A study reports that recent college graduates from Conne

Homeworklib Q&A #1260 - A study reports that recent college graduates from Conne

A study reports that recent college graduates from Connecticut face the highest average debt of $38,510 (forbescom. September 18, 2019). A researcher from Pennsylvania wants to determine how recent undergraduates from that state fare. The accompanying file contains data on debt from 40 recent undergraduates. Assume that the population standard deviation is $5,000. (You may find it useful to reference the 2 table.) EinctureClick here for the Excel Data File 5. Construct the 95% confidence interval for the mean debt ofall undergraduates from Pennsylvania. (Round final answers to 2 decimal places.) 9 Answer is complete but not entirely correct. Confidence interval 29,669 50 6 to 32,392.50 9 b. Use the 95% confidence interval to determine if the debt of Pennsylvania undergraduates differs from that of Connecticut undergraduates. The debt of Pennsylvania undergraduates differs from that of Connecticut undergraduates. O The debt of Pennsylvania undergraduates does not differ from that of Connecticut undergraduates.

A survey reported that approximately 70% of people in the 50 to 64 age bracket have tried some type of alternative therapy (for instance, acupuncture or the use of nutrition supplements). Assume this survey was based on a sample of 400 people. (You may find it useful to reference the z table.) a-1. Identify the relevant parameter of interest for this categorical variable. O The parameter of interest is the mean of all people between the ages of 50-64. The parameter of interest is the proportion of all people between the ages of 50-64. a-2. Compute the point estimate as well as the margin of error with 90% confidence. (Round final answers to 3 decimal places.) x Answer is complete but not entirely correct. Point estimate 0.700 Margin of error 0.020 X b. You decide to redo the analysis with the margin of error reduced to 2%. How large a sample do you need to draw? (Round up final answer to nearest whole number.) Answer is complete and correct. Sample size
1,421

Homeworklib Always Helps You

If you want answers to questions about any subject then you have come to the right place to get free study material because here you will not only get answers but also information about many other such courses, study material, book guides, career guides, courses, and college guides their entire library for free.

You can see answers to many questions on all these topics.
#Chemistry, #Biology, #Math, #Physics, #ElectriaclEngineering, #CivilEngineering, #MechnicalEngineering, #chemicalEngineering, #Accounting, #Business, #Nursing, #ComputerScience, #EarthScience, #English, #History

So now you can easily ask all subjects questions by filling below form to get free Answers, this service is very fast and free on Homeworklib.com

Homeworklib: Your Answer

 $\text{Exercise 8-13 Static}$

$\text{a.}\text{Constructthe95\%confidenceintervalforthemeandebtofallfromPennsylvania.}$ Confidence interval $\underline{23,921.08}$ to $\underline{27,020.12}$   $$\begin{gathered} \text{b.}\text{Usethe95\%confidenceintervaltodetermineifthedebtofPennsylvaniaundergraduatesdifferfromthatof} \\ \text{Connecticutundergraduates.} \end{gathered}$$ $\text{ThedebtofPennsylvaniaundergraduatesdiffersfromthatofConnecticutundergraduates.}$

$\text{Exercise 8-69 Static}$

$\text{a-1.}\text{Identifytherelevantparameterofinterestforthiscategoricalvariable.}$ $\text{Theparameterofinterestistheproportionofallpeoplebetweentheagesof50-64.}$   $\text{a-2.}\text{Computethepointestimateaswellasthemarginoferrorwith90\%confidence.}$     Point estimate $\underline{0.700}$ Margin of error $\underline{0.119}$   $\text{b.}\text{Youdecidetoredotheanalysiswiththemarginoferrorreducedto2\%.Howlargeasampledoyouneedtodraw?}$ Sample size $\underline{1421}$

 

Step-by-step explanation

 

$\text{Exercise 8-13 Static}$

 

Given Data:

 

$\text{c}=95\%\text{(confidence level)}$

$\sigma =\$5,000\text{(population standard deviation)}$

$\text{n }=40\text{(sample size; number of data from recent undergraduates)}$

$24040	$19153	$26762	$31923	$31533	$34207	$14623	$24370	$31016	$20107
$22090	$17089	$16306	$20653	$21673	$14951	$22701	$23521	$26215	$23714
$23906	$23690	$32254	$25115	$24610	$22975	$35402	$29869	$37419	$22266
$33848	$17479	$28236	$30052	$35136	$25118	$22922	$28917	$23634	$29329

 

 

Solution:

 

$\text{a.}\text{Constructthe95\%confidenceintervalforthemeandebtofallfromPennsylvania.}$

 

In the given problem, the sample size is greater than 30 (n = 40 > 30). Additionally, the population standard deviation is known σ = $5,000. (Note: The population standard deviation (σ) is different from the sample standard deviation (s)). In this case, knowing that the sample size (n) is greater than 30 and that the population standard deviation is known, we use a z-distribution.

