Exam 7: Estimating Parameters and Determining Sample Sizes

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

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

Of 118 randomly selected adults, 34 were found to have high blood pressure. Construct a $95 \%$ confidence interval for the true percentage of all adults that have high blood pressure.

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?

In a Gallup poll of 557 randomly selected adults, 284 said that they were underpaid. Identify the best point estimate of the percentage of adults who say they are underpaid. Construct a 95% confidence interval estimate of the percentage of adults who say that that they are underpaid. Can we safely conclude that the majority of adults say that they are underpaid?

Fill in the blank: The critical value $z _ { \alpha / 2 }$ that corresponds to a _________% confidence level is 2.33.

Bert constructed a confidence interval to estimate the mean weight of students in his class. The population was very small - only 30. Ruth constructed a confidence interval for the mean weight of all adult males in the city. She based her confidence interval on a very small sample of only 5. Which confidence interval is likely to give a better estimate of the mean it is estimating? Which is likely to be more of a problem, a small sample or a small population?

The following is a 95% confidence interval of the proportion of female medical school students: 0.449 < p < 0.511, based on data from the American Medical Association. What is the point estimate of the proportion of females in the population of medical school students? Write a brief statement that correctly interprets the confidence interval given.

If the computed minimum sample size n needed for a particular margin of error is not a whole number, round the value of n _______ (up or down)to the next ________ (smaller or larger) Whole number.

A one-sided confidence interval for $\mathrm { p }$ can be written as $p < \hat { p } + \mathrm { E }$ or $p > \hat { p } - \mathrm { E }$ where the margin of error $\mathrm { E }$ 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.

Express the confidence interval $0.039 < p < 0.479$ in the form of $\hat { p } \pm \mathrm { E }$ .

In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$ , a table is used to find the critical values $\chi _ { \mathrm { L } } ^ { 2 }$ and $\chi _ { \mathrm { R } } ^ { 2 }$ for values of $n \leq 101$ . For larger values of $n , \chi _ { \mathrm { L } } ^ { 2 }$ and $\chi _ { \mathrm { R } } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } \pm z _ { \alpha / 2 } + \sqrt { 2 k - 1 } ^ { 2 }$ where $k$ is the number of degrees of freedom and $z _ { \alpha / 2 }$ is the critical $z$ -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.

Which of the following critical values is appropriate for a $98 \%$ confidence level where $n = 7$ ; $\sigma = 27$ and the population appears to be normally distributed.

Find the value of $z _ { \alpha / 2 }$ that corresponds to a confidence level of 89.48%.

Define margin of error. Explain the relation between the confidence interval and the margin of error. Suppose a confidence interval is 9.65 < $\mu$ < 11.35. Redefine the confidence interval into a format using the margin of error and the point estimate of the population mean.

In general, what does "degrees of freedom" refer to? Find the degrees of freedom for the given information, assuming that you want to construct a confidence interval estimate of $\mu$ : Six human skulls from around 4000 B.C. were measured, and the lengths have a mean of $94.2 \mathrm {~mm}$ and a standard deviation of $4.9 \mathrm {~mm}$

To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be $s = 0.16$ . Find the $95 \%$ confidence interval for the population standard deviation $\sigma$ .

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: $0.008$ ; confidence level: $98 \% ; \hat { p }$ and $\hat { q }$ unknown

Six human skulls from around 4000 B.C. were measured, and the lengths have a mean of 94.2 mm and a standard deviation of 4.9 mm. If you want to construct a 95% confidence interval estimate of the mean length of skulls, what requirements must be satisfied?