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A simple random sample of size n = 40 is carried out | Complete Solution

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1.A simple random sample of size n = 40 is carried out. The sample mean x-bar is 35.1 and the sample standard deviation s is 8.7.  Find a 98% confidence interval for the population mean μ.

 

2.An article in a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 522 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 90% confidence level. Express the answer in the form  ± E and round to the nearest thousandth.

 

3.The diameter of the Douglas fir tree is measured at a height of 1.37 meters. The following data represents the diameter in centimeters of a random sample of 12 Douglas firs in the western Washington Cascades.

 

156190147173159181

162130101147113109

Source: L. Winter, “Live Tree and Tree-Ring Records to Reconstruct the Structural Development of an Old-Growth Douglas Fir/Western Hemlock Stand in the Western Washington Cascades.” Corvallis, OR: Forest Science Data Bank, 2005.

 

Assume that the fir tree diameters came from a population that is normally distributed. Find the point estimate for the mean and standard deviation of a Douglas fir tree in the western Washington Cascades. Why did I need to point out that the fir tree diameters came from a normally distributed population?

 

4.What is meant by the term “95% confident when constructing a confidence interval for a mean?

 

5.(10 pts) A random sample of 50 recent college graduates results in a mean time to graduate of 4.58 years with a standard deviation of 1.10 years. a) Find and interpret the 90% confidence interval for time to graduate with a bachelor’s degree. b) Does this evidence contradict the widely held belief that it takes 4 years to complete a bachelor’s degree?

 

6.Find the point estimate of the population mean and margin or error if the confidence interval is (125.6, 152.4)

 

7.(10 pts) A question on the 2006 General Social Survey asked “How many emails do you send in a day?” The results of 928 respondents indicate that the mean number of emails sent in a day is 10.4 with a standard deviation of 28.5.   a) Given the fact that 1 standard deviation to the left of the mean results in a negative number of emails sent, what shape would you expect the distribution of emails to have and why?  b) Find the 90% confidence interval for the mean number of emails sent per day.

 

8.We know that adult  IQ scores based on the Stanford-Binet IQ test are normally distributed with a mean of 100 and a standard deviation of 15. If you were to obtain 200 different simple random samples of size 20 from the population of all adults and determine 95% confidence intervals for each of them, how many of the intervals would you expect to include the IQ  100?

 

9.(20 pts) From a random sample of 678 adult males 20 to 34 years of age, it was determined that 58 of them have hypertension (high blood pressure). Source: The Centers for Disease Control

a.Find a point estimate for the proportion of adult males 20 to 34 years of age who have hypertension.

 

b.Construct a 95% confidence interval for the proportion of adult males 20 to 34 years of age who have hypertension.

 

c.You wish to conduct your own study to determine the proportion of adult males 20 to 34 years old who have hypertension. What sample size would be needed for the estimate to be within 3 percentage points with a 95% confidence if you use the point estimate found in part a?

 

d.You wish to conduct your own study to determine the proportion of adult males 20 to 34 years old who have hypertension. What sample size would be needed for the estimate to be within 3 percentage points with a 95% confidence if you don’t have a prior estimate?

 

 

10.(10 pts) A study of 35 golfers showed that their average score on a particular gold course is 92. The standard deviation of the population is 7.

a.Find the 95% confidence interval of the mean score for all golfers.

 

b.Find the 95% confidence interval of the mean score if a sample of 60 golfers is used instead of 35.

 

c.Which interval is smaller and why?

 

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[Solved] A simple random sample of size n = 40 is carried out | Complete Solution

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  • Submitted On 19 Jul, 2016 02:40:06
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The 98% confidence interval is, = (X ̅±t(0.01,40-1)*S/√n) = (35.1±2.426*8.7/√40) = (31.7628, 38.4372) An article in a Florida newspaper reported on the topics that teenagers most want to discuss wit...
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