Lesson 19: Review for Exam 4

Review:

• Significance Testing
• Exam 4 on Wed 11 Nov 2026
• Review session with Caleb

Exam 2 Topics:

• Covers Significance Testing (80%) and Estimation (20%)
  • Lessons 11-13, 15-18
• Significance Testing
  • Step 1. Set up Ho and Ha
  • Step 2. Calculate Z statistic
  • Step 3. Calculate p-value
  • Step 4. Compare p-value to alpha (0.05) and decide to Reject Ho or Fail to Reject Ho
  • Types
    • Left tail (negative Z, simple table lookup)
    • Right tail (positive Z, 1 – table lookup)
    • Two-tail (positive or negative Z, multiply p-value by 2)
  • Z-table only
    • mean testing, large sample
    • proportion testing, large sample
    • no t-table
• Estimation (20%)
  • Confidence Intervals = Point estimate +/- Margin of error
  • Mean estimation with large sample (n>30)
    • Margin of Error = Z*s/
  • Proportion estimate with large sample (n>30)
  • t-distribution for small samples
  • Z-table only

Activity:

• Practice Exam

Estimation Confidence Interval Reminder:

• If n is sufficiently large (using n≥30 as our rule of thumb), we can
  • Assume the sampling distribution of x̅ is approximately Normal
  • Estimate the population mean, μ = x̅ (this is our point estimate)
  • Calculate margin of error, m = z*SE
    • where z = 1.96 for 95% confidence
    • where standard error = SE = s/(√n)
  • So, estimation of a population mean with confidence interval = x̅ ± z*SE

Lesson 9: Intro to Monte Carlo Simulation

Review:

• Multiple Regression exam results

• 

Presentation:

• Probability
  • Topic worthy of its own course
• Simulation
  • Random phenomenon
    • Individual outcomes are uncertain
    • Regular distribution of outcomes in large number of trials
  • Random number generation and stochastic processes
  • Discrete and Continuous (we’ll start with discrete)
• Coin Toss Simulation
  • Demonstrate Coin Toss Simulation in Sheets
• Dice Roll Simulation
  • Demonstrate Dice Roll Simulation in Sheets in Sheets
  • Lucky 7 game
    • Roll a 7, win bet
    • Roll anything else (other than a 7), lose bet

Assignment:

• Replicate the Coin Toss simulation
  • 100 coin tosses (instead of 32)
  • reproduce the graph (showing 100 tosses)
• Create a “Roll 2 Dice” simulation
  • 100 rolls (instead of 32)
  • Roll 2 Dice instead of only 1
  • Be sure the probability distribution is the same as actually rolling 2 Dice.
• Create a “Lucky 7” simulation
  • Roll 2 Dice
  • If you roll a 7, you win
  • If you don’t roll a 7, you lose
  • Track the number of wins and losses
• Create a “Simple Craps” simulation
  • Roll 2 Dice
  • If you roll a 7 or 11, you win $10
  • If you roll a 2, 3 or 12, you lose $5
  • If you roll anything else (i.e., 4, 5, 6, 8, 9, 10), it’s a “push” or a tie and you don’t lose or win $$.
  • Set up a scoring system to keep track of how much $$ you win or lose.

Lesson 18: Analysis of Colorado Election Returns

Review:

• Exam 4 on Wed 11 Nov

Presentation:

• Download Colorado Election Returns
• Comparison
  • Actual Voting
  • Polling results from University of Colorado
• Were the polls accurate?
• How were the polls biased?

Lesson 17: Significance Testing with Colorado Polling Data

Review:

• Significance Testing, one-tail vs two-tail
• Exam 4 on Wed 11, Nov

Presentation:

• Statewide elections and ballot measures
  • Public Opinion Polling Basics (Pew Research)
• Cast a “ballot” for these 2 statewide races:
  • Governor: Phil Weiser (D) vs Victor Marx (R)
  • Senate: Mark Baisley (R) vs John Hickenlooper (D)
• Proportions are different
  • p-hat approach for significance testing

Practice Problem – Colorado Proposition 134: The University of Colorado Boulder and YouGov surveyed 800 Colorado eligible voters between October 9 and October 20, 2026, about Proposition 134, concerning eligibility policies for school and college sports.

The poll reported: 55% support, 34% oppose, 10% undecided. Is there sufficient statistical evidence that more than half of Colorado eligible voters support Proposition 134?

Step 1. Hypotheses

• Ho : p = 0.50
• Ha : p > 0.50

Step 2. Sample proportion

• p-hat = 0.55

Step 3. Test statistic

Step 4. P-value

• P(Z > 2.83) = 0.0023

Step 5. Conclusion

• Because 0.0023 < 0.05 , we Reject Ho, the null hypothesis.
• If the participants surveyed are an unbiased representative of Colorado voters, Proposition 134 is likely to pass.

