Lesson 20: Pearson Correlation Coefficient and R-Squared

Review:

• Linear Regression
• Exam 2 corrections
  • Only for test takers who took Exam 2 on the scheduled date.
  • You must complete this entire exam.
  • Your exam score will increase up to 10 points or up to 50% of points deducted, whichever is less. For example, if you received an 88 on the exam you can only earn up to 6 points.
  • Your actual points awarded will be based on your exam score. For example, if you get an 80% on the new exam, you will earn 8/10 points (or .8 x 50% of points deducted).

Presentation:

• Pearson Correlation Coefficient
  • r = SSxy/√(SSxx*SSyy)
  • -1 < r < 1
• R-Squared
  • Calculate r and then square the result, i.e., =r^2
  • 0 < r^2 < 1
• Let’s solve this problem from last time, including calculating r and r^2, from beginning to end.

Problem 2. A local brewery tracks weekly social media ad spending and kegs sold. Calculate Sum of Squares and the equation of the regression line.

Week	Ad Spend (x, $100’s)	Kegs Sold (y)
1	2	11
2	3	15
3	4	17
4	5	20
5	7	26
6	9	30

Let’s also estimate kegs sold when the ad spend is $650 (x=6.5)

Video: Fitting Lines to Data

Activity:

Now, here’s some real data, gathered from USDA National Water and Climate Center and the USGS Water Data Center. The snowpack data (snow-water equivalent inches) is for the Freemont Pass location. The streamflow data (ft^3/SEC) is for the Arkansas River station in Pueblo.

Water Year	Snowpack (x)	Streamflow (y)
2018	20.4	577.5
2019	28.1	1599.3
2020	20.2	901.0
2021	15.5	659.4
2022	17.6	730.4
2023	17.1	1016.0
2024	22.2	1455.7

Calculate Sum of Squares, the Linear Equation, the Pearson Correlation Coefficient and R^2.

Using the regression equation, estimate Streamflow for 2025 assuming Snowpack was 18.2.

Assignment:

• Complete the corrections exam or enjoy a pleasant weekend.

Lesson 19: Linear Regression

Review:

• Exam 2 review
• Scatterplots
• Sum of Squares

Presentation:

• Linear regression (simple, bivariate)
  • Calculate equation of the regression line
    • y-hat = b1*x + bo
      • b1 = “slope” of the line
      • b0 = “y-intercept”
    • Calculate Sum of Squares: SSxx, SSyy, SSxy
    • b1 = SSxy/SSxx
    • b0 = (∑y/n) – b1*(∑x/n)
• Example 1: Beer Sales by Temp
  • Use temperature to predict kegs sold
  • Historical Data (Temp degrees F, # of kegs sold)
    • Day 1: 60°, 12
    • Day 2: 70°, 14
    • Day 3: 80°, 16
    • Day 4: 90°, 18
    • Day 5: 100°, 20
  • Scatter Plot
  • Table setup
• Sum of Squares calculations
  • SSxx = ∑x² – (∑x)²/n
  • SSyy = ∑y² – (∑y)²/n
  • SSxy = ∑xy – (∑x*∑y)/n
• Calculate Slope of regression line
  • b1 = SSxy/SSxx
• Calculate the y-intercept
  • b0 = (∑y/n) – b1*(∑x/n)
• Video: Fitting Lines to Data

Activity:

• Example 2: Beer Sales by Price
  • Historical Data (Price $ # of kegs sold)
    • Day 1: $90, 22
    • Day 2: $100, 20
    • Day 3: $110, 18
    • Day 4: $120, 16 
    • Day 5: $130, 14
  • Calculate Sum of Squares
  • Calculate the slope and y-intercept
  • Write the linear equation

Assignment:

First, go back to the problems assigned last week. Use your Sum of Squares calculations to find the equation of the regression line for each problem.

1. Advertising vs. Sales (n = 8)
   • SSxx = 55.5
   • SSyy = 1,385.875
   • SSxy = 275.75
2. Training Hours vs. Productivity (n = 8)
   • SSxx = 42
   • SSyy = 327.5
   • SSxy = 117
3. Price vs. Units Sold (n = 8)
   • SSxx = 1050
   • SSyy = 29,400
   • SSxy = –5,500

Problem 1. A small coffee shop tracks daily high temperature (°F) and the number of iced coffees sold. Calculate Sum of Squares and the equation of the regression line.

