ch16 #10. (16.08) We have the survey data on the body mass index (BMI) of 646 yo

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ch16

#10. (16.08) We have the survey data on the body mass index (BMI) of 646 young women. The mean BMI in the sample was x=25.7x¯=25.7. We treated these data as an SRS from a Normally distributed population with standard deviation ==7.1.

a. Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence.

90% = ___ to ____

95% = ____ to ____

b. Conf. Level Interval (±±0.01) margins of error (±±0.0001) 90%, 95%, 99%

90% =

95% =

99% =

#11

We have the survey data on the body mass index (BMI) of 645 young women. The mean BMI in the sample was x=26.6x¯=26.6. We treated these data as an SRS from a Normally distributed population with standard deviation ==8.

Find the margins of error for 90% confidence based on SRSs of N young women.

= ??

#12

An SRS of 450 high school seniors gained an average of xhat = 23 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 48.

a. Find a 90% confidence interval for based on this sample.

Confidence interval (±±0.01) is between ____ and _____

b. What is the margin of error (±±0.01) for 90%?

c. Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±±0.01) for 90% confidence?

#13

A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 462 students was xx¯ = 14 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 8 hours in the population of all first-year students at this university.

What is the 99% confidence interval (±±0.001) for the population mean?

Confidence interval is from ____ to _____ hours.

#14

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Here are data from students doing a laboratory exercise:            

We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 2800 pounds.

Give a 95% confidence interval for the mean load required to pull the wood apart (give your answer with three decimal places).

a. The sample mean is ______ (give three decimal places)

b. 95% confidence interval for the mean load is _____ to _______ .

= ??

Explanation / Answer

Since you have posted multiple questions,i am attaching answer to first problem.Can please post next set of problems one at a time

=7.1

n=646

For 90% confidence

Margin of Error,MOE =Z/2 × (/n)

                              = 1.645 x(7.1/646)

                              =0.4595

CI =x ± MOE

   = 25.7 ± 0.4595

   = 25.2405 to 26.1595

For 95% confidence

Margin of Error,MOE =Z/2 × (/n)

                              = 1.96 x(7.1/646)

                              =0.5475

CI =x ± MOE

   = 25.7 ± 0.5475

   = 25.1525 to 26.2475

For 99% confidence

Margin of Error,MOE =Z/2 × (/n)

                              = 2.576 x(7.1/646)

                              =0.7196

CI =x ± MOE

   = 25.7 ± 0.7196

   = 24.9804 to 26.4196

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