Exam 7: Estimating Parameters and Determining Sample Sizes

Use the given information to find the minimum sample size required to estimate an unknown population mean $\mu$ -How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98% confidence that the sample mean is within $400 of the population mean, and the Population standard deviation is known to be $1400.

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

Express a confidence interval defined as (0.432, 0.52) in the form of the point estimate ________ ± the margin of error ________. Express both in three decimal places.

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 $n \leq 101$ . For larger values of $n , \chi { } _ { L } ^ { 2 }$ and $\chi \frac { 2 } { R }$ 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. 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 interval limits to the nearest hundredth.

Use the given information to find the minimum sample size required to estimate an unknown population mean $\mu$ -How many business students must be randomly selected to estimate the mean monthly earnings of business students at one college? We want 95% confidence that the sample mean is within $135 of the population mean, And the population standard deviation is known to be $538.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion $p$ . $n = 56 , x = 30 ; 95 \%$ 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 daily intakes of milk (in ounces) for ten randomly selected people were: 23.3 28.4 10.5 16.4 26.4 18.1 20.4 17.3 27.4 13.2 Find a $99 \%$ confidence interval for the population standard deviation $\sigma$ .

Find the critical value $\mathrm { z } _ { \alpha / 2 }$ that corresponds to a 93% confidence level.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -n = 195, x = 162; 95% confidence

Solve the problem. -The following confidence interval is obtained for a population proportion, p: $( 0.688,0.724 )$ . Use these confidence interval limits to find the point estimate, $\hat { p }$ .

Solve the problem. -Find the critical value $\chi _ { \mathrm { R } } ^ { 2 }$ corresponding to a sample size of 19 and a confidence level of 99 percent.

Express the confidence interval using the indicated format. -Express the confidence interval $0.38 < \mathrm { p } < 0.54$ in the form of $\hat { \mathrm { p } } \pm \mathrm { E }$ .

Solve the problem. -The following confidence interval is obtained for a population proportion, p: 0.753 < p < 0.797. Use these confidence interval limits to find the point estimate, $\hat { p } .$

Use the given data to find the minimum sample size required to estimate the population proportion. -Margin of error: 0.009; confidence level: 99%; $\hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }$

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. - $\mathrm { n } = 10 , \overline { \mathrm { x } } = 8.1$ , s $= 4.8,95 \%$ confidence

Find the critical value $z _ { \alpha / 2 }$ that corresponds to a 91% confidence level.

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 amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.4 15.8 15.4 15.1 15.8 15.9 15.8 15.7 Construct a $98 \%$ confidence interval for the mean amount of juice in all such bottles.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.

Solve the problem. -When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population Correction factor: $\sqrt { \frac { N - n } { N - 1 } }$ Find the 95% confidence interval for the mean of 200 weights if a sample of 34 of those weights yields a mean of 155.7 lb and a standard deviation of 24.1 lb.

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. $95 \% \text { confidence; } n = 2388 , x = 1672$