Exam 7: Estimating Parameters and Determining Sample Sizes

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. - $91 \% ; \mathrm { n } = 45 ; \sigma$ is known; population appears to be very skewed.

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. -To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s = 0.16. Find the 95% confidence interval For the population standard deviation $\sigma .$

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.008; confidence level: 98%; $\hat { p } \text { and } \hat { q } \text { unknown }$

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 eggs: 95% confidence; n = 22, $\bar { x }$ = 1.65 oz, s = 0.47 oz

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

The following is a 95% confidence interval of the proportion of female medical school students: 0.449 < p < 0.511, based on data from the American Medical Association. What is the point estimate of the proportion of females in the population of medical school students? Write a brief statement that correctly interprets the confidence interval given.

Solve the problem. Round the point estimate to the nearest thousandth. -50 people are selected randomly from a certain population and it is found that 12 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall?

In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$ , a table is used to find the critical values $\chi \frac { 2 } { L }$ and $\chi \frac { 2 } { R }$ for values of $n \leq 101$ . For larger values of $n , \chi _ { L } ^ { 2 }$ and $\chi _ { R } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } =$ $\frac { 1 } { 2 } [ \pm z / 2 / 2 + \sqrt { 2 k - 1 } ] 2$ where $k$ is the number of degrees of freedom and $z \alpha / 2$ is the critical $z$ -score. Construct the $90 \%$ confidence interval for $\sigma$ using the following sample data: a sample of size $n = 232$ yields a mean weight of $154 \mathrm { lb }$ and a standard deviation of $25.5 \mathrm { lb }$ . Round the confidence interyal limits to the nearest hundredth.

Which critical value is appropriate for a 99% confidence level where n = 17; $\sigma$ is unknown and the population appears to be normally distributed?

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. -Test scores: $\mathrm { n } = 104 , \overline { \mathrm { x } } = 95.3 , \sigma = 6.5 ; 99 \%$ confidence

Do one of the following, as appropriate: (a) Find the critical value ${ } ^ { Z } \alpha / 2$ , (b) find the critical value ${ } ^ { t } \alpha / 2$ 99%; n = 17; $\alpha$ is unknown; population appears to be normally distributed.

Solve the problem. -Suppose that n trials of a binomial experiment result in no successes. According to the "Rule of Three", we have 95% confidence that the true population proportion has an upper bound of 3/n $3 / n$ . If a manufacturer randomly Selects 25 computers for quality control and finds no defective computers, what statement can you make by Using the rule of three, about the proportion p, of all its computers which are defective?

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. -The principal randomly selected six students to take an aptitude test. Their scores were: 88.0 84.1 74.9 83.2 83.7 85.5 Determine a 90% confidence interval for the mean score for all students.

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; the sample size is 9900, of which 30% are successes

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. -The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 7.0 10.8 9.5 8.0 11.5 7.5 6.4 11.3 10.2 12.6 Determine a $95 \%$ confidence interval for the mean time for all players.

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

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.

Use the given information to find the minimum sample size required to estimate an unknown population mean $\mu$ -How many commuters must be randomly selected to estimate the mean driving time of Chicago commuters? We want 90% confidence that the sample mean is within 4 minutes of the population mean, and the population Standard deviation is known to be 12 minutes.

Solve the problem. -A one-sided confidence interval for $\mathrm { p }$ can be written as $\mathrm { p } < \hat { \mathrm { p } } + \mathrm { E }$ or $\mathrm { p } > \hat { \mathrm { p } } - \mathrm { E }$ where the margin of error $\mathrm { E }$ is modified by replacing $\mathrm { z } _ { \alpha / 2 }$ with $\mathrm { z } _ { \alpha }$ . If a teacher wants to report that the fail rate on a test is at most $x$ with $90 \%$ confidence, construct the appropriate one-sided confidence interval. Assume that a simple random sample of 74 students results in 8 who fail the test.

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 football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 9 7 15 6 15 12 8 6 14 5 Find a $95 \%$ confidence interval for the population standard deviation $\sigma$ .