Exam 7: Estimates and Sample Sizes

Find the margin of error. -The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.7 15.6 15.3 15.3 15.1 15.6 15.3 15.4 Find a 98 percent confidence interval for the population standard deviation $\sigma$ .

Solve the problem. -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: $\chi ^ { 2 } = \frac { 1 } { 2 } \left[ \pm z _ { \alpha / 2 } + \sqrt { 2 \mathrm { k } - 1 } \right] ^ { 2 }$ 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 $\chi _ { \mathrm { L } } ^ { 2 }$ and $\chi _ { \mathrm { R } } ^ { 2 }$ for a situation in which you wish to construct a $95 \%$ confidence interval for $\sigma$ and in which the sample size is $\mathrm { n } = 298$ .

Solve the problem. -Of 132 adults selected randomly from one town, 33 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke.

Find the margin of error. -The principal randomly selected six students to take an aptitude test. Their scores were: 77.9 89.1 80.7 78.6 74.4 82.0 Determine a 90 percent confidence interval for the mean score for all students.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: $0.045$ ; confidence level: $96 \% ; \hat { p }$ and $\hat { q }$ are unknown

Find the margin of error. - $95 \%$ confidence interval; $\mathrm { n } = 12 ; \overline { \mathrm { x } } = 45.6 ; \mathrm { s } = 4.2$

Find the chi-square value $\chi _ { L } ^ { 2 }$ corresponding to a sample size of 13 and a confidence level of 98 percent.

Do one of the following, as appropriate: (a) Find the critical value $z_{?/2}$ , (b) find the critical value $t_{?/2}$ , (c) state that neither the normal nor the t distribution applies - $95 \% ; \mathrm { n } = 11 ; \sigma$ is known; population appears to be very skewed.

Solve the problem. -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 $\mathrm { n } , \chi _ { \mathrm { L } } ^ { 2 }$ and $\chi _ { \mathrm { R } } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } \left[ \pm z _ { \alpha / 2 } + \sqrt { 2 k - 1 } \right] ^ { 2 }$ where $\mathrm { k }$ is the number of degrees of freedom and $\mathrm { z } \alpha / 2$ is the critical z score. Construct the $90 \%$ confidence interv: for $\sigma$ using the following sample data: a sample of size $n = 261$ yields a mean weight of $154 \mathrm { lb }$ and a standard deviation of $25.3 \mathrm { lb }$ . Round the confidence interval limits to the nearest hundredth.

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.

Find the margin of error. -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 10 14 15 15 5 12 15 11 11 Find a 95 percent confidence interval for the population standard deviation $\sigma$ .

Solve the problem. -The confidence interval below for the population variance is based on the following sample statistics: $n = 25$ , $\bar { x } =$ $31.1$ , and $s = 4.2$ . $9.29 < \sigma ^ { 2 } < 42.82$ 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 = 400. In a random sample of 82 students, selected without replacement, there are 11 left handers.

Do one of the following, as appropriate: (a) Find the critical value $z_{?/2}$ , (b) find the critical value $t_{?/2}$ , (c) state that neither the normal nor the t distribution applies - $91 \% ; \mathrm { n } = 45 ; \sigma$ is known; population appears to be very skewed.

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

Find the margin of error. - $99 \%$ confidence interval; $\mathrm { n } = 201 ; \overline { \mathrm { x } } = 217 ; \mathrm { s } = 34$

The following confidence interval is obtained for a population proportion, p: (0.348, 0.382) Use these confidence interval limits to find the margin of error, E.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: $0.011$ ; confidence level: $92 \% ; \hat { \mathrm { p } }$ and $\hat { \mathrm { q } }$ unknown

Solve the problem. -A simple random sample of women aged 18-24 is selected, and the systolic blood pressure of each woman is measured. The results (in $\mathrm { mmHg }$ ) are as follows: $\overline { \mathrm { x } } = 122.1 , \mathrm {~s} = 14.4$ . The sample size is less than 20. A $99 \%$ confidence interval for the population mean is found to be $\mu = 122.1 \pm 10.20$ . Find the sample size.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: $0.005$ ; confidence level: $97 \%$ ; $\mathrm { p }$ and $\mathrm { q }$ are unknown