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QUAN 2010 Quiz Examples and Exam Information: Fall 2025

Lauren M. Nelsen

Section Exam 2 Information

Subsection Exam 2 Information

Exam 2 will be available on Canvas to take during class on the day listed in the syllabus. (You must be in person in the classroom to take the exam.)
  • The exam will cover Chapters 7(Part 2), 8, 9, and 13 from the textbook, which are Chapters 8-11 in the course notes: Link to course notes
     8 
    laurennelsen.github.io/QUAN2010/QUAN_2010_Notes.html
  • During the exam you may use your notes, the online course notes, the quiz/exam information site, a basic calculator (no phone) and a blank Excel file. Some problems will require you to upload an Excel file with all your work/calculations in order to receive credit. During quizzes you may NOT search the internet, use generative AI or use other resources that are not allowed.
  • You are NOT allowed to use other people’s work or search internet sites while taking the test. If you share you answers and others use them, both you and the students you shared with may receive a zero for the exam and may fail the course.
  • Even though you are allowed to use your notes and course resources during the exam, I strongly recommend making a “cheat sheet” to use while you take the exam! It can help you prepare for the exam and be ready for the types of questions that will be on the exam. (If you don’t prepare and practice the example problems, you might find yourself spending a long time searching and trying to figure out how to do problems while you’re taking the exam.)
  • Unless otherwise indicated, you must show steps/include all of your Excel calculations and formulas that demonstrate the process you take to get to the final answer.
    For problems involving probability distributions (such as the binomial or normal distribution), you need to use the Excel files to find any relevant values and not use values from a table.
    For problems that require showing work, credit will be given for not only the final answer, but the work/justification/formulas that supports that answer.
  • If you have a question during the exam, ask me! It is much better to ask right away than to stay stuck for a long time.

Subsection Exam 2 Format

The exam may have some or all of the following problems/types of problems. (The exam may have all of these types of problems or a subset of these types of problems. Some of these problems might not have all the parts that were on corresponding quiz/homework problems.) Examples of these problems (except for the multiple choice problems) are shown below under Exercises .
Note: On the exam, the problems may not all be labeled with the corresponding quiz objective/homework problem.
For the multiple choice problems, I recommend looking back at conceptual homework assignments, reviewing concepts from the notes/the book and making sure you understand the definitions and topics we’ve talked about.

Subsection Suggestions for Study

Look over all of your class notes and make sure you understand everything we have talked about. Look back at these chapters/problems in the textbook. Review the MyLab homework and the quiz problems. (I do not guarantee that all exam questions will look exactly like one of those problems.)

Exercises Exercises

Note that the exam questions will not look exactly like the problems below. For example, the example problem might be a right-tailed test, and on the exam you could have a problem that is a left-tailed or a two-tailed test. Make sure you understand each step of the example problems so that you would be able to do a slightly different version of a similar problem.

1.

Quiz outcome \(\boxed{5a}\text{:}\)
A company wondered if people liked the user interface of their upgraded website. They survey a random selection of their customers.
Suppose that for \(n=45\) surveyed customers, 32 preferred the new user interface.
(a)
Find a \(95\%\) confidence interval for the proportion of all the company’s customers that prefer the new interface.
Answer.
\(\alpha=1-0.95=0.05\text{,}\) \(\alpha/2=0.025\)
\(\overline{p}=\frac{32}{45}\approx 0.7111\)
\(z_{\alpha/2}=NORM.S.INV(.025+.95)\approx 1.96\)
Convidence interval formula: \(\overline{p}\pm (z_{\alpha/2})\cdot\left(\sqrt{\frac{\overline{p}(1-\overline{p})}{n}}\right)\)
The confidence interval is approximately:
\begin{equation*} .7111\pm (1.96)\left(\sqrt{\frac{.7111(1-.7111)}{45}}\right)\rightarrow \boxed{ (0.5787, 0.8435) } \end{equation*}
(b)
Suppose the company claims that at least \(85\%\) of all the company’s customers prefer the new interface. Does the confidence interval you found support this claim? Explain why or why not.
Answer.
No, it does not. We are \(95\%\) confident that \(p\) is between \(57.87\%\) and \(84.35\%\text{,}\) and \(85\%\) is larger than \(84.35\%\text{.}\)

2.

Exercises 8.2.2 and 8.2.3 from the Course Notes: (https://laurennelsen.github.io/QUAN2010/sec-Central-Limit-Theorem.html)
(a)
For a normal population with a mean equal to 87 and a standard deviation equal to 16, determine the probability of observing a sample mean of 90 or less from a sample of size 15.
Answer.
\begin{equation*} P(\overline{X}\leq 90) = NORM.DIST(90,87,16/SQRT(15),1) \approx 0.7661 \end{equation*}
(b)
For a population that is left-skewed with a mean of 21 and a standard deviation equal to 15, determine the probability of observing a sample mean of 18 or more from a sample of size 33.
Answer.
\begin{equation*} P(\overline{X}\geq 18)=P\left(Z\geq \frac{18-21}{15/\sqrt{33}}\right)\approx 1-P(Z\lt -1.1489)=1-NORM.S.DIST(-1.1489,1)\approx 0.8747 \end{equation*}

3.

