1 / 28

# - PowerPoint PPT Presentation

Inference about Mean (σ Unknown). When σ is known, the sampling distribution for a sample mean is normal if conditions are satisfied.

Related searches for

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about '' - kaemon

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Inference about Mean(σ Unknown)

When σ is known, the sampling distribution for a sample mean is normal if conditions are satisfied.

For many years, it was thought that when σ was unknown, this was still the case. However, because of the increased variability introduced by not knowing σ, the sampling distribution for a sample mean with unknown σ is not normal.

This was discovered by W. S. Gosset, an Irish Brewery quality control inspector in 1908. He also discovered that when σ is unknown, we can still do inference using a sampling distribution model he discovered, the t-distribution.

Section 9.1, Page 184

• The t-Distribution is unimodal and symmetric

• It has more area in that tails and is flatter in the middle than the normal distribution

• It is a family of distributions, with a different curve for each μ, σ, and df (degrees of freedom) df=n-1, where n is the sample size.

Critical Values for 95% Confidence Interval

Known σ: 1.960 from Normal Distribution

Unknown σ with 3 df: 3.18 from t-Distribution

Unknown σ with 10 df: 2.23 from t-Distribution

Unknown σ with 1000 df: 1.962 from t-Distribution

Section 9.1, Page 185

1. Sample is large enough if n ≥ 30

2. If sample is not large enough, no analysis can be done.

Section 9.1, Page 185

Confidence Interval(Unknown σ)

Confidence Interval =

Sample Mean ± Margin of Error

± Critical Value × Standard Error of

where s is the standard deviation of the sample and n is the sample size, and df is the degrees of freedom, n-1.

Conditions: The population must be normal or the sample is large (n ≥ 30)

Section 9.1, Page 187

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Add-in Programs

C.I. =Sample Mean ± Critical Value * Standard Error

PRGM – CRITVAL – ENTER

2: T-DIST

CONF LEVEL = .95

df = 19

PRGM – STDERROR-ENTER

4: 1-MEAN

n=20

Sx=1.76

C.I. = 6.87 ± 2.0930*.3935 = 6.87 ± .8236

=(6.87 - .8236, 6.87 + .8236) = (6.05, 7.69)

Section 9.1, Page 187

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Black Box Program

STAT – TESTS – 8:Tinterval

Inpt: Stats

6.87

Sx: 1.76

n: 20

C-Level: .95

Calculate

Section 9.1, Page 187

• Find the 98% confidence interval.

• Find the critical value

• Find the margin of error.

• Find the standard error.

• What assumption must we make about the the population to have a t-sampling distribution.

• What are the proper words to describe the confidence interval?

• If you wanted to have a margin of error of one minute and the 98% confidence interval for this data, how large must the sample be?

Problems, Page 205

• Find the 93% confidence interval.

• Find the critical value

• Find the margin of error.

• Find the standard error.

• What assumption must we make about the population to have a t-sampling distribution.

• What are the proper words to describe the confidence interval?

• If you wanted to have a margin of error of 25 lbs. and a confidence level of 93% for this data, how many students would you have to monitor?

Section 9.1, Page 205

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Add-In

The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution.

Ho: μ = 4.9 (≤) (no higher than)

Ha: μ > 4.9 (higher than)

t-Distributiondf = 21

P-value = .2158

PRGM – TDIST – ENTER

LOWER BOUND = 5.1

UPPER BOUND = 2ND EE99

MEAN = 4.9

df = 21

Answer: .2158 (Evidence is not sufficient to support the EPA’s claim.)

Section 9.1, Page 188

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Black Box

The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution.

Ho: μ = 4.9 (no higher than)

Ha: μ > 4.9 (higher than)

There is not sufficient evidence to support the EPA claim.

Section 9.1, Page 188

• Find the p-value.

• State your conclusion.

• What is the name of the probability model used for the sampling distribution

• What is the mean of the sampling distribution?

• What is the value of the standard error?

• If your conclusion is in error, what type of error is it?

Problems, Page 209

• Find the p-value.

• State your conclusion.

• What is the name of the probability model used for the sampling distribution

• What is the mean of the sampling distribution?

• What is the value of the standard error?

• If your conclusion is in error, what type of error is it?

Problems, Page 205

• State the appropriate hypotheses.

• Find the p-value.

• State your conclusion.

• What is the name of the probability model used for the sampling distribution

• What is the mean of the sampling distribution?

