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

Proportion pMean, μStandard Deviation, σ,

or variance σ²

1) n ≤ 0.05N2) np(1-p) ≥ 10

1) normal population2) no outliers

Assumptions Met?

1) apx normal population2) no outliers

Compute Z-interval Compute t-interval

Compute χ²-interval Compute Z-interval

Nonparametics

σ known?

n≥30 n≥30yes yes

yes no

no

no no

yes & σ yes & s

yes yes

Chapter 9 – Section 1

If the sample mean is 9, which of these could reasonably be a confidence interval for the population mean?

1) 92

2) (3, 6)

3) (7, 11)

4) (0, ∞)

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) 1

2) 2

3) 5

4) 25

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?

1) A t-distribution with 14 degrees of freedom

2) A t-distribution with 15 degrees of freedom

3) A general normal distribution

4) A nonparametric method

Chapter 9 – Section 2

What issue do we have in calculatingσ / √ n when the population standard deviation is not known?

1) There are no issues

2) We do not know which value to use for n

3) We do not know how to calculate the sample mean

4) We do not know which value to use for σ

Chapter 9 – Section 3

A study is trying to determine what percentage of students drive SUVs. The population parameter to be estimated is

1) The sample mean

2) The population proportion

3) The standard error of the sample mean

4) The sample size required

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

1) 200 students

2) 400 students

3) 600 students

4) 900 students

Chapter 9 – Section 4

Which probability distribution is used to compute a confidence interval for the variance?

1) The normal distribution

2) The t-distribution

3) The α distribution

4) The chi-square distribution

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

1) (4, 6)

2) (8, 18)

3) (160, 360)

4) Cannot be determined from the information given

Chapter 9 – Section 5

Which of the following methods are used to estimate the population mean?

1) The margin of error using the normal distribution

2) The margin of error using the t-distribution

3) Nonparametric methods

4) All of the above

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?

1) The population mean

2) The population proportion

3) The population variance

4) The population standard deviation

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

Homework1: (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