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1.Find for each of the following:

a. = .10

b. = .01

c. = .05

d. = .20

2.A random sample of n measurements was selected from a population with unknown mean and standard deviation =20. Calculate a 95% confidence interval for for each of the following situation:

a.N = 75, = 28

b.N = 200, = 102

c.N = 100, = 15

d.N = 100, = 4.05

e.Is the assumption that the underlaying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a-d? explain.

3.The mean and standard deviation of a random sample of n measurements are equal to 33.9 and 3.3, respectively.

a.Find a 95% confidence interval for if n = 100.

b.Find a 95% confidence interval for if n = 400.

c.Find the widths of the confidence intervals you calculated in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?

4.Explain the differences in the sampling distributions of for large and small samples under the following assumptions:

a.The variable of interest, x, is normally distributed.

b.Nothing is known about the distribution of the variable x.

5.Let be a specific value of t. Use technology of table III in appendix B to find values such that the following statements are true:

a.P( t = .025, where df = 10.

b.P( t = .01, where df = 17.

c.P( t = .005, where df = 6

d.P( t = .05, where df = 13

6.Pitch memory of amusiacs. A team of psychologist and neuroscientists tested the pitch memory of individuals diagnosed with amusia ( a disorder that impacts ones perception of music) and reported their results in advances in cognitive psychology). Each in a sample of 17 amusiacs listened to a series of tone pairs and then were asked to determine if the tones were the same or different. In a one trial, the tones were separated by 1 second. In a second trial, the tones were separated by 5 seconds. Scores of the two trials were compared for each amusiac. The mean score difference was .11 with a standard deviation of .19. Use this information to form a 90% confidence interval for the true mean score difference for all amusiacs. Interpret the result. What assumption about the population of score difference must hold true for the interval to be valid?

7.Antigens for a parasite roundworm in birds. Ascaridia galli is a parasitic roundworm that attacks the intestines of birds, especially chickens and turkeys. Scientists are working on a synthetic vaccine (antigen) for the parasite. The success of the vaccine hinges on the characteristics of DNA in peptide (protein) produced by the antigen. In the journal Gene Therapy and molecular biology (June 2009), scientists tested alleles of antigen-produced protein for level of peptide. For a sample of 4 alleles, the mean peptide score was 1.43 and the standard deviation was .13.

a.Use this information to construct a 90% confidence interval for the true mean peptide score in alleles of the antigen-produced protein.

b.Interpret the interval for the scientists.

c.What is meant by the phrase “90% confidence?”

8.Explain the meaning of the phrase “ is an unbiased estimator of p.”

9.A random sample of size n = 196 yielded = .64.

a.Is the sample size large enough to use the methods of this section to construct a confidence interval for p? Explain.

b.Construct a 95% confidence interval for p.

c.Interpret the 95% confidence interval.

d.Explain what is meant by the phrase “95% confidence interval.”

10.National Firearms Survey. Refer to the Harvard School of Public Health survey to determine the size and composition of privately held firearm stock in the United States, presented in exercise 3.85 (pg 142). Recall that, in a representative household telephone survey of 2,770 adults, 26% reported that they owned at least 1 gun. The researchers want to estimate the true percentage of adults in the united states that own at least one gun.

a.Identify the population of interest to the researchers:

b.Identify the parameter of interest to the researchers:

c.Compute an estimate of the population parameter:

d.Form a 99% confidence interval around the estimate.

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