Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that t
0 0
Add a comment Improve this question Transcribed image text
Answer #1

> 23.90227735 = upper
22.89772264 = lower

otaku haru Thu, Nov 4, 2021 3:44 PM

Add a comment
Know the answer?
Add Answer to:
An economist wants to estimate the mean per capita income (in thousands of dollars) for a...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • An economist wants to estimate the mean per capita income (in thousands of dollars) for a...

    An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Californ Suppose that the mean income is found to be $21.3 for a random sample of 2817 people. Assume the populatio standard deviation is known to be $79. Construct the 98 % confidence interval for the mean per capita income i thousands of dollars. Round your answers to one decimal place. Keypad 1 Point Answer(How to Enter) Lower endpoint Upper endpoint:

  • An economist wants to estimate the mean per capita income (in thousands of dollars) for a...

    An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $24.3$⁢24.3 for a random sample of 758758 people. Assume the population standard deviation is known to be $9.2$⁢9.2. Construct the 90%90% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

  • An economist wants to estimate the mean per capita income (in thousands of dollars) for a...

    An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $⁢24.3, and the variance is known to be $⁢94.09. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $⁢0.43? Round your answer up to the next integer.

  • An economist wants to estimate the mean per capita income (in thousands of dollars) for a...

    An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $23.8, and the variance is known to be $29.16. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $0.59? Round your answer up to the next integer.

  • Question 18 ? of 20 Step 1 of 1 An economist wants to estimate the mean...

    Question 18 ? of 20 Step 1 of 1 An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $23.4. and the variance is known to be $129.96. How large of a sample would be required in order to estimate the mean per capita income at the 98 % level of confidence with an error of at most $0.55? Round your answer up...

  • A research scholar wants to know how many times per hour a certain strand of virus...

    A research scholar wants to know how many times per hour a certain strand of virus reproduces. The mean is found to be 7.6 reproductions an the population standard deviation is known to be 2.2. If a sample of 201 was used for the study, construct the 80 % confidence interval for the true mean number of reproductions per hour for the virus. Round your answers to one decimal place. Answer/How to Enter) 10 Points Keyr Keyboard Short Lower endpoint...

  • The American Sugar Producers Association wants to estimate the annual mean sugar consumption per capita. A...

    The American Sugar Producers Association wants to estimate the annual mean sugar consumption per capita. A sample of 347 individuals had a mean of 61 pounds consumed per year, with a standard deviation of 19 pounds. Construct and interpret a 90 percent confidence interval for the true population mean of annual sugar consumption per capita.

  • The mean amount of money spent on lunch per week for a sample of 95 students...

    The mean amount of money spent on lunch per week for a sample of 95 students is $19. Ifthe margin of error for the population mean with a 95 % confidence interval is 1 0, construct a 95 % confidence interval for the mean amount of money spent on lunch per week for all students Answer How to Enter) 2 Points m Tables Keypad Lowee endpoint Upper endpoint

  • An environmentalist wants to find out the fraction of oil tankers that have spills each month....

    An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2: Suppose a sample of 211 tankers is drawn. Of these ships, 146 did not have spills. Using the data, construct the 80% confidence interval for the population proportion of oil tankers that have spills each month. Round your answers to three decimal places Tables Answer How to enter your answer Keypad Previous step answe Lower endpoint: Upper endpoint:

  • Question 15 of 15 Step 4 of 4 02:02:20 The following measurements (in picocuries per liter)...

    Question 15 of 15 Step 4 of 4 02:02:20 The following measurements (in picocuries per liter) were recorded by a set of hydrogen gas detectors installed in a research facility: 689.1, 699.2, 707.8, 729.8 Using these measurements, construct a 99 % confidence interval for the mean level of hydrogen gas present in the facility. Assume the population is approximately normal. Copy Data Step 4 of 4: Construct the 99 % confidence interval. Round your answer to two decimal places. AnswerHow...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT