Exam 7: Estimating Parameters and Determining Sample Sizes

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union Members who favor the Republican candidate.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -n = 130, x = 69; 90% confidence

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. 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 amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.2 15.1 15.9 15.5 15.6 15.1 15.8 15.0 Find a $98 \%$ confidence interval for the population standard deviation $\sigma$ .

Provide an appropriate response. -The confidence interval, $3.56 < \sigma < 5.16$ , for the population standard deviation is based on the following sample statistics: $\mathrm { n } = 41 , \overline { \mathrm { x } } = 30.8$ , and $\mathrm { s } = 4.2$ . What is the degree of confidence?

When determining the sample size needed to achieve a particular error estimate you need to know σ. What are two methods of estimating σ if σ is unknown?

Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean. -A random sample of 187 full-grown lobsters had a mean weight of 19 ounces and a standard deviation of 3.3 ounces. Construct a 98% confidence interval for the population mean $\mu _ {. }$

Solve the problem. -A 99% confidence interval (in inches) for the mean height of a population is $65.92 < \mu < 67.48$ This result is based on a sample of size 144. If the confidence interval $66.13 < \mu < 67.27$ is 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).

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -n = 85, x = 49; 98% confidence

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

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 185 milligrams with s = 17)6 milligrams. A confidence interval of $173.8 \mathrm { mg } < \mu < 196.2 \mathrm { mg }$ constructed for the true mean cholesterol Content of all such eggs. It was assumed that the population has a normal distribution. What confidence level Does this interval represent?

Fifty people are selected randomly from a certain population and it is found that 12 people in the sample are over six feet tall. What is the best point estimate of the proportion of people in the population who are over 6 Feet tall?

Use the given information to find the minimum sample size required to estimate an unknown population mean $\mu$ -How many women must be randomly selected to estimate the mean weight of women in one age group. We want 90% confidence that the sample mean is within 2.1 lb of the population mean, and the population standard Deviation is known to be 19 lb.

Use the given data to find the minimum sample size required to estimate the population proportion. - $\text { Margin of error: } 0.028 \text {; confidence level: } 99 \% ; \hat { p } \text { and } \hat { q } \text { unknown }$

Solve the problem. -The following confidence interval is obtained for a population proportion, p: 0.724 < p < 0.752. Use these confidence interval limits to find the margin of error, E.

Find $\mathrm { z } _ { \alpha / 2 } \text { for } \alpha = 0.07$

Use the given data to find the minimum sample size required to estimate the population proportion. -Margin of error: 0.04; confidence level: 95%; from a prior study, $\hat { p }$ is estimated by the decimal equivalent of 60%.

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. -Weights of men: 90% confidence; n = 14, $\overline { \mathrm { x } }$ = 161.5 lb, s = 13.7 lb

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. - $\mathrm { n } = 30 , \overline { \mathrm { x } } = 84.6 , \mathrm {~s} = 10.5,90 \%$ confidence

Do one of the following, as appropriate: (a) Find the critical value $\mathrm { z } _ { \alpha / 2 }$ , (b) find the critical value $t _ { \alpha / 2 }$ , (c) state that neither the normal nor the t distribution applies. - $99 \% ; \mathrm { n } = 17 ; \sigma$ is unknown; population appears to be normally distributed.

Solve the problem. -Find the critical value $x _ { L } ^ { 2 }$ corresponding to a sample size of 24 and a confidence level of 95 percent.