• What is the value of the standard error?

• If your conclusion is in error, what type of error is it?

Problems, Page 206

A binomial experiment is a experiment that has only two outcomes. Binomial experiments relate to categorical variables. Consider the variable, Supports Obama. The variable has two categories, yes and no. We will consider the yes category as a “success” and the no category a “failure”.

Suppose we have a population of 1000 voters, and 550 support Obama. We define proportion of success for the population, p = 550/1000 = .55. The proportion of failures is q = 1-p = .45.

Suppose we take a sample to estimate p, the true proportion of voters that support Obama. We take a random sample of 100 voters and 53 support Obama. Our sample proportion is p’ = 53/100=.53. Our sample q’ = 1-p’ = .47.

Section 9.2, Page 192

• In practice, the following conditions will insure the sampling distribution for a proportion is normal.

• The sample size is greater than 20

• The product np and nq are both greater than 5. Where we do not know p, we substitute p’, np’ = # success in sample must be > 5 and nq’ =n(1-p’) = # failures in sample must be > 5.

• The sample consists of less than 10% of the population.

Section 9.2, Page 192

Given a sample proportion p’, that has a normal sampling distribution, the confidence interval for the true population proportion, p, is:

sample proportion ± margin of error =

sample proportion ± critical value × standard error of p’

where q’ = proportion of failures = 1-p’ and n is the sample size.

Section 9.2, Page 193

Confidence Interval for ProportionIllustrative Example TI-83 Add-in Programs

In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles.

Check conditions for normal sampling condition:

n = 200 > 20

# successes = 17 > 5

# failures = 200 – 17 = 183 > 5

Conditions are satisfied.

PRGM: CRITVAL 1 – CONF LEVEL = .90

PRGM: STDERROR – 1:1 PROP –

p’ = 17/200; n = 200

Confidence Interval = 17/200 ± 1.6449*.0197 =

0.0850 ± 0.0324 = (.0850 - .0324, .0850 + .0324)=

(.0526, .1174)

Section 9.2, Page 193

Confidence Interval for ProportionIllustrative Example TI-83 Black Box Program

In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles.

STAT – TESTS – A:1 – PROPZInt

x: 17 (The number of success in p’: must be integer)

n: 200

C-Level: .90

Calculate

We are 90% confident that the true proportion of convertibles is in the interval.

Section 9.2, Page 193

Problems, Page 206

Problems, Page 206

Problems, Page 206

Confidence Interval for ProportionRequired Sample Size

ME = Critical Value × Standard Error

If we have a good estimate of p, we use it. If we have no good estimate of p, we estimate p = .5. This will produce the largest sample for the given conditions. If p’ turns out to be different that p, our ME will be less than we initially required.

Section 9.2, Page 194

Required Sample SizeIllustrative Problem TI-83

Consider a manufacturer that purchases bolts from a supplier who claims the bolts are approximately 5% defective. How large a sample do we need to estimate the true proportion to be within ± .02 with 90% confidence?

PRGM – SAMPLSIZ – 1:PROPORTION

CONF LEVEL = .90

ME = .02

p EST = .05 (Problem estimate)

Section 9.2, Page 195

Problems, Page 208

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Add-In

Ho: p = .61

Ha: p > .61

p-value

Sampling Distribution for p’

p’ = 235/350 = .6714

p=.61

PRGM – NORMDIST – 1

LOWER BOUND = .6714 – UPPERBOUND = 2ND EE99

MEAN = .61 ;

Answer: p-value = .0093 (Reject Ho, We proved that more than 61% sleep more that 7 or more hours)

Section 9.2, Page 196

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Black Box

Ho: p = .61

Ha: p > .61

STAT – TESTS – 5:1-PropZTest

po: .61

x: 235 (# of successes in p’; must be integer)

n: 350

prop > po

Calculate

Answer: p-value = .0092 (Ho is rejected)

Section 9.2, Page 196

• Check the conditions for a normal sampling distribution.

• State the hypotheses.

• Find the p-value.

• State your conclusion

• If you make an error in your conclusion, what type is it?

• Find the mean of the sampling distribution.

• Find the standard error of the sampling distribution.

Problems, Page 207

• Check the conditions for a normal sampling distribution.

• State the hypotheses.

• Find the p-value.

• State your conclusion

• If you make an error in your conclusion, what type is it?

• Find the mean of the sampling distribution.

• Find the standard error of the sampling distribution.

Problems, Page 207