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# Confidence interval estimation

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Using the Central Limit Theorem. In Exercises 5-8, assume that women's height are normally distributed with a mean given by µ = 63.6 in. and a standard deviation given by &#963; = 2.5 (based on data from the National Health Survey).

Sect. 6-5, p. 287 (For these problems, see examples on p. 283, 284
#5. a. If 1 woman is randomly selected, find the probability that her height is less than 64 in.
b. If 36 women are randomly selected, find the probability that they have a mean height less than 64 in.

8a. If 1 woman is randomly selected, find the probability that her height is between 60 in. and 65 in.

b.) If 16 women are randomly selected, find the probability that they have a mean height between 60 in. and 65 in.

Sect. 7-2, p. 332
Constructing Confidence Intervals. In Exercises 21-24, use the sample data and confidence level to construct the confidence interval estimate of the population proportion p. (For these problems see example on pg. 327)

#21. n = 500, x = 200, 95% confidence.

24. n = 4500, x = 2925, 90% confidence.
Determining Sample Size. In Exercises 25-28, use the given data to find the minimum sample size required to estimate a population proportion or percentage. (For these problems see example on pg. 329)
25. Margin of error: 0.020; confidence level: 95%; sample proportion and sample proportion equal to 1- sample proportion.

28. Margin of error: five percentage points: confidence level: 90%; from prior study, sample proportion is estimated by the decimal equivalent of 27%.

#34a (See example on pg. 327) Misleading Survey Responses. In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.
a. Find a 99% confidence interval estimate of the proportion of people who say that they voted.

Sect. 7-3, p. 345 (For these problems see example on pg. 341)
Finding a Confidence Interval. In Exercises 13-16, use the given confidence level and sample data to find a confidence interval for estimating the population mean &#956;.

#16. Amounts lost by gamblers who took a bus to an Atlantic City casino; 99% confidence; n = 40, mean of the values in a sample = \$189, and &#963; is known to be \$87.

Finding Sample Size. In Exercises 17-20, use the given margin of error, confidence level, and population standard deviation &#963; to find the minimum sample size required to estimate an unknown population mean &#956;.

#18 (See example on pg. 344) Margin of error: 0.25 sec, confidence level: 99%, &#963; = 5.40 sec.

Sect. 7-4, p. 359(See example problems in textbook and Instructor's Notes - Section 5, Example #20.)

Constructing Confidence Intervals. In Exercises 17-26, construct the confidence interval.

#17. Birth Weights. A random sample of the birth weights of 186 babies has a mean of 3103 g and a standard deviation of 696 g (based on data from "Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure," by Singer et al., Journal of the American Medical Association, Vol. 291, No.20). These babies were born to mothers who did not use cocaine during their pregnancies. Construct a 95% confidence interval estimate of the mean birth weight for all such babies. Compare the result to the confidence interval obtained in the example in this section that involved birth weights of babies born to mothers who used cocaine during pregnancy. Does cocaine use appear to affect the birth weight of a baby?

19. Forecast and Actual Temperatures. Data Set in Appendix B includes a list of actual high temperatures and the corresponding list of three-day-forecast high temperatures. If the difference for each day is found by subtracting the three-day-forecast high temperature from the actual high temperature, the result is a list of 35 values with a mean of -1.3% and a standard deviation of 4.7 degrees.
A. Construct a 99% confidence interval estimate of the mean difference between all actual high temperatures and three-day-forecast high temperatures.
b. Does the confidence interval include 0 degrees? If a meteorologist claims that three-day-forecast high temperatures tend to be too high because the mean difference of the sample is -1.3 degrees, does that claim appear to be valid? Why or why not?

https://brainmass.com/statistics/confidence-interval/confidence-interval-estimation-241312

#### Solution Summary

The solution provides step by step method for the calculation of confidence interval for . Formula for the calculation and Interpretations of the results are also included.

\$2.19

## Confidence Interval Estimate, Hypotheses and Test Statistics

1. Mimi was the 5th seed in 2012 UMUC Tennis Open that took place in August. In this tournament, she won 83 of her 100 serving games. Based on UMUC Sports Network, she wins 80% of the serving games in her 5-year tennis career.

(a) Find a 95% confidence interval estimate of the proportion of serving games Mimi won. (Show work)

(b) Based on the confidence interval estimate you got in part (a), is this tournament result consistent with her career record of 80%? Why or why not? Please explain your conclusion.

(c) A sport reporter commented that Mimi's performance in the tournament is better than usual. You decide to test if the reporter's claim is valid by using hypothesis testing that you just learned from STAT 200 class. What are your null and alternative hypotheses?

(d) What is the test statistic? (Show work)

(e) What is the P-value? (Show work)

(f) What is the critical value? (Show work)

(g) What is your conclusion of the testing at 0.05 significance level? Why?

2. A simple random sample of 120 SAT scores has a mean of 1540. Assume that SAT scores have a population standard deviation of 333.

(a) Construct a 95% confidence interval estimate of the mean SAT score. (Show work)

(b) Is a 99% confidence interval estimate of the mean SAT score wider than the 95% confidence interval estimate you got from part (a)? Why? [You don't have to construct the 99% confidence interval]

3. The playing times of songs are normally distributed. Listed below are the playing times (in seconds) of 10 songs from a random sample. Use a 0.05 significance level to test the claim that the songs are from a population with a standard deviation less than 1 minute.

448 231 246 246 227 213 239 258 255 257

(a) What are your null hypothesis and alternative hypothesis?

(b) What is the test statistic? (Show work)

(c) What is your conclusion? Why? (Show work)

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