Section 1 Homework
Use the values on the number line to find the sampling error. (Since I don't have Quizlet+, I can't insert the image of the actual infinite number line; ergo, I pasted the description.) An infinite number line, labeled from 4 to 6, has tick marks in increments of 0.2. From left to right, a point labeled *"x̄" = 4.7* is plotted at 4.7, and a point labeled *"μ" = 4.96* is plotted at 4.96. The sampling error is *___*.
Correct Answer: *-0.26*
Find the margin of error for the given values of *c*, *σ*, and *n*. (Round answer to *three* decimal places.) *c = 0.90, σ = 2.1, n = 36* *Level of Confidence* = *z*∨*c* *90%* = *1.645* *95%* = *1.96* *99%* = *2.575* *E* = *____*
Correct Answer: *0.576*
Find the margin of error for the given values of *c*, *σ*, and *n*. (Round answer to *three* decimal places.) *c = 0.95, σ = 3.8, n = 36* *Level of Confidence* = *z*∨*c* *90%* = *1.645* *95%* = *1.96* *99%* = *2.575* *E* = *____*
Correct Answer: *1.241*
Use the values on the number line to find the sampling error. (Since I don't have Quizlet+, I can't insert the image of the actual infinite number line; ergo, I pasted the description.) An infinite number line, labeled from 24 to 27, has tick marks in increments of 0.5. Labeled points are plotted at *"μ" = 24.88* and *"x̄" = 26.18*. The sampling error is *___*.
Correct Answer: *1.30*
Find the critical value *z*∨*c* necessary to form a confidence interval at the level of confidence shown below. (Round answer to *two* decimal places.) *"c" = 0.86* *z*∨*c* = *___*
Correct Answer: *1.48*
Find the critical value *z*∨*c* necessary to form a confidence interval at the level of confidence shown below. (Round answer to *two* decimal places.) *"c" = 0.93* *z*∨*c* = *___*
Correct Answer: *1.81*
Determine the minimum sample size required when you want to be 95% confident that the sample mean is within one unit of the population mean and *σ = 13.8*. Assume the population is normally distributed. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) A 95% confidence level requires a sample size of *___*.
Correct Answer: *732*
Which statistic is the best unbiased estimator for *μ*? The best unbiased estimated for *μ* is *_.*
Correct Answer: *x̄.*
When estimating a population mean, are you more likely to be correct when you use a point estimate or an interval estimate? Explain your reasoning. Choose the correct answer below. A.) You are more likely to be correct using a point estimate because an interval estimate is too broad and contains many possible values. B.) If *n* ≤ 30, an interval estimate is more accurate. If *n* > 30, a point estimate is more accurate. C.) You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean. D.) There is no difference between an interval estimate and a point estimate in terms of accuracy.
Correct Answer: C.) You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean.
When a population is finite, the formula that determines the standard error of the mean *σ*∨*x̄* needs to be adjusted. If *N* is the size of the population and *n* is the size of the sample (where *n* ≥ 0.05*N*), then the standard error of the mean is *σ*∨*x̄* = ((*σ*)/√(*n*)) × √((*N* − *n*)/(*N* − 1)). The expression √((*N* − *n*)/(*N* − 1)) is called the finite population correction factor. Use the finite population correction factor to construct the confidence interval for the population mean described below. (Round answer(s) to *two* decimal places.) *c = 0.95* *x̄ = 18.5* *σ = 4.3* *N = 900* *n = 81* The 95% confidence interval for the population mean is (*__(1)__*, *__(2)__*).
Correct Answers: *(1):* *17.61* *(2):* *19.39*
For each level of confidence c below, determine the corresponding normal confidence interval. Assume each confidence interval is constructed for the same sample statistics. *17.)* c = 0.88 *18.)* c = 0.90 *19.)* c = 0.95 *20.)* c = 0.98 Drag (match) each normal confidence interval given above to (with) the level of confidence *c.* (Since I don't have Quizlet+, I can't insert the images of the normal confidence intervals; ergo, I pasted each of their sample means (*x̄*), and their *lower* & *upper* limits. (Also, MyMathLab doesn't show the confidence intervals *BEFORE* I completed the problem.)) 17.) 18.) 19.) 20.)
