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# Sample size, confidence interval and hypothesis testing

1. A student surveys 100 classmates by asking each if they have outstanding loans. After finding the sample of proportion for this sample of n = 100 subjects, can the methods of this section be used to estimate the proportion of all adults who have outstanding loans? Why or why not?
2. Express the confidence interval 0.600 < p <0.800 in the form of p ± E.
3. Express the confidence interval 0.337 ± 0.0050 in the form p - E < p < p + E.
4. Find the point estimate p and the margin of error E.
a. 0.0144 < p < 0.0882

5. Find the margin of error E that correspond to the given statistics and confidence level.
a. 90% confidence; the sample size is 2107, of which 65% are successes.
6. Construct the confidence interval estimate of the population proportion p.
a. N = 4500, x = 292, 90% confidence

7. Find the minimum sample size required to estimate a population portion or percentage.
a. Margin of error: five percentage points; confidence level: 90%; from a prior study, p is estimated by the decimal equivalent of 65%.

8. An important issue facing Americans is the large number of medical malpractice lawsuits and the expense that they generate. In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856 of them were later dropped of dismissed. Construct a 99% confidence interval estimate of the proportion of medical malpractice lawsuits that are dropped or dismissed. Does it appear that the majority of such suits are dropped or dismissed?
9. Find the minimum sample size required to estimate a population proportion or percentage.
a. Toyota provides an option of a sunroof an side air bag package for it Corolla mode. This package costs \$1400 (\$1159 invoice price). Assume that prior to offering this option package, Toyota wants to determine the percentage of Corolla buyers who would pay \$1400 extra for the sunroof and side air bags. How many Corolla buyers must be surveyed if we want to be 95% confident that the same percentage is within four percentage points of the true percentage for all Corolla buyers?

10. researcher calculates the sample size needed to estimate the force that can be exerted by legs of people on amusement park rides, and the sample size of 120 obtained. If the researcher cannot obtain a random sample and must rely instead on a convenience sample consisting of friends and relatives, can he or she compensate and get good results by using a much larger sample size?
11. Find the critical value a&#8734;/2 that corresponds to 96%
12. Verifying requirements and find the margin of error.
a. The confidence level is 99%, the sample size is n = 12, &#963; is not known, and the original populations is normally distributed.

13. A study of the ages of motorcyclists killed in crashes involves the random selection of 150 drivers with a mean of 37.1 years. Assuming that &#963; = 12.0 years, construct a 99% confidence interval estimate of the man age of all motorcyclists killed in crashes. If the confidence interval limits do not include ages below 20 years, does it mean that motorcyclists under the age of 20 rarely die in crashes?
14. Find the indicated sample size. Media research wants to estimate the man amount of time (in minutes) that full time college students spend watching television each weekday. Find the sample size necessary to estimate that mean with a a 15 min margin of error. Assume that a 96% confidence level is desired. Also assume that a pilot study showed that the standard deviation is estimated to be 112.2 min.
15. Construct the confidence interval. Because cardiac deaths appear to increase after heavy snowfalls, an experiment was designed to compared cardiac demands of snow shoveling to those of using an electric snow thrower. Ten subjects cleared tracts of snow using both methods, and their maximum heart rates (beats per minutes) were recorded during both activities. The following results were obtained.
Manual snow shoveling maximum heart rates: n = 10, x = 175, s = 15.
Electric snow thrower maximum heart rates: n = 10, x = 124, s = 18.
a. Find the 95% confidence interval estimate of the population mean for those people who shovel snow manually.
b. Find the 95% confidence interval estimate of the population mean for thos people who use the electric snow thrower.
c. If you are a physician with concerns about cardiac deaths fostered by manual snow shoveling, what single value in the confidence interval from part (a) would be of greatest concern?
d. Compare the confidence intervals from parts a and b and interpret findinds.

16. Find Chi-Square critical value:
a. 95%; n = 7
b. 90%; n= 91

17. Finding confidence intervals:
a. Speeds of drivers ticketed in a 55 mi/h zone: 95% confidence; n = 90, x = 66.2 mi/h, s = 3.4 mi/h.
b. Amounts lost by gamblers who took a bus to an Atlantic City casino: 99% confidence; n =40, x=\$189, s = \$87.

18. Determining sample size. Assume that each sample is a simple random sample obtained from a normally distributed population.
a. Find the minimum sample size needed to be 95% confident that the sample standard deviation s is within 20% or &#963;.
b. Find the minimum sample size needed to be 95% confident that the sample variance is within 30% of the population variance.

#### Solution Preview

No . The selected sample consist of only students which is not a representative of population. To obtain a good estimate of the proportion of all adults who have outstanding loans, the sample should consist of all representatives adults from all sections of society. Estimate based on the students will be biased.

2. Express the confidence interval 0.600 < p <0.800 in the form of p ± E.

Confidence Interval 0.700 ± 0.100

3. Express the confidence interval 0.337 ± 0.0050 in the form p - E < p < p + E.

Confidence Interval 0.337 - 0.005 < p < 0.337 + 0.005
0.332 <p< 0.342

4. Find the point estimate p and the margin of error E.

a. 0.0144 < p < 0.0882
Point Estimate = (.0144+0.0882) /2 = 0.0513

Margin of Error = Radius of Confidence Interval = 0.0882 - 0.0513 = 0.0369

5. Find the margin of error E that correspond to the given statistics and confidence level.
a. 90% confidence; the sample size is 2107, of which 65% are successes.

Margin of Error =

6. Construct the confidence interval estimate of the population proportion p.
a. N = 4500, x = 292, 90% confidence

The confidence interval is given by

= [0.0588489 < p < 0.0709289]

7. Find the minimum sample size required to estimate a population portion or percentage.
a. Margin of error: five percentage points; confidence level: 90%; from a prior study, p is estimated by the decimal equivalent of 65%.

We have margin of error =
Given that
Margin of Error = 0.05
p = 0.65

Thus n is given by

Minimum sample Size = 247

8. An important issue facing Americans is the large number of medical malpractice lawsuits and the expense that they generate. In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856 of them were later dropped of dismissed. Construct a 99% confidence interval estimate of the proportion of medical malpractice lawsuits that are dropped or dismissed. Does it appear that the majority of such suits are dropped or dismissed?

The confidence interval is given by

Given
n =1228
x = 856
= 856/1228 = 0.697068
p = 0.5

The confidence interval for p is [0.463245, 0.536755]

Since the sample ...

#### Solution Summary

The solution contain step by step procedure for sample size determination, hypothesis testing and confidence interval estimation. Null hypothesis, alternative hypothesis, significance level, critical value, p value and test statistic are given with interpretations.

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