Activity:

Conduct Hypothesis Tests on the US Senate and Governor polling to determine if classroom polls are significantly different from the statewide polling conducted by the University of Colorado political climate research.

Lesson 16: One-Tail vs Two-Tail Significance Testing

Review:

• Significance Testing

Presentation:

• Hypothesis testing: 1-tail vs 2-tail
  • 1-tail
    • left tail
    • right tail
  • 2-tail
• Compare >, < and ≠
  • What is the p-value for Z = -1.73
    • Ha: μ < μo [left tail, use P(Z)]
    • Ha: μ > μo [right tail, use 1 – P(Z)]
    • Ha: μ ≠ μo [two tail, multiply by 2]

Activity:

Determine whether it’s a 1-sided or 2-sided hypothesis test and solve. Report p-values and determine if you can reject or must fail to reject the null hypothesis.

1. A test of the null hypothesis Ho yields test statistic, Z = 1.34.
   1. What is the P-value if the alternative is Ha: μ > μo
   2. What is the P-value if the alternative is Ha: μ < μo
   3. What is the P-value if the alternative is Ha: μ ≠ μo
2. The college bookstore tells students the average textbook price is $52 with a standard deviation of $4.50. A group of students thinks the average price is higher. In order to test the bookstore’s claim, the students select a random sample of size n=100 and find a sample mean price of $52.80. Perform a hypothesis test to determine if the price difference is significantly higher for α = 0.05.
3. A certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million) and standard deviation σ = 8 ppm. A group of factory representatives whose companies discharge liquids into the river is now claiming they have lowered the average with improved filtration devices. A group of environmentalists will test to see if this is true. Assume their sample of size n=50 gives a mean of 32.5 ppm. Perform a hypothesis test to determine if the pollution levels are significantly lower for α = 0.05.
4. A manufacturing process produces ball bearings with diameters that have a normal distribution with mean, μ = 0.50 centimeters and known standard deviation, σ = .04 centimeters. Ball bearings with diameters that are too small or too large are problematic.
   1. Assume a random sample n=25 with a sample mean diameter = 0.51 cm. Perform a hypothesis test at α = 0.05.
   2. Assume a random sample n=25 with a sample mean diameter = 0.48 cm. Perform a hypothesis test at α = 0.05.
5. A battery brand claims its mean lifetime = 70 hours with a standard deviation of 12 hours. A random sample of n = 64 batteries lasts an average of 73.1 hours. Conduct a Hypothesis Test to determine if batteries last longer than 70 hrs.
6. A machine fills boxes with cereal averaging 500 g with standard deviation of 8 g. A sample of n = 36 boxes has a mean of 497.8 g. Conduct a Hypothesis Test to determine whether the machine is under-filling boxes.
7. A coffee distributor claims mean weight = 454 grams with a standard deviation of 5 grams. A sample of n = 25 bags averages 456.5 grams. Conduct a Hypothesis Test to determine whether the average is different from 454 g.
8. A manufacturer claims its hybrid car averages 50 mpg with standard deviation of 4 mpg. A sample of n = 25 vehicles yields an average of 51.6 mpg. Conduct a Hypothesis Test to determine if the mean mpg is greater than 50.

Lesson 15: Introduction to Significance Testing

Review:

• Exam 3

• 

Presentation:

• Significance Testing
  • aka Hypothesis Testing
  • Purpose: to evaluate data for evidence of significant agreement or disagreement
• Testing Process
  • Step 1. Setup the null hypothesis (Ho) and alternate hypothesis (Ha)
  • Step 2. Calculate the appropriate test statistic using the t-distribution
  • Step 3. Find the P-value (probability of obtaining result by chance)
  • Step 4. Compare P-value to α = 0.05 ; if P-value < 0.05, “Reject Ho” else “Fail to Reject Ho”
• Significance testing is like a courtroom trial.
  • Null Hypothesis (Ho): The defendant is innocent until proven guilty.
    → We start by assuming there’s no effect, no difference, no relationship.
  • Alternative Hypothesis (Ha): The defendant is guilty.
    → We’re testing whether there’s enough evidence to reject the assumption of innocence.
  • Evidence (Data): The prosecution presents data—witness testimony, forensics, etc.
    → In statistics, we gather a sample and compute a test statistic (like a z-score).
  • Judge or Jury (Decision Rule): They set a high standard for conviction (e.g., beyond a reasonable doubt).
    → In statistics, this is our significance level (α), often 0.05, meaning we accept at most a 5% chance of wrongly rejecting H₀.
  • Verdict (Statistical Decision):
    • If the evidence is strong enough (p < α), we reject H₀ — “guilty” verdict.
    • If not (p ≥ α), we fail to reject H₀ — “not guilty” (but not necessarily innocent).
  • Type I Error: Convicting an innocent person.
    → Rejecting a true null hypothesis (false positive).
  • Type II Error: Letting a guilty person go free.
    → Failing to reject a false null hypothesis (false negative).

