Question

A 99% confidence interval for the mean lifespan of Michellin ZX tires is 48500 miles to 56950 miles. Find the value of the sa

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Z0.005=2.58

mean=T=a

Standard error=\frac{s}{\sqrt{n}}=b

CI=T +/- Z0.005  \frac{s}{\sqrt{n}}

CI=a +/- 2.58 *b

lower limit=48500

a-2.58 b=48500 ---------(1)

upper limit=56950

a+2.58 b=56950 -------------(2)

Adding 1 and 2 we get:

a+a=48500+56950

2a=105450

a=52725=mean(T)

putting this in 1:

52725 -2.58 b=48500

52725-48500=2.58 b

4225= 2.58 b

b=1637.5969=Standard Error(\frac{s}{\sqrt{n}})

Hence, mean(T)=52725 and Standard Error(\frac{s}{\sqrt{n}})=1637.6

please rate my answer and comment for doubts.

Add a comment
Know the answer?
Add Answer to:
A 99% confidence interval for the mean lifespan of Michellin ZX tires is 48500 miles to...
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
  • a sample size of _ is needed So there a 99% confidence interval will have a...

    a sample size of _ is needed So there a 99% confidence interval will have a margin of error of three.so there a 99% confidence interval will have a margin of error of three. 1. simple random sample of 100 2. mean was 125 hours 3. standard deviation is 20 hours.

  • (3 points) A news report states that the 99% confidence interval for the mean number of...

    (3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1860, 2060). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 16 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...

  • (3 points) A news report states that the 99% confidence interval for the mean number of...

    (3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1750, 1880). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 20 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...

  • what is the margin of error and the confidence interval? Question Help In a random sample...

    what is the margin of error and the confidence interval? Question Help In a random sample of seven people, the mean driving distance to work was 24.7 miles and the standard deviation was 6.6 miles. Assuming the population is normally distributed and using the I-distribution, a 90% confidence interval for the population mean is (15.5, 33.9) (and the margin of error is 9.2). Through research, it has been found that the population standard deviation of driving distances to work is...

  • In a random sample of twelve ​people, the mean driving distance to work was 20.8 miles...

    In a random sample of twelve ​people, the mean driving distance to work was 20.8 miles and the standard deviation was 5.4 miles. Assume the population is normally distributed and use the​t-distribution to find the margin of error and construct a 99​% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.

  • A confidence interval of .99 can correctly be interpreted to mean that: a. 99% of the...

    A confidence interval of .99 can correctly be interpreted to mean that: a. 99% of the time in repeated sampling intervals using an appropriate formula will contain the sample value. b. 99% of the time in repeated sampling intervals using an appropriate formula will contain the relevant population parameter c. 99% of the time in repeated sampling intervals using an appropriate formula will contain the sample value as the midpoint of the interval d. 99% of the time in repeated...

  • A 95% confidence interval of 19.8 months to 47.4 months has been found for the mean...

    A 95% confidence interval of 19.8 months to 47.4 months has been found for the mean duration of imprisonment, u, of political prisoners of a certain country with chronic PTSD. a. Determine the margin of​ error, E. b. Explain the meaning of E in this context in terms of the accuracy of the estimate. c. Find the sample size required to have a margin of error of 12 months and a 99% confidence level. (use standard deviation=38 months.) d) Find...

  • Assuming that the population is normally distributed, construct a 99% confidence interval for the population mean,...

    Assuming that the population is normally distributed, construct a 99% confidence interval for the population mean, based on e ollowing sample sizeof 1, 2, 3, 4, 5, 6, 7, and 25 In the given data, replace the value 25 with 8 and recalculate the confidence interval. Using these results, describe the effect of an outlier (that is, an extreme value) on the confidence interval, in general. Find a 99% confidence interval for the population mean, using the formula or technology....

  • Two researchers plan to construct a 99% confidence interval for the mean μ of a Normal...

    Two researchers plan to construct a 99% confidence interval for the mean μ of a Normal population with (known) standard deviation s.  Researcher A uses a random sample of 50 individuals.  Researcher B uses a random sample of 800 individuals. How does the margin of error for Researcher A’s estimate compare to that of Researcher B’s estimate?

  • Use the t-distribution to find a confidence interval for a mean μ given the relevant sample...

    Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 99% confidence interval for μ using the sample results x¯=93.5, s=34.3, and n=15 Round your answer for the point estimate to one decimal place, and your answers for the margin of...

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