Exam 7: Estimating Parameters and Determining Sample Sizes

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

In constructing a confidence interval for $\sigma \text { or } \sigma ^ { 2 }$ a table is used to find the critical values $\chi _ { \mathrm { L } } ^ { 2 } \text { and } \chi _ { \mathrm { R } } ^ { 2 }$ for values of $n \leq 101$ For larger values of n , $\chi _ { \mathrm { L } } ^ { 2 } \text { and } \chi _ { \mathrm { R } } ^ { 2 }$ approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } \pm z _ { \alpha / 2 } + \sqrt { 2 k - 1 } ^ { 2 }$ 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 lb and a standard deviation of 25.5 lb . Round the confidence interval limits to the nearest hundredth.

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% confidence; n = 2388, x = 1672

Describe the process for finding the confidence interval for a population proportion.

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.

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

Describe the steps for finding a confidence interval.

Identify the correct distribution (z, t, or neither)for each of the following. Sample Size Standard Deviation Shape of the Distribution z or t or neither n=35 s=4.5 Somewhat skewed n=20 s=4.5 Bell shaped n=35 \sigma=4.5 Bell shaped n=20 \sigma=4.5 Extremely skewed

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?

Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students Who plan to work in a rural community.

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

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

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?

Do one of the following, as appropriate: (a) Find the critical value $z _ { \alpha / 2 }$ (b) find the critical value $t _ { \alpha / 2 }$ 99% ; n=17 ; $\alpha$ is unknown; population appears to be normally distributed.

Identify the distribution that applies to the following situation: In constructing a confidence interval of $\sigma$ you have 50 sample values and they appear to be From a population with a skewed distribution.

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

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?

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. $n = 30 , \bar { x } = 84.6 , s = 10.5,90 \%$ confidence

Fifty people are selected randomly from a certain population and it is found that 12 people in the sample are over six feet tall. What is the best point estimate of the proportion of people in The population who are over 6 feet tall?

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.