Correct Answers: *17.)* *Lower:* *57.2* *x̄* = *58.7* *Upper:* *60.2* *18.)* *Lower:* *57.1* *x̄* = *58.7* *Upper:* *60.3* *19.)* *Lower:* *56.8* *x̄* = *58.7* *Upper:* *60.6* *20.)* *Lower:* *56.4* *x̄* = *58.7* *Upper:* *61*
When a population is finite, the formula that determines the standard error of the mean needs to be adjusted. If *N* is the size of the population and *n* is the size of the sample (where *n* ≥ 0.05*N*), the standard error of the mean is shown below. The finite population correction factor is given by the expression √((*N* − *n*)/(*N* − 1)). Determine the finite population correction factor for each of the following (round all answers to *three* decimal places). *Part 1 (a):* *N = 1,400* and *n = 1,050* *Part 2 (b):* *N = 1,400* and *n = 140* *Part 3 (c):* *N = 1,400* and *n = 100* *Part 4 (d):* *N = 1,400* and *n = 70* *Part 5 (e):* What happens to the finite population correction factor as the sample size *n* decreases, but the population size *N* remains the same? *Part 1 (a):* The finite population correction factor is *__*. *Part 2 (b):* The finite population correction factor is *____*. *Part 3 (c):* The finite population correction factor is *____*. *Part 4 (d):* The finite population correction factor is *____*. *Part 5 (e):* What happens to the finite population correction factor as the sample size *n* decreases, but the population size *N* remains the same? A.) The finite population correction factor does not approach any set value. B.) The finite population correction factor approaches 0. C.) The finite population correction factor approaches 1. D.) The finite population correction factor approaches √(1/2) ≈ 0.707.
Correct Answers: *Part 1 (a):* *0.5* *Part 2 (b):* *0.949* *Part 3 (c):* *0.964* *Part 4 (d):* *0.975* *Part 5 (e):* C.) The finite population correction factor approaches 1.
Use the confidence interval to find the estimated margin of error. Then find the sample mean. *A biologist reports a confidence interval of (3.0, 4.4) when estimating the mean height (in centimeters) of a sample of seedlings.* *Part 1:* The estimated margin of error is *__*. *Part 2:* The sample mean is *__*.
Correct Answers: *Part 1:* *0.7* *Part 2:* *3.7*
An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.6 years of the population mean. Assume the population of ages is normally distributed. *Part 1 (a):* Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.7 years. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* The sample mean is 21 years of age. Using the minimum sample size with a 90% level of confidence, does it seem likely that the population mean could be within 9% of the sample mean? Within 10% of the sample mean? Explain. (Round all answers to *two* decimal places.) *(I didn't copy the accompanying Standard Normal Tables because they're too big.)* *Part 1 (a):* The minimum sample size required to construct a 90% confidence interval is *_* students. *Part 2 (b):* The 90% confidence interval is (*__(1)__*, *__(2)__*). It *__(3)__* likely that the population mean could be within 9% of the sample mean because the interval formed by the values 9% away from the sample mean *______(4)______* the confidence interval. It *__(5)__* likely that the population mean could be within 10% of the sample mean because the interval formed by the values 10% away from the sample mean *______(6)______* the confidence interval.
Correct Answers: *Part 1 (a):* *4* *Part 2 (b):* *(1):* *19.60* *(2):* *22.40* *(3):* *seems* *(4):* *entirely contains* *(5):* *seems* *(6):* *entirely contains*
A cheese processing company wants to estimate the mean cholesterol content of all one-ounce servings of a type of cheese. The estimate must be within 0.67 milligram of the population mean. *Part 1 (a):* Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 3.18 milligrams. (Round answer *up* to the *nearest whole number* (*zero* decimal places).) *Part 2 (b):* The sample mean is 35 milligrams. Using the minimum sample size with a 95% level of confidence, does it seem likely that the population mean could be within 2% of the sample mean? within 0.2% of the sample mean? Explain. (Round all answers to *two* decimal places.) *(I didn't copy the accompanying Standard Normal Tables because they're too big.)* *Part 1 (a):* The minimum sample size required to construct a 95% confidence interval is *__* servings. *Part 2 (b):* The 95% confidence interval is (*__(1)__*, *__(2)__*). It *__(3)__* likely that the population mean could be within 2% of the sample mean because the interval formed by the values 2% away from the sample mean *______(4)______* the confidence interval. It *__________(5)__________* [seem] likely that the population mean could be within 0.2% of the sample mean because the interval formed by the values 0.2% away from the sample mean *____________(6)____________* the confidence interval.