• Example A: Is the soda company short-changing customers?
  • A soda company claims that its cans contain 12.0 oz on average.
  • You take a sample of n=30 cans and find:
  • sample mean = 11.8 oz
  • sample standard deviation = 0.3 oz
• Step 1. Set up hypotheses
  • Ho: μ = 12.0 (the cans are filled correctly)
  • Ha: μ < 12.0 (the cans are underfilled)
• Step 2. Compute the Test Statistic
  • Z-test = (x̅ – μ)/(σ/√n)
  • (11.8 – 12.0)/(0.3/√30) = -0.2/0.05477 = -3.65
• Step 3. Find the p-Value
  • p=P(Z<−3.65)≈0.00013
• Step 4: Interpret Results
  • 0.00013 ≅ 0.013% probability of getting this result by chance
  • 0.00013 < 0.05
  • Reject Ho
  • The soda company seems to be ripping us off!

• Video: Tests of Significance
• Example B: Do Math SAT scores improve significantly with coaching?
  • National Math SAT scores are normally distributed with mean = 505 and stdev = 62.
  • Sample of 1,000 students who received coaching.
  • Sample mean score was 509.
  • Are these results significantly better than the national average?
• Step 1: Setup the hypothesis test
  • μ = 505, σ = 62, x̅ = 509, n = 1,000
    • Ho: μ = 505
      Ha: μ > 505
• Step 2: Calculate the test statistic
  • Z-test = (x̅ – μ)/(σ/√n)
  • (509 – 505)/(62/√1000) = 4/1.96 = 2.04
• Step 3: Find P-value
  • P = 1 – P(Z<2.04) = 1 – 0.9793 = 0.0207
• Step 4: Interpret results
  • 0.0207 ≅ 2.07% probability of getting this result by chance
  • 0.0207 < 0.05
  • Reject Ho
  • Coaching seems to improve scores significantly

Activity:

• Problem 1.1. More than 200,000 people worldwide take the GMAT examination each year as they apply for MBA programs. Their scores vary Normally with mean about μ = 525 and standard deviation about σ = 100. One hundred students, n = 100, go through a rigorous training program designed to raise their GMAT scores. The students who go through the program have an average score of x̅ = 541.4. Is there evidence to suggest the training program significantly improves GMAT scores?
• Problem 1.2. A newly installed rooftop solar system has been producing energy for n = 100 days. Average energy production is 41.8 kWh per day with a standard deviation of 13.9 kWh. The solar panel manufacturer claims the panels typically produce 40 kWh per day. Is the newly installed system producing significantly more energy than estimated by the manufacturer?

* Most example and activity problems presented above are derived from Moore, D.S., McCabe, G.P., and Craig, B.A., 2009. Introduction to the Practice of Statistics, 6th Edition. New York: W.H. Freeman and Company. 

Lesson 8: Categorical Data and Indicator Variables

Review:

• Multicollinearity
• Interaction Variables
• Exam 2 on Wed 28 Oct – focus on Multiple Regression, more to follow

Presentation:

• Pivot tables to aggregate by category
• Indicator Variables (aka “dummy” variables)
  • incorporate categorical variables
  • binary – value is either 1 or 0
  • if 0, then the coefficient has no effect on y-hat
  • if 1, then the coefficient is added to the estimate
• Demonstration with Sample Real Estate data
  • Use a Pivot table to summarize price by neighborhood
  • create a dummy variable for a neighborhood
  • create a dummy variable for multiple neighborhoods

Activity:

• Use the Pueblo Real Estate model (from last lesson)
  • Incorporate one or more Indicator variables
  • Use an indicator variable to represent high cost or low cost neighborhoods
  • Can you improve R^2 and/or F with an indicator variable?

Lesson 14: Review for Exam 3

Review:

• Proportions
• Exam 3 on Wed

Presentation:

• Topics
  • Census
  • Sampling
  • Sampling Distributions
  • Central Limit Theorem
  • Estimation
    • Confidence Intervals = Point estimate +/- Margin of error
    • Z-table
    • t-distribution for small samples
    • Inference for Proportions
• Problem Types
  • Mean estimation with large sample (n>30)
  • Mean estimation with small sample (n<30)
  • Proportion estimation

Activity:

• Practice Exam

Lesson 7: Interaction Variables in Multiple Regression

Review:

• Regression Output Interpretation
• Variable Selection

Presentation:

