Statistics 19 mcqs | Statistics homework help

 The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 94 inches, and a standard deviation of 14 inches. What is the probability that the mean annual snowfall during 49 randomly picked years will exceed 96.8 inches? (Points : 5)

6. Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.
With n = 18 and p = 0.30, estimate P(6). (Points : 5)

7. Use the normal distribution to approximate the desired probability.
A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 14 tosses. What is the probability of being correct 14 or more times by guessing? Does this probability seem to verify her claim? (Points : 5)
0.4418, no
0.0582, no
0.4418, yes
0.0582, yes

8. Solve the problem.
The following confidence interval is obtained for a population proportion, p: 0.689 < p < 0.723. Use these confidence interval limits to find the margin of error, E. (Points : 5)

9. Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.
95% confidence; n = 320, x = 60 (Points : 5)

10. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
n = 51, x = 27; 95% confidence (Points : 5)
0.414 < p < 0.644
0.392 < p < 0.666
0.391 < p < 0.667
0.413 < p < 0.645

11. Use the given data to find the minimum sample size required to estimate the population proportion.
Margin of error: 0.004; confidence level: 95%; unknown (Points : 5)

12. Solve the problem. Round the point estimate to the nearest thousandth.
Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 304 people, 20 people had hearing aids. (Points : 5)

13. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. (Points : 5)
0.471 < p < 0.472
0.435 < p < 0.508
0.438 < p < 0.505
0.444 < p < 0.500

14. Solve the problem.
A newspaper article about the results of a poll states: “In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States.” Find the sample size suggested by this statement. (Points : 5)

16. Use the confidence level and sample data to find a confidence interval for estimating the population mu. Round your answer to the same number of decimal places as the sample mean.
Test scores: n = 92, mean = 90.6, sigma = 8.9; 99% confidence (Points : 5)
88.4 < mu < 92.8
88.8 < mu < 92.4
88.2 < mu < 93.0
89.1 < mu < 92.1

17. Use the given information to find the minimum sample size required to estimate an unknown population mean mu.
Margin of error: $120, confidence level: 95%, sigma = $593 (Points : 5)

18. Assume that a sample is used to estimate a population mean mu. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation.
95% confidence; n = 91; x-bar = 16, s = 9.1 (Points : 5)

19. Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu. Assume that the population has a normal distribution.
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 225 milligrams with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs. (Points : 5)
215.0 mg < mu < 235.0 mg
216.9 mg < mu < 233.1 mg
214.9 mg < mu < 235.1 mg
215.1 mg < mu < 234.9 mg


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