Day	Temp (x)	Iced Coffees Sold (y)
1	60	42
2	65	48
3	70	52
4	75	61
5	80	65

Problem 2. A local brewery tracks weekly social media ad spending and kegs sold. Calculate Sum of Squares and the equation of the regression line.

Week	Ad Spend (x, $100’s)	Kegs Sold (y)
1	2	11
2	3	15
3	4	17
4	5	20
5	7	26
6	9	30

Problem 3. A realtor wants to estimate the relationship between square footage (x) and listing price (y in $1000s). Calculate Sum of Squares and the equation of the regression line.

House	Sq Ft (x)	Listing Price (y)
A	1,600	315
B	1,850	344
C	2,000	365
D	2,300	399
E	2,500	425

Lesson 18: Least Squares Regression

Review:

• Scatterplots

Presentation:

• Least Squares Regression
  • Find a line of best fit for a set of points
  • Least Squares Regression Proof: MethodLeastSquares
  • Example 1: Beer Sales by Temp
    • Use temperature to predict kegs sold
    • Historical Data (Temp degrees F, # of kegs sold)
      • Day 1: 60°, 12
      • Day 2: 70°, 14
      • Day 3: 80°, 16
      • Day 4: 90°, 18 
      • Day 5: 100°, 20
    • Scatter Plot
    • Table setup
  • Sum of Squares calculations
    • SSxx = ∑x² – (∑x)²/n
    • SSyy = ∑y² – (∑y)²/n
    • SSxy = ∑xy – (∑x*∑y)/n
  • Example 2: Beer Sales by Price
    • Experiment with price to predict demand
    • Historical Data (Price $ # of kegs sold)
      • Day 1: $90, 22
      • Day 2: $100, 20
      • Day 3: $110, 18
      • Day 4: $120, 16 
      • Day 5: $130, 14
    • Students complete calculations
  • We will use these Sum of Squares calculations (next time) to determine
    • linear equation
    • Pearson correlation coefficient
    • R-squared

Activity:

Problem 1 – Advertising and Sales (n = 8)

A retail chain studied the link between weekly advertising spend and sales (in $1,000s).

Week	Advertising (x)	Sales (y)
1	6	64
2	8	77
3	10	85
4	4	55
5	7	72
6	9	82
7	3	50
8	11	88

Task:
Calculate SSxx, SSyy and SSxy

Problem 2 – Training Hours and Productivity (n = 8)

An HR analyst examined whether more employee training hours lead to higher productivity (daily packages processed per worker).

Dept	Training Hours (x)	Productivity (y)
A	5	46
B	8	54
C	4	42
D	6	49
E	10	59
F	3	40
G	9	57
H	7	51

Task:
Calculate SSxx, SSyy and SSxy

Problem 3 – Price and Units Sold (n = 8)

A manufacturer tested different price points to estimate demand.

Trial	Price (x)	Units Sold (y)
1	40	210
2	35	250
3	45	190
4	50	160
5	30	280
6	55	150
7	25	310
8	60	130

Task:
Calculate SSxx, SSyy and SSxy

Lesson 17: Scatterplots

Review:

• Exam 2 Results

• 

Presentation:

• Scatterplots
  • Visualize relationship between 2 variables
    • independent (x) and dependent (y)
    • horizontal axis (x) and vertical axis (y)
  • Visual Correlation
    • Direction: Positive or Negative
    • Strength: Strong, Moderate, Weak
  • Examples
    • Ice Cream Sales
    • Colorado TCAP Scores and % Free-Reduced Lunch
    • Gender Wage Gap
  • Alternative Uses
    • Predicted vs Actual
    • Hot/Crazy
  • Unit 10 Scatterplots Video

Activity:

Complete the problems below twice, once by hand and once in Sheets or Excel.

Problem 1. A study was conducted on mercury (Hg) concentrations in fish taken from Lake Natoma in California. The researchers were concerned that mercury concentration levels in sample fish tissue might differ depending on the lab testing the fish. Fish tissue samples from 10 largemouth bass were sent to two labs, Columbia Environmental Research Center (CERC) and University of California, Davis (UC Davis), for inter-laboratory comparison. Mercury concentration is measured in micrograms of mercury per gram of fish tissue (dry weight). Use the mercury concentration data below to create a Scatterplot and characterize the nature of any correlation.