 

Now, to determine the confidence interval, we can do as follows:

Step 1: Calculate the sample mean  $\overline{\text{x}}$.

 

The sample mean $\overline{\text{x}}$ is the average of the data, so we add all the data up and then divide the sum by the number of data. In this case,

 

$\overline{\text{x}}=\frac{\sum x}{n}$

 

$\overline{\text{x}}=\frac{\begin{array}{c}\$24,040+\$19,153+\$26,762+\$31,923+\$31,533+\$34,207+\$14,623+\$24,370\\ +\$31,016+\$20,107+\end{array}}{40}$

$\begin{array}{c}\$22,090+\$17,089+\$16,306+\$20,653+\$21,673+\$14,951\\ +\$22,701+\$23,521+\$26,215+\$23,714+\$23,906+\end{array}$

$\begin{array}{c}\$23,690+\$32,254+\$25,115\\ +\$24,610+\$22,975+\$35,402+\$29,869+\$37,419+\$22,266+\$33,848+\$17,479\\ +\$28,236+\end{array}$

$\begin{array}{c}\$30,052+\$35,136+\$25,118+\$22,922+\$28,917+\$23,634+\$29,329\end{array}$

 

$\overline{\text{x}}=\frac{\$1,018,824}{40}$

 

$\overline{\text{x}}=\$25,470.6$

 

Therefore, the sample mean is equal to:

 

$\overline{\text{x}}=\$25,470.6$

 

 

Step 2: Determine the significance level α.

 

By definition, the significance level (α) can be calculated by subtracting 1 with the given confidence level of 95%. Hence,

 

$\alpha =1-0.95=0.05\text{(significance level)}$

 

 

Step 3: Calculate the critical value z and draw a graph.

 

The value of the critical value z can be determined by knowing the area represented by the significance level. In this case, the area represented by the significance level are the areas at the left end and the right end of the distribution curve. Knowing this, we divide the significance level by two to get $\left(\frac{\alpha }{2}=0.025\right)$ the area in the left and in the right. This can be illustrated clearly in the following distribution curve:

 

Using the z-distribution table, we can determine the corresponding critical value z for a probability area of $\frac{\alpha }{2}=0.025$. Doing so, we get the absolute value of the critical value z as:

 

$z=|z_{\frac{\alpha }{2}}|=|\pm 1.96|=1.96$.

Step 4: Calculate the margin of error.

 

The formula for the margin of error is expressed as:

 

$E=z(\sqrt{n}$

​σ​)

 

Where z is the value of the critical value, σ is the population standard deviation, and n is the sample size. Substituting the values, we get:

 

$E=1.96(\sqrt{40}$

​$5,000​)=$1549.516053   Hence, the margin of error at a 95% confidence level is:   $E=\$1549.516053$     Step 5: Determine the 95% confidence interval.   The confidence interval can be calculated using the formula expressed as:   $\text{Confidence Interval}=\overline{x}\pm E$   Where x̄ is the sample mean and E is the margin of error calculated in the previous section. Substituting the values, we get:   $\text{Confidence Interval}=\$25,470.6\pm \$1549.516053$   Hence, the confidence interval of the population mean is:   $\text{Confidence Interval}=\left(\$23,921.08395,\$27,020.11605\right)$   Rounding to two decimal places, we get:   $\text{Confidence Interval}=\left(\$23,921.08,\$27,020.12\right)$     Therefore, the 95% confidence interval is: Confidence interval $\underline{23,921.08}$ to $\underline{27,020.12}$

 

 

 

$$\begin{gathered} \text{b.}\text{Usethe95\%confidenceintervaltodetermineifthedebtofPennsylvaniaundergraduatesdifferfromthatof} \\ \text{Connecticutundergraduates.} \end{gathered}$$

