1. Do one of the following, as appropriate: (a) Find the critical valuez_/2, (b) find the critical valuet_/2, (c) state that neither the normal nor the t distribution applies. 95%; n = 11;is known; population appears to be very skewed.

Group of answer choices

z_/2= 1.96

t_/2= 2.228

z_/2= 1.812

Neither the normal nor the t distribution applies.

2. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a two-tailed test is z = 1.95.

Group of answer choices

0.9744; fail to reject the null hypothesis

0.0512; fail to reject the null hypothesis

0.0512; reject the null hypothesis

0.0256; reject the null hypothesis

3. Solve the problem. A 99% confidence interval (in inches) for the mean height of a population isThis result is based on a sample of size 144. If the confidence interval65.66-u-66.74is obtained from the same sample data, what is the degree of confidence? (Hint: you will first need to find the sample mean and sample standard deviation).

Group of answer choices

95%

93%

92%

94%

4. Solve the problem. A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to exaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition? His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed.

Group of answer choices

96.6%

98.2%

96.5%

98.4%

5. Express the confidence interval using the indicated format. 0.668, 0.822 in the form of p +- E

Group of answer choices

0.668 0.154

0.668 0.077

0.745 0.154

0.745 0.077

6. Use the confidence level and sample data to find a confidence interval for estimating the population. Round your answer to the same number of decimal places as the sample mean. A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 192 milligrams witho = 16.5milligrams. Construct a 95% confidence interval for the true mean cholesterol content,, of all such eggs.

Group of answer choices

187 mg << 196 mg

187 mg << 195 mg

188 mg << 196 mg

189 mg << 197 mg

7. 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.

Group of answer choices

0.444 < p < 0.500

0.435 < p < 0.508

0.438 < p < 0.505

0.471 < p < 0.472

8. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. The daily intakes of milk (in ounces) for ten randomly selected people were: 12.7 26.0 16.5 18.9 28.8

30.0 12.3 12.2 29.1 22.0 Find a 99% confidence interval for the population standard deviation.

Group of answer choices

4.37 oz << 14.94 oz

0.96 oz << 3.53 oz

4.51 oz << 14.94 oz

4.51 oz << 16.63 oz

Flag question: Question 9

The claim is that the proportion of accidental deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics includedeaths of the elderly with 15% of them attributable to residential falls.

Group of answer choices

3.96

4.71

-3.96

-4.71

Flag question: Question 10

Find the P-value for the indicated hypothesis test. An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 18 were no-shows. Find the P-value for a test of the airline's claim.

Group of answer choices

0.1492

0.3508

0.1230

0.0746

Flag question: Question 11

Solve the problem. Round the point estimate to the nearest thousandth. 313 randomly selected light bulbs were tested in a laboratory, 69 lasted more than 500 hours. Find a point estimate of the proportion of all light bulbs that last more than 500 hours.

Group of answer choices

0.780

0.217

0.220

0.181

Flag question: Question 12

Use the given data to find the minimum sample size required to estimate the population proportion. margin of error = 0.04, confidence level = 90%; from prior study, p is estimated by 0.16

Group of answer choices

202

228

684

9

Flag question: Question 13

Solve the problem. The following confidence interval is obtained for a population proportion, p: 0.835 < p < 0.861. Use these confidence interval limits to find the margin of error, E.

Group of answer choices

0.026

0.013

0.848

0.014