• MLB Baseball model
  • 2013 MLB Team statistics 
  • Walk through model development
• Applied Regression Analysis
  • Demonstrate with Sample Real Estate data
  • Problem: Multicollinearity
  • Solution: Interaction Variables
    • combine separate variables mathematically
    • add, multiply, etc
    • other ways to avoid problems with multicollinearity
  • Other solution: eliminate variables
  • Estimate selling price for a 1500 sqft, 3 bd, 2 ba home

Activity:

• Pueblo Real Estate with Interaction
  • Use this larger Pueblo real estate data set
  • Build a regression model to estimate selling price
    • You must use 1 interaction variable
    • Experiment with multiple combinations of variables to find the best model, maximizing R^2 and the F Statistic
  • Use your regression model to estimate selling prices for these 4 hypothetical properties
    • 3 BR, 2 BA, 1600 SqFt
    • 4 BR, 3 BA, 2400 SqFt
    • 3 BR, 1 BA, 1200 SqFt
    • 2 BR, 1 BA, 800 SqFt

Lesson 13: Inference for Proportions

Review:

• Inference for a population mean with Z-distribution
• Inference for small sample size with t-statistic
• Confidence Intervals

Presentation:

• How to Estimate a Population Proportion
  • Conduct a Simple Random Sample (SRS) of size n
  • Record the count, x, of some attribute attributed to a portion of the population (e.g., number of voters favoring a candidate)
  • Calculate the sample proportion, p-hat = x/n
  • If n is sufficiently large (≥30), we can assume p-hat is Normally distributed
  • Estimate of the population proportion mean, μ = p-hat
  • Estimate of the population proportion std dev, σ = √(p*(1-p)/n)
  • Estimate margin of error, m = z*σ  (use z = 1.96 for 95% confidence)
  • Estimate with 95% confidence interval is p-hat ± m
  • Same procedure for a small sample size (<30) using the t distribution and substituting t* for z*

• Binge Drinking Survey
  • n = 13,819
  • x = 3,140
  • p-hat = 3140/13819 = 0.227
  • standard deviation = √(p-hat*(1-p-hat)/n) = 0.00356
  • p-hat ± z*√(p-hat*(1-p-hat)/n) = 0.227 ± 1.96*(0.00356) = 0.227 ± 0.007
  • [0.220,0.234]
• Video – Against All Odds Unit 28
• Consumer Survey – Online vs in-store shopping.
  • A marketing firm surveyed 200 consumers, and 126 said they prefer shopping online to in-store.
  • Construct a 95% confidence interval for the true proportion of consumers who prefer online shopping.
• Click here for Handwritten Notes

Activity:

1. A random sample of 2,454 12th-grade students were asked the following question: Taking all things together, how would you say things are these days – would you say you’re happy or not too happy? Of the responses, 2,098 students selected happy. Determine the sample proportion of students who responded they were happy and calculate a 95% confidence interval for the population proportion of 12th-grade students who are happy.
2. A phone survey contacted 1,910 households in which a computer was owned and respondents were asked if they could access the Internet from their home. A total of 1,816 of the households responded yes. Calculate a 95% confidence interval to estimate the proportion of American households with internet access.
3. Currently, mothers in North America are advised to put babies to sleep on their backs. This recommendation has reduced the number of cases of sudden infant death syndrome (SIDS). However, it is a likely cause of another problem, i.e., flat spots on babies’ heads. A study of 440 babies aged 7 – 12 weeks found that 46.6% had flat spots on their heads. Calculate a 95% confidence interval for the proportion of babies in this age group that have flat spots.

Assignment:

1. A marketing firm surveyed 200 consumers, and 126 said they prefer shopping online to in-store. Construct a 95% confidence interval for the true proportion of consumers who prefer online shopping.
2. A start-up wants to estimate how many potential customers would buy its new smartwatch. Out of a random sample of 300 adults, 174 said they would consider buying one. Construct a 95% confidence interval for the true proportion of adults likely to buy the smartwatch.
3. A factory produces smartphone screens. Out of 600 screens sampled, 18 were defective. Find a 95% confidence interval for the true defect rate.
4. In a statewide poll of 1,200 registered voters, 516 said they support Candidate A. Compute a 95% confidence interval for the true proportion supporting Candidate A.
5. Out of 250 students enrolled in an online statistics course, 205 finished all assignments. Estimate the true course-completion rate with a 95% confidence interval.
6. A retail store examines 400 purchases and finds 28 items were later returned. Construct a 95% confidence interval for the true return proportion.
7. A data-analytics firm records that 310 of 500 Instagram posts received at least one comment. Compute a 95% confidence interval for the population proportion of posts with engagement.
8. A company offers a new health-benefit plan. Out of 180 employees, 132 enroll. Calculate a 95% confidence interval for the true proportion who will enroll company-wide. Can the company claim that “over 70% of employees will sign up.”