Exam 7: Estimating Parameters and Determining Sample Sizes

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 group of 59 randomly selected students have a mean score of $29.5$ with a standard deviation of $5.2$ on a placement test. What is the $90 \%$ confidence interval for the mean score, $\mu$ , of all students taking the test?

Find the appropriate minimum sample size. -You want to be 99% confident that the sample standard deviation s is within 5% of the population standard deviation.

A one-sided confidence interval for p can be written as $p < \hat { p } + \mathrm { E } \text { or } p < \hat { p } - \mathrm { E }$ E where the margin of error E is modified by replacing $z _ { \alpha / 2 }$ with ${ } ^ { Z } \alpha$ . 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 construct a confidence interval for the population proportion p. -Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke.

Provide an appropriate response. -In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$ , a table is used to find the critical values $\chi _ { L } ^ { 2 }$ and $\chi _ { R } ^ { 2 }$ for values of $\mathrm { n } \leq 101$ . For larger values of $n , \chi _ { L } ^ { 2 }$ and $\chi \frac { 2 } { \mathrm { R } }$ can be approximated by using the following formula: $\chi ^ { 2 } =$ $\frac { 1 } { 2 } \left[ \pm \mathrm { z } _ { \alpha / 2 } + \sqrt { 2 \mathrm { k } - 1 } \right] ^ { 2 }$ where $\mathrm { k }$ is the number of degrees of freedom and $\mathrm { z } _ { \alpha / 2 }$ is the critical $\mathrm { z }$ score. Estimate the critical values $x _ { L } ^ { 2 }$ and $x _ { R } ^ { 2 }$ for a situation in which you wish to construct a $95 \%$ confidence interval for $\sigma$ and in which the sample size is $n = 232$ .

Find the appropriate minimum sample size. -To be able to say with 95% confidence level that the standard deviation of a data set is within 10% of the population's standard deviation, the number of observations within the data set must be greater than or equal to What quantity?

Provide an appropriate response. -The confidence interval, $17.12 < \sigma ^ { 2 } < 78.88$ , for the population variance is based on the following sample statistics: $n = 25 , \bar { x } = 42.6$ and $s = 5.7$ . What is the degree of confidence?

Solve the problem. -Suppose we wish to construct a confidence interval for a population proportion p. If we sample without replacement from a relatively small population of size N, the margin of error E is modified to include the finite Population correction factor as follows: $E = z _ { \alpha / 2 } \sqrt { \frac { \hat { p}\hat{ q } } { n } } \sqrt { \frac { N - n } { N - 1 } }$ Construct a 90% confidence interval for the proportion of students at a school who are left handed. The number Of students at the school is N = 330. In a random sample of 86 students, selected without replacement, there are 8 left handers.

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. - $\mathrm { n } = 12 , \overline { \mathrm { x } } = 23.6 , \mathrm {~s} = 6.6,99 \%$ confidence

Under what circumstances can you replace σ with s in the formula $E = z _ { \alpha / 2 } \cdot \frac { \sigma } { \sqrt { n } }$ ?

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. -98% confidence; the sample size is 800, of which 40% are successes

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. -90% confidence; n = 300, x = 140

Identify the correct distribution (z, t, or neither) for each of the following. Sample Size Standard Deviation Shape of the Distribution z or t or neither n=35 s=4.5 Somewhat skewed n=20 s=4.5 Bell shaped n=35 \sigma=4.5 Bell shaped n=20 \sigma=4.5 Extremely skewed

A 99% confidence interval (in inches) for the mean height of a populatio $65.7 < \mu < 67.3$ This result is based on a sample of size 144. Construct the 95% confidence interval. (Hint: you will first need to find the sample Mean and sample standard deviation).

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 898 people, 46 people had hearing aids.

Express the confidence interval using the indicated format. -Express the confidence interval $( 0.432,0.52 )$ in the form of $\hat { p } \pm \mathrm { E }$ .

Find the appropriate minimum sample size. -You want to be 95% confident that the sample variance is within 40% of the population variance.

Solve the problem. -The following confidence interval is obtained for a population proportion, p: (0.399, 0.437). Use these confidence interval limits to find the margin of error, E.

In a Gallup poll, 1011 adults were asked if they consume alcoholic beverages and 64% of them said that they did. Construct a 90% confidence interval estimate of the proportion of all adults who consume alcoholic beverages. Can we safely conclude that the majority of adults consume alcoholic beverages?

You want to estimateσ for the population of waiting times at McDonald's drive-up windows, and you want to be 95% confident that the sample standard deviation is with 20% of $\sigma$ Find the minimum sample size. Is this sample size practical?