Question

The average cost of owning and operating an automobile is $8121 per 15000 miles. A random...

The average cost of owning and operating an automobile is $8121 per 15000 miles. A random survey of 40 automobile owners revealed an average cost of $8350 with a population standard deviation of $750. At the one percent level of significance is there sufficient evidence to conclude that the average is greater than $8121? Use critical values to perform your test.

Homework Answers

Know the answer?
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
Similar Questions
Operating Costs of an Automobile (Population – Critical Value) The average cost of owning and operating...
Operating Costs of an Automobile (Population – Critical Value) The average cost of owning and operating an automobile is $9122 per year. A random survey of 40 owners revealed an average cost of $9350 with a population standard deviation of $750. Is there sufficient evidence to conclude that the average is greater than $9122. Use α = 0.01. Step 1: State the hypotheses. Step 2: Find the critical value. Step 3: Compute the test value. Step 4: Make the decision....
It is necessary for an automobile producer to estimate the number of miles per gallon achieved...
It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 100 cars is 28.6 miles and assume the standard deviation is 3.9 miles. Now suppose the car producer wants to test the hypothesis that ?, the mean number of miles per gallon, is 27 against the alternative hypothesis that it is not 27. Conduct a test using ?=.05 by giving the...
t is necessary for an automobile producer to estimate the number of miles per gallon achieved...
t is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 150 cars is 30.2 miles and assume the standard deviation is 3.6 miles. Now suppose the car producer wants to test the hypothesis that μ, the mean number of miles per gallon, is 30.5 against the alternative hypothesis that it is not 30.5. Conduct a test using α=.05 by giving the...
It is necessary for an automobile producer to estimate the number of miles per gallon achieved...
It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 110 cars is 30 miles and assume the standard deviation is 2.4 miles. Now suppose the car producer wants to test the hypothesis that μ, the mean number of miles per gallon, is 33 against the alternative hypothesis that it is not 33. Conduct a test using α=.05 by giving the...
(1 point) It is necessary for an automobile producer to estimate the number of miles per...
(1 point) It is necessary for an automobile producer to estimate the number of miles per gallon (mpg) achieved by its cars. Suppose that the sample mean for a random sample of 50 cars is 30.2 mpg and assume the standard deviation is 2.8 mpg. Now suppose the car producer wants to test the hypothesis that μ, the mean number of miles per gallon, is 29.9 against the alternative hypothesis that it is not 29.9. Conduct a test using a...
A random sample of 40 senators from various states had an average age of 55.4 years,...
A random sample of 40 senators from various states had an average age of 55.4 years, and the population standard deviation is 6.5 years. α = .05, is there sufficient evidence to conclude that senators are on average younger than 60.35? What is the value of the standardized test statistic?
Many people believe that the average number of Facebook friends is 130. The population standard deviation...
Many people believe that the average number of Facebook friends is 130. The population standard deviation is 36.9. A random sample of 39 high school students in a particular county revealed that the average number of Facebook friends was 150. At =α0.01, is there sufficient evidence to conclude that the mean number of friends is greater than 130?
An automobile manufacturer has given its van a 31.8 miles/gallon (MPG) rating. An independent testing firm...
An automobile manufacturer has given its van a 31.8 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 210 vans, they found a mean MPG of 31.7. Assume the population standard deviation is known to be 2.6. Is there sufficient evidence at the 0.05 level to support the testing firm's claim? Step 5 of 6 :...
The data to the right show the average retirement ages for a random sample of workers...
The data to the right show the average retirement ages for a random sample of workers in Country A and a random sample of workers in Country B. Complete parts below. Country A Country B Sample mean 62.2 years 64.7 years Sample size 40 40 Population standard deviation 4.1 years 5.2 years part a.) Perform a hypothesis test using alphaαequals=0.05 to determine if the average retirement age in Country B is higher than it is in Country A. part b.)...
According to Best Value Auto Insurance, the average annual cost of automobile insurance was $1248 in...
According to Best Value Auto Insurance, the average annual cost of automobile insurance was $1248 in Nebraska in 2012. An insurance broker wants to know if the current mean annual rate of automobile insurance in Nebraska is different than $1248. He took a random sample of 100 insured automobiles from the state of Nebraska and found the mean annual automobile insurance rate of $1375 with a standard deviation of $128. Using a 1% significance level and the critical value approach,...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT