Category: f19-busad265

Review for Exam 2

Review:

  • Quiz 5 and Extra Credit
  • Significance Testing
  • Exam 2 on Wed Oct 23

Exam 2 Topics:

  • Census
  • Sampling
  • Sampling Distributions
  • Central Limit Theorem
  • Estimation
    • Confidence Intervals = Point estimate +/- Margin of error
    • 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

Activity:

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Lesson 13: One-Tail vs Two-Tail Significance Testing

Review:

  • Quiz 5
  • Significance Testing
  • Exam 2 on Wed, Oct 23

Presentation:

  • Hypothesis testing: 1-tail vs 2-tail
  • 2-Tail Tests
    • Example 6.15 (p. 383-384):
      • Ho: μ = 168, Ha: μ ≠ 168, x-bar = 173.7, n=71, σ = 27, α = 0.05
    • Use Your Knowledge 6.43 (p. 385-386):
      • Ho: μ = 25, Ha: μ ≠ 25, x-bar = 27, n=25, σ = 5, α = 0.05
  • Compare >, < and ≠
    • Z = -1.73
    • What is the p-value for
      • Ha: μ > μo [use 1 – P(Z)]
      • Ha: μ < μo [use P(Z)]
      • Ha: μ ≠ μo [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: μ = μo 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 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 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.
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Lesson 12: Tests of Significance

Review:

  • Estimation with Confidence Intervals
  • Small Sample Estimation
  • Inference for Proportions
  • Quiz 4
  • Exam 2 on Wed, Oct 23

Presentation:

  • Significance Testing
    • aka Hypothesis Testing
    • Purpose: to evaluate data for evidence of significant agreement or disagreement
  • Significance testing is like paternity testing.
    • When you check father-child DNA for a match you can prove one person is or is not the father.
    • The same test does not prove another person is the father.
    • You’re evaluating only one possibility at a time.
  • Significance Testing video
  • 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. Interpret results, compare P-value to α = 0.05 ; if P-value < 0.05, “Reject Ho” else “Fail to Reject Ho”
  • Example A: Do Math SAT scores improve significantly with coaching?
    • National Math SAT scores are normally distributed with mean score = 505 and std. dev = 62
    • Sampled 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 testμ = 505, σ = 62
      x̅ = 509, n = 1,000
      Ho: μ = 505
      Ha: μ > 505
    • Step 2: Calculate the appropriate 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

Assignment:

  • 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. 

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Lesson 11: Inference for Proportions

Review:

  • Quiz 4
  • t-distribution for small samples

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*
  • Example 8.1 on p. 489 – 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}

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.
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Lesson 10: The t-distribution for Small Sample Inference

Review:

  • Estimation with Confidence Intervals (n>=30)

Presentation:

  • 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
      • test statistic and p-value for significance testing (don’t worry, we’ll cover  these topics later)
    • t distribution critical values
      • print the t distribution table for the next exam
        • use the “0.025” column for estimation with 95% confidence intervals
      • 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 (top label) or 95% confidence level (bottom 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* = 2.179
    • t-statistic formulas
      • for confidence intervals
        • x-bar ± t * s/√n
        • instead of the Z equivalent:  x-bar ± z * σ/√n
    • Video
    • Examples:
      • Identify appropriate t distribution critical values
        • n=10 with 95% confidence, t* = 2.262
        • n=20 with 95% confidence, t* = 2.093
      • 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.

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Lesson 9: Estimation with Confidence Intervals

Review:

  • Quiz 3
  • 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 video
  • Example: 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.

Assignment:

  • Read pp. 353-362, Introduction to Inference and Estimation with Confidence
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Lesson 8: Sampling Distributions and the Central Limit Theorem

Review:

  • Exam 1 Results
  • Design of Experiments
  • Census and Sampling
  • Sampling and Surveys

Presentation:

  • Sampling Distributions
    • Take multiple Simple Random Samples, sample size = n, and calculate each sample mean
    • “Sampling Distribution” is the distribution of sample means
    • Standard deviation of sample means = s/√n
    • Figure 5.8 on p. 336: 
    • Example 5.18 on p. 338
  • Central Limit Theorem
    • Simple random samples of size n from any population with mean μ and finite standard distribution σ. When n is large (typically ≥ 30), the sampling distribution of the sample mean is approximately Normal.
    • Example 5.19 on p. 339
  • Video

Activity:

  • Test the Central Limit Theorem
  • Calculate sample means (use Vehicle Year)
  • Report results

Study:

  • For practice complete Exercises 5.36, 5.37, 5.38, 5.39 on pp. 336-340
  • Read pp. 335-344, Sampling Distribution
Mean2007.65
Standard Error0.63
Median2006.50
Mode2013.00
Standard Deviation5.77
Range23.00
Minimum1996.00
Maximum2019.00
Sum168643.00
Count84.00
Largest(1)2019.00
Smallest(1)1996.00
sampling distribution mean2007.65
sampling distribution stdev0.63
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Lesson 7: Census, Sampling, Surveys and Inference

Review:

Presentation:

  • Census and Sampling
    • Census = attempt to count entire population
      • US Census every 10 years
      • Most costly non-military federal government operation (except bailing out Wall Street)
    • Sample = gather info from a portion of the population
      • Sample Types
        • Voluntary response sample
        • Simple random sample
          • select from a population
          • each individual has equal chance of being selected
          • Example: random selection of subset of enrolled students
        • Stratified random sample
          • divide population into groups or strata
          • random sampling, equal chance of selection within each group
          • Example: random selection within major (30% CIS, 35% Mgmt, 35% Econ)
    • Video: Census and Sampling
  • Samples and Surveys
    • Toward Statistical Inference
      • Population parameters
      • Sample statistics
      • Sampling distribution
      • Bias and Variability
        • to reduce bias, use random sampling
        • to reduce variability, use larger sample sizes (p. 215)
      • Margin of Error Calculator
    • Video: Samples and Surveys

Activity:

Assignment:

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