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Lesson 9 - R

Lesson 9 - R. Chapter 9 Review. Objectives. Summarize the chapter Define the vocabulary used Complete all objectives Successfully answer any of the review exercises. Vocabulary. None new. Determine the Appropriate Confidence Interval to Construct. Which parameter are we estimating.

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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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