An online used car company sells second-hand cars. For 30 randomly selected tran

Phoebe 2021-09-20 Answered
An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price is 2800 dollars.
a) Assuming a population standard deviation transaction prices of 130 dollars, obtain a \(\displaystyle{99}\%\) confidence interval for the mean price of all transactions.
b) Which of the following is the correct interpretation for your answer in part (a)?
A) If we repeat the study many times, \(\displaystyle{99}\%\) of the calculated confidence intervals will contain the mean price of all transactions.
B) There is a \(\displaystyle{99}\%\) change that the mean price of all transactions lies in the interval
C) We can be \(\displaystyle{99}\%\) confident that the mean price for this sample of 30 transactions lies in the interval
D) None of the above

Expert Community at Your Service

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

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Nola Robson
Answered 2021-09-21 Author has 16534 answers
Step 1
The provided information are:
Sample size \(\displaystyle{\left({n}\right)}={30}\)
Sample mean \(\displaystyle{\left(\overline{{{x}}}\right)}={2800}\)
Population standard deviation \(\displaystyle{\left(\sigma\right)}={130}\)
Confidence level \(\displaystyle={99}\%\)
Step 2
a) Using the standard normal table, the z-critical value at \(\displaystyle{99}\%\) confidence level is 2.58.
The confidence interval is:
\(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={2800}\pm{2.58}\times{\frac{{{130}}}{{\sqrt{{{30}}}}}}\)
\(\displaystyle={\left({2738.9},\ {2861.1}\right)}\)
Thus, the \(\displaystyle{99}\%\) confidence interval is \(\displaystyle{\left({2738.9},\ {2861.1}\right)}\)
Step 3
b) If many confidence interval of same size \(\displaystyle={\left({30}\right)}\) is taken and confidence interval is calculated for each of the sample, then about \(\displaystyle{99}\%\) of them will contain the true population parameter.
Thus, the correct option is (A).
Not exactly what you’re looking for?
Ask My Question
10
 

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 2021-08-09
Customers randomly selected at a grocery store included 172 women and 45 men. 74 of the women and 12 of the men used coupons. Find the \(\displaystyle{95}\%\) confidence interval.
asked 2021-10-02

It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of \(\displaystyle{95}\%\) confidence level for \(\displaystyle\pi\) ratio of smokers in the population.
a) \(\displaystyle{P}{\left({0.45}{<}{p}{<}{0.78}\right)}={0.95}\)
b) \(\displaystyle{P}{\left({0.11}{<}{p}{<}{0.28}\right)}={0.95}\)
c) \(\displaystyle{P}{\left({0.51}{<}{p}{<}{0.78}\right)}={0.95}\)
d) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.59}\right)}={0.95}\)
e) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.39}\right)}={0.95}\)

asked 2021-08-06
In a sample of 7 cars, each car was tested for nitrogen-oxide emissions and the following results were obtained: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, 0.15. Assuming that the sample is representative of all cars in use and that they are normally distributed, construct a \(\displaystyle{98}\%\) confidence interval estimate for the mean amount of nitrogen-oxide emission for all cars.
Use the sample data from #4 to construct a \(\displaystyle{98}\%\) confidence interval estimate for the standard deviation of the amount nitrogen-oxide emission for all cars.
asked 2021-05-11
Car Repaors An automobile lealesrship has found that for every 140 cars sold, 23 will be brought back to the dealer for major repairs. If the dealership sells 980 cars this year, approximately how many cars will be brought back for major repairs?
asked 2021-12-30
Here are summary statistic for randomly selected weights of newborn girls: \(\displaystyle{n}={211},\ \overline{{{x}}}={33.7}{h}{g},\ {s}={6.8}{h}{g}.\) Construct a confidence intercal estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval \(\displaystyle{32.1}{h}{g}{ < }\mu{ < }{34.3}{h}{g}\) with only 16 sample values, \(\displaystyle\overline{{{x}}}={33.2}{h}{g}\), and \(\displaystyle{s}={2.6}{h}{g}\)?
What is the confidence interval for the population mean \(\displaystyle\mu\)?
Are the results between the two confidence intervals very different?
a) Yes, because the confidence interval limits are not similar.
b) No, because each confidence interval contains the mean of the other confidence interval.
c) No, because the confidence interval limits are similar
d) Yes, because one confidence interval does not contain the mean of the other confidence interval.
asked 2021-09-29

The monthly incomes for 12 randomly selected​ people, each with a​ bachelor's degree in​ economics, are shown on the right. Complete parts​ (a) through​ (c) below.
Assume the population is normally distributed.
a) Find the sample mean.
b) Find the sample standard deviation
c) Construct a \(\displaystyle{95}\%\) confidence interval for the population mean \(\displaystyle\mu\). A \(\displaystyle{95}\%\) confidence interval for the population mean is
\(\begin{array}{|c|c|}\hline 4450.42 & 4596.96 & 4366.46 \\ \hline 4455.62 & 4151.52 & 3727.77 \\ \hline 4283.26 & 4527.94 & 4407.68 \\ \hline 3946.49 & 4023.61 & 4221.73\\ \hline \end{array}\)

asked 2021-08-09
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 200 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.3 seats. \(\displaystyle{s}_{{{x}}}=?,\ {n}=?,\ {n}-{1}=?\). Construct a \(\displaystyle{92}\%\) confidence interval for the population average number of unoccupied seats per flight. (State the confidence interval. (Round your answers to two decimal places.)
...