Comparing the constructed confidence interval to the average debt ($38,510) of college graduates from Connecticut, we can see that the average debt of college graduates from Connecticut is higher than the constructed confidence interval. Hence, we can say that there is a difference. From the given choices, the answer is: $\text{ThedebtofPennsylvaniaundergraduatesdiffersfromthatofConnecticutundergraduates.}$

$\text{Exercise 8-69 Static}$

Given Data:   $\hat{p}$

​=70%(sample proportion)

$\text{n}=400\text{(sample size; number of people in the survey)}$

 

$\text{Confidence Level}=90\%=0.90$

 

 

Solution:

 

$\text{a-1.}\text{Identifytherelevantparameterofinterestforthiscategoricalvariable.}$

From the given choices, the answer is: $\text{Theparameterofinterestistheproportionofallpeoplebetweentheagesof50-64.}$

 

 

 

$\text{a-2.}\text{Computethepointestimateaswellasthemarginoferrorwith90\%confidence.}$

The sample proportion serves as the point estimate for the true population proportion. So, in this case, the point estimate is the sample proportion of people in the 50 to 64 age bracket who have tried some type of alternative therapy. $\text{Point Estimate}=\hat{p}$

​=70%=0.70

 

 

Now, to determine the margin of error with 90% confidence, we can do as follows:

Step 1: Calculate for the significance level $\alpha$.   By definition, the significance level $(\alpha )$ can be calculated by subtracting 1 with the confidence level of 90%. Hence,   $\alpha =1-0.90=0.10\text{(significance level)}$     Step 2: Calculate for the z-value and draw a graph.   The value for the z-value can be determined by knowing the area represented by the significance level. In this case, the area represented by the significance level are the areas at the left end and the right end of the distribution curve. Knowing this, we divide the significance level by two to get $\left(\frac{\alpha }{2}=0.050\right)$ the area in the left and in the right.

 

 

 

Using the z distribution table we can determine the corresponding z-value for an area of $\frac{\alpha }{2}=0.050$. Doing so, we get the absolute value of the z-value as:

 

$z=|z_{\frac{\alpha }{2}}|=|\pm 1.645|=1.645$

 

 

Step 3: Calculate the margin of error.

 

The formula for the margin of error of a proportion is expressed as:

 

$E=z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}}{n}}$

​(1−p​)​

​

 

Substituting the values, we get:

 

$E=1.645\sqrt{\frac{0.70(1-0.70)}{40}}$

​=0.1191915737   Rounding to three decimal places, we get:   $E\approx 0.119$   Therefore, margin of error with 90% confidence is: $E=0.119$     Therefore, the point estimate as well as the margin of error with 90% confidence is:     Point estimate $\underline{0.700}$ Margin of error $\underline{0.119}$

 

 

 

$\text{b.}\text{Youdecidetoredotheanalysiswiththemarginoferrorreducedto2\%.Howlargeasampledoyouneedtodraw?}$

Additional given data:   $E=2\%\text{(margin of error)}$     In part a-2, we have determined that for a 90% confidence, the critical value z is:   $z=|z_{\frac{\alpha }{2}}|=|\pm 1.96|=1.96$ Now, the sample size can be determined using the formula for the margin of error of a proportion. The formula for the margin of error of a proportion is expressed as:   $E=z\sqrt{\frac{\hat{p}}{n}}$

​(1−p​)​

​

 

Substituting the values, we get:

 

   $E=z\sqrt{\frac{\hat{p}}{n}}$

​(1−p​)​

​

 

 $2\%=1.645\sqrt{\frac{0.7(1-0.7)}{n}}$

​

 

Solving for the value of the sample size n, we get:

 

  $0.02=1.645\sqrt{\frac{0.7(0.3)}{n}}$

​

 

         $\frac{0.02}{1.645}=\sqrt{\frac{0.21}{n}}$

​

 

${\left(\frac{0.02}{1.645}\right)}^2=\frac{0.21}{n}$

 

  $n=\frac{0.21}{{\left(\frac{0.02}{1.645}\right)}^2}$

 

  $n=1420.663125$

  $n\approx 1421$

 

Therefore, the sample size is:

Sample size

$\underline{1421}$

 

How to get free answer questions for all subjects from Homeworklib?

Homeworklib will give you answers to all kinds of subject questions from our social networks like Telegram, Email address, Whatsapp group, Facebook, and Instagram.
You can also send us your questions through our google form which is given below link.

Submit More Question Link