Problem 2. Satellites are one of the many tools used for predicting flash floods, heavy rainfall, and large amounts of snow. Geostationary Operational Environmental Satellites (GOES) collect data on cloud top brightness temperatures (measured in degrees Kelvin (°K)). It turns out that colder cloud temperatures are associated with higher and thicker clouds, which could be associated with heavier precipitation. Data consisting of cloud top temperature measured by a GOES satellite and rainfall rate measured by ground radar appear in the table below. Because ground radar can be limited by location and obstructions, having an alternative for predicting the rainfall rates can be useful. Use the temperature and radar rain rate data below to create a Scatterplot and characterize the nature of any correlation.

Problems copied from the Against All Odds video series guide for Unit 10.

Lesson 16: Review for Exam 2

Review:

• Significance Testing
• Exam 2
  • Thu, Oct 23
  • Covers Lessons 8–15
• Review with Rylee tomorrow (Wed) 1:30-4p in HSB 2nd floor conference

Exam 2 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
• 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
• Problem Types
  • Estimation with confidence intervals
    • Mean estimation with large sample (n>30)
    • Mean estimation with small sample (n<30)
    • Proportion estimation
  • Significance Testing
    • One-tail
      • Left tail (negative Z, simple table lookup)
      • Right tail (positive Z, 1 – table lookup)
    • Two-tail
      • Left tail (negative Z, simple table lookup), multiply x2
      • Right tail (positive Z, 1 – table lookup), multiply x2

Activity:

• Practice Exam

Lesson 15: One-Tail vs Two-Tail Significance Testing

Review:

• Significance Testing
  • Solutions below
• Exam 2 on Thu, Oct 23 (review on Tue, Oct 21)

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.

Assignment:

1. A battery brand claims its mean lifetime is 70 hours with a standard deviation of 12 hours. A random sample of n = 64 batteries lasts an average of 73.1 hours.
Q: Test whether batteries last longer than 70 hours.

2. 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.
Q: Test whether the machine is under-filling boxes.

3. A coffee distributor claims mean weight 454 g with a standard deviation of 5 g. A sample of n = 25 bags averages 456.5 g.
Q: Test whether the average is different from 454 g.

4. 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.
Q: Test if the mean mpg is greater than 50.

Lesson 14: Significance Testing

Review:

• Small Sample Estimation
• Exam 2 on Thu, Oct 23

Presentation:

• Significance Testing
  • aka Hypothesis Testing
  • Purpose: to evaluate data for evidence of significant agreement or disagreement
• 4-step Testing Process
  • Step 1. Setup the null hypothesis (Ho) and alternate hypothesis (Ha)
  • Step 2. Calculate the appropriate test statistic
  • 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!

• Significance Testing video (unit 25)
• 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 13: The t-distribution for Small Sample Inference

Review:

• Estimation with Confidence Intervals
• Inference for proportions

Presentation:

• What if the best you can do is a sample size of n<30?
• t-distribution
  • For larger samples, n ≥ 30 use Z-table
  • For small samples, n < 30 use t-distribution
  • Only impacts calculation of the test statistic
    • margin of error for confidence intervals
    • p-value for significance testing (don’t worry, we cover  these topics later)
  • t distribution critical values
    • print the t distribution table for the next exam
      • use the 95% confidence or “0.025” column
    • select the row corresponding to df = n – 1
      • df = degrees of freedom
      • n = number of records in sample data
    • find the t* critical value
      • row = df = n – 1
      • column = .025 (bottom label) or 95% confidence level (top label)
      • e.g., if n = 13
        • df = n – 1 = 12
        • go to the row where df = 12 and move to the .025 column
        • you should find t* = 2.18
  • t-statistic formulas
    • for confidence intervals
      • x-bar ± t * s/√n
      • instead of the Z equivalent:  x-bar ± z * σ/√n
  • Video – Against All Odds Unit 26
  • Examples:
    • Identify appropriate t distribution critical values
      • n=10 with 95% confidence, t* = 2.26
      • n=20 with 95% confidence, t* = 2.09
    • A sample of size n = 25 produced the sample mean, x-bar = 36.0 and standard deviation, s = 9.0. Construct a 95% confidence interval to estimate the population mean.
      • n = 25, x-bar = 36.0, s = 9.0
      • df = 24, t* = 2.064
      • m = 2.064*(9/(√25)) = 3.7152
      • Estimate for μ = 36.0 ± 3.7152
      • 95% confidence interval: [32.2848, 39.7152]
        • previous lesson interval with Z=1.96 and n=400: [35.118, 36.882]
        • sample size can make a big difference

