Exam 7: Estimates and Sample Sizes

Use the confidence level and sample data to find a confidence interval for estimating the population µ. -Test scores: $\mathrm { n } = 109 , \overline { \mathrm { x } } = 79.1 , \sigma = 6.9 ; 99$ percent

Solve the problem. -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.

Provide an appropriate response. -Draw a diagram of the chi-square distribution. Discuss its shape and values.

Provide an appropriate response. -Interpret the following 95% confidence interval for mean weekly salaries of shift managers at Guiseppe's Pizza and Pasta. $325.80 < \mu < 472.30$

Solve the problem. -Find the critical value $\chi _ { L } ^ { 2 }$ corresponding to a sample size of 9 and a confidence level of 90 percent.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: 0.04; confidence level: 99%; from a prior study, ^p is estimated by 0.08

Find the appropriate minimum sample size -You want to be 95% confident that the sample variance is within 30% of the population variance.

Solve the problem. -464 randomly selected light bulbs were tested in a laboratory, 424 lasted more than 500 hours. Find a point estimate of the true proportion of all light bulbs that last more than 500 hours.

Find the margin of error for the 95% confidence interval used to estimate the population proportion -In a clinical test with 2353 subjects, 1136 showed improvement from the treatment.

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 - $99 \% ; \mathrm { n } = 17 ; \sigma$ is unknown; population appears to be normally distributed.

Find the margin of error. - $95 \%$ confidence interval; $\mathrm { n } = 21 ; \overline { \mathrm { x } } = 0.44 ; \mathrm { s } = 0.44$

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 35 of those weights yields a mean of $\mathrm { lb }$ and $\mathrm { a }$ standard deviation of $23.2 \mathrm { lb }$ .

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -n = 182, x = 135; 95 percent

Find the margin of error. - $n = 30 , \bar { x } = 86.5 , s = 10.3,90$ percent

The following confidence interval is obtained for a population proportion, p: 0.802 < p < 0.828 Use these confidence interval limits to find the point estimate, $\hat { p }$ .

Solve the problem. -Find the critical valu $\mathrm { z } _ { \alpha / 2 }$ that corresponds to a degree of confidence of 98%.

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: $0.002$ ; confidence level: $93 \% ; \hat { \mathrm { p } }$ and $\hat{\mathrm { q }}$ unknown

Use the confidence level and sample data to find the margin of error E -The sample size is $n = 11 , \sigma$ is not known, and the original population is normally distributed.

Determine whether the given conditions justify using the margin of error $\mathrm { E } = \mathrm { z } _ { \alpha / 2 } \sigma / \sqrt { \mathrm { n } }$ when finding a confidence interval estimate of the population mean $\mu$ . -The sample size is n = 286 and $\sigma = 15$

Solve the problem. -A $99 \%$ confidence interval (in inches) for the mean height of a population is $65.3 < \mu < 66.9$ . 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).