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Lesson 9 - R

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  1. Lesson 9 - R Chapter 9 Review

  2. Objectives • Summarize the chapter • Define the vocabulary used • Complete all objectives • Successfully answer any of the review exercises

  3. Vocabulary • None new

  4. Determine the Appropriate Confidence Interval to Construct Which parameter are we estimating Standard Deviation, σ, or variance σ² Mean, μ Proportion p Assumptions Met? 1) normal population 2) no outliers 1) n ≤ 0.05N 2) np(1-p) ≥ 10 yes yes σ known? yes no Compute Z-interval Compute χ²-interval n≥30 n≥30 yes no yes no yes & σ yes & s 1) apx normal population 2) no outliers no Nonparametics Compute Z-interval Compute t-interval

  5. Chapter 9 – Section 1 If the sample mean is 9, which of these could reasonably be a confidence interval for the population mean? • 92 • (3, 6) • (7, 11) • (0, ∞)

  6. Chapter 9 – Section 1 If the population standard deviation σ = 5 and the sample size n = 25, then the margin of error for a 95% normal confidence interval is • 1 • 2 • 5 • 25

  7. Chapter 9 – Section 2 A researcher collected 15 data points that seem to be reasonably bell shaped. Which distribution should the researcher use to calculate confidence intervals? • A t-distribution with 14 degrees of freedom • A t-distribution with 15 degrees of freedom • A general normal distribution • A nonparametric method

  8. Chapter 9 – Section 2 What issue do we have in calculatingσ / √nwhen the population standard deviation is not known? • There are no issues • We do not know which value to use for n • We do not know how to calculate the sample mean • We do not know which value to use for σ

  9. Chapter 9 – Section 3 A study is trying to determine what percentage of students drive SUVs. The population parameter to be estimated is • The sample mean • The population proportion • The standard error of the sample mean • The sample size required

  10. Chapter 9 – Section 3 A study of 100 students to determine a population proportion resulted in a margin of error of 6%. If a margin of error of 2% was desired, then the study should have included • 200 students • 400 students • 600 students • 900 students

  11. Chapter 9 – Section 4 Which probability distribution is used to compute a confidence interval for the variance? • The normal distribution • The t-distribution • The α distribution • The chi-square distribution

  12. Chapter 9 – Section 4 If the 90% confidence interval for the variance is (16, 36), then the 90% confidence interval for the standard deviation is • (4, 6) • (8, 18) • (160, 360) • Cannot be determined from the information given

  13. Chapter 9 – Section 5 Which of the following methods are used to estimate the population mean? • The margin of error using the normal distribution • The margin of error using the t-distribution • Nonparametric methods • All of the above

  14. Chapter 9 – Section 5 A professor wishes to compute a confidence interval for the average percentage grade in the class. Which population parameter is being studied? • The population mean • The population proportion • The population variance • The population standard deviation

  15. Summary and Homework • Summary • We can use a sample {mean, proportion, variance, standard deviation} to estimate the population {mean, proportion, variance, standard deviation} • In each case, we can use the appropriate model to construct a confidence interval around our estimate • The confidence interval expresses the confidence we have that our calculated interval contains the true parameter • Homework:pg 497 – 501; 1, 3, 8, 9, 15, 23, 24

  16. Homework 1: (α/2=0.005, df=18-1=17) read from table: 2.898 3: (α/2=0.025, df=22-1=21) read from table: 10.283, 35.479 8: a) n>30 large sample (σ known)b) (α/2=0.03) Z=1.88 [315.15, 334.85]c) (α/2=0.01) Z=2.326 [312.81, 337.19]d) (α/2=0.025) Z=1.96 n > 147.67 9: a) large sample size allows for x-bar to be normally distributed from a non-normal (skewed data distribution: mean vs median) b) (α/2=0.05) t=1.646 MOE=0.298 [12.702, 13.298] 15: a) x-bar = 3.244, s = 0.487 b) yes c) (α/2=0.025) [2.935, 3.553] d) (α/2=0.005) [2.807, 3.681] e) (α/2=0.005) [0.312, 1.001] 23: same because t-dist is symmetric 24: t-dist, because the tails are larger

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