Activity:

• Problem 1. Use the t distribution table, assuming 95% confidence, to find the value of t* for each of the following sample sizes.
  • n = 12
  • n = 18
  • n = 24
  • n = 28
• Problem 2. The weights (in lbs) from a random sample of n=16 four-year-old children who took part in a study on childhood obesity are provided below. Estimate the mean weight of four-year-old children using a 95% confidence interval. The sample standard deviation equals 4.2.
  • {37.1, 26.7, 36.1, 36.2, 40.3, 43.9, 36.2, 40.7, 42.5, 34.8, 37.9, 34.5, 31.1, 36.4, 35.7, 33.4}
• Problem 3. After purchasing a new fuel-efficient vehicle, the owner calculates miles per gallon (mpg) the first n=12 times he fills the fuel tank. His calculations are provided below. Estimate vehicle fuel efficiency using a 95% confidence interval. You will need to calculate standard deviation.
  • {42, 43, 37, 36, 34, 45, 48, 43, 38, 42, 43, 46}

These examples and problems are from the Against All Odds video series and Introduction to the Practice of Statistics (6th Ed.) by Moore et al.

Lesson 12: Inference for Proportions

Review:

• Inference for a population mean
• Confidence Intervals
• Lesson 11 problem solutions

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.

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.”

Lesson 11: Estimation of Population Parameters with Confidence Intervals

Review:

• Sampling Distributions

Presentation:

• How to Estimate a Population Mean (μ)
  • Conduct a Simple Random Sample (SRS) of size n
  • Calculate the sample mean, x-bar = (∑x)/n
  • Calculate the sample standard deviation, s = √(∑(x – x-bar)²/(n-1)
  • If n is sufficiently large (≥30), we can
    • Assume x-bar is Normally distributed
    • Estimate the population mean, μ = x-bar (point estimate)
    • Calculate margin of error, m = z*σ
      • where z = 1.96 for 95% confidence
      • where σ = s/(√n)
  • Estimation with confidence interval for μ = x-bar ± m
• Watch Against All Odds Video Unit 24, Confidence Intervals
• Calculation Demonstration
  • A sample of size n = 400 produced the sample mean, x-bar = 36.0 and standard deviation, s = 9.0. Construct a 95% confidence interval to estimate the population mean.
    • n = 400
    • x-bar = 36.0
    • s = 9.0
    • m = 1.96*(9/(√400)) = 0.882
    • Estimate for μ = 36.0 ± 0.882
    • 95% confidence interval: [35.118, 36.882]

Activity:

• Problem 1. A sample of size n = 100 produced the sample mean, x-bar = 16.0 and standard deviation, s = 3.0. Construct a 95% confidence interval to estimate the population mean.
• Problem 2. An operation manager at a large plant observed 120 workers assembling an electronic component. The average time needed for assembly was 16.2 minutes with a standard deviation of 3.6 minutes. Construct a 95% confidence interval to estimate the mean assembly time.
• Problem 3. A computer technician installs new hard drives on 64 different computers. The average installation time is 42 minutes with a standard deviation of 5 minutes. Construct a 95% confidence interval for the mean installation time.
• Problem 4. A research firm conducted a survey of regular smokers to estimate the average amount spent per week on cigarettes. A sample of 49 regular smokers revealed average spending on cigarettes to be $21.55 with a standard deviation of $5.21. Construct a 95% confidence interval to estimate mean weekly cigarette spending.
• Problem 5. Tonight on Thursday Night Football the Los Angeles Rams will host the San Francisco 49ers. The “over-under” is set at 45.5 points. According to pro-football-reference.com, in all NFL games (n=1,086 games) the total score is normally distributed with a mean total score of 52.9 and a standard deviation of 20.2. Download the data here. What’s the probability that the total score will be less than 45.5? Will this information help with the over-under bet? What other data would be helpful?

Assignment:

• Repeat Problems 1-5 (above) in Google Sheets or Microsoft Excel