Construct the indicated confidence intervals for (a) the population variance
\(\displaystyle\sigma^{{{2}}}\)
and (b) the population standard deviation
\(\displaystyle\sigma\)
Assume the sample is from a normally distributed population.
\(\displaystyle{c}={0.95},{s}^{{{2}}}={11.56},{n}={30}\)
asked 2021-09-26
A simple random sample size of 100 is selected from a population with \(\displaystyle{p}={0.40}\)
What is the expected value of \(\displaystyle\overline{{{p}}}\)?
What is the standard error of \(\displaystyle\overline{{{p}}}\)?
Show the sampling distribution of \(\displaystyle\overline{{{p}}}\)?
What does the sampling distribution of \(\displaystyle\overline{{{p}}}\) show?
asked 2021-09-19
A simple random sample is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 12.6
A. Construct a \(\displaystyle{90}\%\) confidence interval for \(\displaystyle\sigma\) if the sample size, n, is 20. (Hint: use the result obtained from part (a)) 7.94 (LB) & 23.66 (UB))
B. Construct a \(\displaystyle{90}\%\) confidence interval for \(\displaystyle\sigma\) if the sample size, n, is 30. (Hint: use the result obtained from part (b)) 8.59 (LB) & 20.63 (UB)
asked 2021-08-09
Your client from question 4 decides they want a \(\displaystyle{90}\%\) confidence interval instead of a \(\displaystyle{95}\%\) one. To remind you - you sampled 36 students to find out how many hours they spent online last week. Your sample mean is 46. The population standard deviation, \(\displaystyle\sigma\), is 7.4. What is the \(\displaystyle{90}\%\) confidence interval for this sample mean? Keep at least three decimal places on any intermediate calculations, then give your answer rounded to 1 decimal place, with the lower end first then upper.
(_______ , ___________ )
...