Correct Answers: *Part 1 (a):* *87* *Part 2 (b):* *(1):* *34.33* *(2):* *35.67* *(3):* *seems* *(4):* *entirely contains* *(5):* *does not seem* *(6):* *overlaps but does not entirely contain*
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean (round answers to *two* decimal places). Interpret the results and compare the widths of the confidence intervals. *From a random sample of 48 business days, the mean closing price of a certain stock was $119.69. Assume the population standard deviation is $10.94.* *Part 1:* The 90% confidence interval is (*___(1)___*, *___(2)___*). *Part 2:* The 95% confidence interval is (*___(1)___*, *___(2)___*). *Part 3:* Which interval is wider? Choose the correct answer below. A.) The 95% confidence interval B.) The 90% confidence interval *Part 4:* Interpret the results. A.) You can be certain that the population mean price of the stock is either between the lower bounds of the 90% and 95% confidence intervals or the upper bounds of the 90% and 95% confidence intervals. B.) You can be certain that the closing price of the stock was within the 90% confidence interval for approximately 43 of the 48 days, and was within the 95% confidence interval for approximately 46 of the 48 days. C.) You can be 90% confident that the population mean price of the stock is outside the bounds of the 90% confidence interval, and 95% confident for the 95% interval. D.) You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
Correct Answers: *Part 1:* *(1):* *117.09* *(2):* *122.29* *Part 2:* *(1):* *116.60* *(2):* *122.78* *Part 3:* A.) The 95% confidence interval *Part 4:* D.) You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean (round answers to *two* decimal places). Interpret the results and compare the widths of the confidence intervals. *From a random sample of 60 dates, the mean record high daily temperature in a certain city has a mean of 82.77°F. Assume the population standard deviation is 14.06°F.* *Part 1:* The 90% confidence interval is (*__(1)__*, *__(2)__*). *Part 2:* The 95% confidence interval is (*__(1)__*, *__(2)__*). *Part 3:* Which interval is wider? Choose the correct answer below. A.) The 90% confidence interval B.) The 95% confidence interval *Part 4:* Interpret the results. A.) You can be certain that the mean record high temperature was within the 90% confidence interval for approximately 54 of the 60 days, and was within the 95% confidence interval for approximately 57 of the 60 days. B.) You can be 90% confident that the population mean record high temperature is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval. C.) You can be certain that the population mean record high temperature is either between the lower bounds of the 90% and 95% confidence intervals or the upper bounds of the 90% and 95% confidence intervals. D.) You can be 90% confident that the population mean record high temperature is outside the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
Correct Answers: *Part 1:* *(1):* *79.78* *(2):* *85.76* *Part 2:* *(1):* *79.21* *(2):* *86.33* *Part 3:* B.) The 95% confidence interval *Part 4:* B.) You can be 90% confident that the population mean record high temperature is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
Use the confidence interval to find the margin of error and the sample mean. *(0.208, 0.350)* *Part 1:* The margin of error is *____*. *Part 2:* The sample mean is *____*.
Correct Answers: *Part 1:* *0.071* *Part 2:* *0.279* INcorrect Answer(s): *Part 1:* *0.279*
Use the confidence interval to find the margin of error (round answer to *two* decimal places), and the sample mean (type answer as either an *integer* or a *decimal*). *(1.73, 1.97)* *Part 1:* The margin of error is *___*. *Part 2:* The sample mean is *___*.
Correct Answers: *Part 1:* *0.12* *Part 2:* *1.85*
The equation for determining the sample size *n* = ((*z*∨*c* × *σ*)/*Ε*)² can be obtained by solving the equation for the margin of error *Ε* = *z*∨*c* × *(σ/√n)*. Show that this is true and justify each step. *Part 1:* How can the equation *Ε* = *z*∨*c* × *(σ/√n)* be used to obtain *n* = ((*z*∨*c* × *σ*)/*Ε*)²? A.) Rewrite the equation by solving for *n*. B.) Rewrite the equation by solving for *Ε*. C.) Rewrite the equation by solving for *z*∨*c*. D.) Rewrite the equation by solving for *σ*. Solve the equation for the specified variable. State the operation to be performed and simplify both sides of the equation. *Ε* = *z*∨*c* × *(σ/√n)* ~~ *Write the original equation.* *[i]* = part of equation *left* of *=* *[ii]* = part of equation *right* of *=* *[iii]* = description (*in words*) of *[i]* and *[ii]* *Part 2:* *[i]* = *[ii]* ~~ *[iii]* *Part 3:* *[i]* = *[ii]* ~~ *[iii]* *Part 4:* *n* = ((*z*∨*c* × *σ*)/*Ε*)² ~~ *[i]*
Correct Answers: *Part 1:* A.) Rewrite the equation by solving for *n*. *Part 2:* *[i]:* *Ε* × *√(n)* *[ii]:* *z*∨*c* × *σ* *[iii]:* *Multiply both sides by √(n).* *Part 3:* *[i]:* *√(n)* *[ii]:* *z*∨*c* × *σ*/*Ε* *[iii]:* *Divide both sides by Ε.* *Part 4:* *[i]:* *Square both sides of the equation.*