Quiz outcome \(\boxed{6a}\text{:}\)
A CEO is curious if the proportion of employees that are satisfied with their job is greater than \(30\%\text{.}\) They conduct a survey of employees. Suppose that for \(n=380\) surveyed employees, 125 were satisfied with their job. Use \(\alpha=0.05\) and test the hypothesis that the proportion of all employees that are satisfied with their job is greater than \(30\%\text{.}\)
(a)
State the null hypothesis.
Answer.
\(H_0:\; p\leq 0.30\)
(b)
State the alternative hypothesis.
Answer.
\(H_1:\; p\gt 0.30\) \(\leftarrow \) (right-tail test)
(c)
Find the appropriate test statistic.
Answer.
\begin{equation*} z=\frac{\overline{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{\frac{125}{380}-0.30}{\sqrt{\frac{0.30(1-0.30)}{380}}}\approx 1.2314 \end{equation*}
(d)
What is the p-value?
Answer.
\(\text{p-value}\approx 1-NORM.S.DIST(1.2314,1)\approx \boxed{ 0.1091 }\)
(e)
What do we conclude and why? Do we have enough evidence to conclude that the proportion of employees that are satisfied with their job is greater than \(30\%\text{?}\) Why or why not?
Answer.
Since the p-value is bigger than \(\alpha=0.05\text{,}\) we fail to reject the null hypothesis. So there is not enough evidence to conclude that the proportion of employees that are satisfied with their job is greater than \(30\%\text{.}\)

4.

Quiz outcome \(\boxed{6b}\text{:}\)
A farmer is curious about the effects of a new fertilizer on the height of her corn stalks. She measures the heights of a random selection of her stalks. Out of 55 measured corn stalks, the sample average height is \(\overline{x}=2.24\) meters, with sample standard deviation \(s=0.15\) meters. Use \(\alpha=0.05\) and test the hypothesis that the average height of her corn stalks is not equal to \(2.3\) meters.
(a)
State the null hypothesis.
Answer.
\(H_0:\; \mu=2.3\)
(b)
State the alternative hypothesis.
Answer.
\(H_1:\; \mu\neq 2.3\)
(c)
Find the appropriate test statistic.
Answer.
\begin{equation*} t_{\overline{x}}=\frac{\overline{x}-\mu}{(s/\sqrt{n})}=\frac{2.24-2.3}{(0.15/\sqrt{55})}\approx -2.9665 \end{equation*}
(d)
What is the p-value?
Answer.
(This is a two-tail test.)
\(\text{p-value}\approx T.DIST.2T(2.9665,55-1)\approx 0.0045\)
(or: \(2\cdot (T.DIST.(-2.9665,55-1,1))\))
(e)
What do we conclude and why? Do we have enough evidence to conclude that the average height of her corn stalks is not equal to 2.3 meters?
Answer.
Since the p-value is less than \(\alpha\text{,}\) we reject the null hypothesis. So we do have enough evidence to conclude that the average height is not equal to 2.3 meters.

5.

Week 5-B HW Question 6:
Consider the hypothesis statment given below:
\begin{equation*} H_0:\; \sigma^2\leq 7.0 \end{equation*}
\begin{equation*} H_1:\; \sigma^2\gt 7.0 \end{equation*}
Assume that \(s=3.2\text{,}\) \(n=40\text{,}\) and \(\alpha=0.05\text{.}\)
(a)
Calculate the appropriate test statistic.
Answer.
\begin{equation*} \chi^2=\frac{(n-1)\cdot s^2}{\sigma^2}=\frac{(40-1)\cdot (3.2)^2}{7.0}\approx 57.0514 \end{equation*}
(b)
Determine the appropriate critical value(s).
Answer.
(This is a right-tailed test.)
\(CHISQ.INV.RT(.05,40-1)\approx 54.5722\)
(c)
Should you reject the null hypothesis or fail to reject the null hypothesis?
Answer.
Since the test statistic is in the rejection region, we should reject the null hypothesis. (It might help to draw a picture to see this.)
(d)
Is there sufficient evidence to conclude that the population variance is greater than 7.0?
Answer.
Yes! Since we rejected the null hypothesis, there is enough evidence to conclude that \(\sigma^2\gt 7.0\text{.}\)

6.

Week 5-B HW Question 8:
Suppose a student has two possible routes for their commute to school. Because they never want to be late for class, they want to choose the route that provides a more consistent commute time. Being a hopeless statistician, the student meticulously drove the first route 20 times and the second route 25 times and calculated the standard deviations for each. The following table summarizes sample data collected from two different routes.
Route A Route B
sample standard deviation 6.5 minutes 10.4 minutes
sample size 25 20
(a)
Using \(\alpha=0.05\text{,}\) we want to determine if Route A provides a more consistent commute time than Route B. First, state the null alternative hypotheses. (Let \(\sigma_1^2\) represent the larger sample variance and \(\sigma_2^2\) represent the smaller sample variance.)
Answer.
(Since Route B has the bigger sample standard deviation, Route B is “Population 1” and Route A is “Population 2”.)
\(H_0:\; \sigma_1^2\leq \sigma_2^2\)
\(H_1:\; \sigma_1^2\gt \sigma_2^2\)
(b)
Calculate the appropriate test statistic.
Answer.
\begin{equation*} F=\frac{s_1^2}{s_2^2}=\frac{(10.4)^2}{(6.5)^2}=\boxed{2.56} \end{equation*}
(c)
Calculate the appropriate critical value (using \(\alpha=0.05\)).
Answer.
\begin{equation*} F.INV.RT(.05,20-1,25-1)\approx \boxed{2.0399} \end{equation*}
(d)
What can we conclude? Do we have enough evidence to conclude that Route A provides a more consistent commute time than Route B? Why or why not?
Answer.
Since the test statistic is in the rejection region, we reject the null hypothesis and have enough evidence to conclude that Route A provides a more consistent commute time than Route B.
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