Exam 7: Estimates and Sample Sizes

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 $\mathrm { n } = 5 , \sigma = 12.7$ and the original population is normally distributed.

Find the margin of error. -Thirty randomly selected students took the calculus final. If the sample mean was 78 and the standard deviation was 7.5, construct a 99 percent confidence interval for the mean score of all students.

Solve the problem. -Find the value of ${ } ^ { Z } \alpha / 2$ that corresponds to a level of confidence of 98.22 percent.

Use the confidence level and sample data to find a confidence interval for estimating the population µ. -The duration of telephone calls directed by a local telephone company: $\sigma = 4.1$ minutes, $\mathrm { n } = 540$ , 97 percent confidence.

Find the margin of error. - $\mathrm { n } = 10 , \overline { \mathrm { x } } = 12.8 , \mathrm {~s} = 4.9,95$ percent

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.

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution. -Weights of men: $90 \%$ confidence; $\mathrm { n } = 14 , \overline { \mathrm { x } } = 160.9 \mathrm { lb } , \mathrm { s } = 12.6 \mathrm { lb }$

Provide an appropriate response. -Define confidence interval and degree of confidence. Make up an example of a confidence interval and interpret the result.

Solve the problem. -The confidence interval below for the population standard deviation is based on the following sample statistics: =41,=43.7 , and =4.3 . 3.64<\sigma<5.28 What is the degree of confidence?

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p -Margin of error: $0.006$ ; confidence level: $96 \% ; \hat { p }$ and $\mathrm { q }$ are unknown

Use the confidence level and sample data to find the margin of error E -Weights of eggs: $95 \%$ confidence; $\mathrm { n } = 47 , \overline { \mathrm { x } } = 1.44 \mathrm { oz } , \sigma = 0.39 \mathrm { oz }$

Use the confidence level and sample data to find a confidence interval for estimating the population µ. -34 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of $25.9$ pounds and a standard deviation of $3.8$ pounds. What is the 95 percent confidence interval for the true mean weight, $\mu$ , of all packages received by the parcel service?

Solve the problem. -Find the critical value $\mathrm { z } _ { \alpha / 2 }$ that corresponds to a degree of confidence of 91%.

Find the margin of error for the 95% confidence interval used to estimate the population proportion -n = 133, x = 82; 90 percent

Find the margin of error. - $95 \%$ confidence interval; $\mathrm { n } = 51 ; \overline { \mathrm { x } } = 388 ; \mathrm { s } = 204$

Find the margin of error. -The amounts (in ounces)of juice in eight randomly selected juice bottles are: 15.0 15.9 15.3 15.3 15.5 15.9 15.9 15.0 Construct a 98 percent confidence interval for the mean amount of juice in all such bottles.

Provide an appropriate response. -Explain the difference between descriptive and inferential statistics.

Find the margin of error -95% confidence interval; $\mathrm { n } = 91 ; \overline { \mathrm { x } } = 72 , \mathrm {~s} = 11.4$

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. If a manufacturer randomly selects 21 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?