Topics 19 - 20Unit 4 – Inference from Data: Principles

TOPIC 19 CONFIDENCE INTERVALS: MEANS

Topic 19 - Confidence Interval: Mean, σ is unknown

The estimated standard deviation of the sample statistic X-bar is called the standard error

n

S

The purpose of confidence intervals is to use the sample statistic to construct an interval of values that you can be reasonably confident contains the actual, though unknown, parameter.

n

StX *error ofmargin estimate

Confidence Interval for a population proportion :

where n >= 30

t * is calculated based on level of confidence

When running for example 95% Confidence Interval: 95% is called Confidence Level and we are allowing possible 5% for error, we call this alpha (α )= 5% where α is the significant level

Topic 19 - Confidence Interval: Mean, σ is unknown

Use if the sample data is given, use the Stat, Edit and enter data in the calculator before running the Confidence Interval

L1 is where data is entered by you

C-Level: is the level you are running the Confidence Interval

Use if the information about sample data is given.

X-Bar mean of sample dataSx is Standard deviation of the samplen is sample size

C-Level: is the level you are running the Confidence Interval

Activity 19-3: M&M Consumption

Travel time to work. 

• A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean is   = 31.25 minutes and the standard deviation is s = 21.88 minutes. What is the standard error of the mean?

• s/√n = 21.88/√20 = 4.8925 minutes.

Ancient air. 

The composition of the earth’s atmosphere may have changed overtime. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen:

63.4 65.0 64.4 63.3 54.8 64.5 60.8 49.1 51.0

Assume (this is not yet agreed on by experts) that these observations are an SRS from the late Cretaceous atmosphere. Use a 90% confidence interval to estimate the mean percent of nitrogen in ancient air.

Enter data for L1.

95% confidence Interval:Using TI83, under Stat, TEST,Choose option 8:TInterval

Mean of the sample = 59.6Standard deviation = 6.26Degree of freedom = df= 8 Confidence interval for mean percent of nitrogen is between 54.8 and 64.4.

Ancient air. 

Exercise 19-15 Page 414Exercise 19-23 Page 417Exercise 19-24 Page 417

TOPIC 20TEST OF SIGNIFICANCE: MEANS

Topic 20 – Test of Significant: MeanThe purpose of Test of Significant is when we do know the population Parameter but we do not necessary agree with it or we have question about it. To do the test we need to run a sample and we use the statistic to test its validity.

Step 2: we initiate hypothesis regarding the question – we can not run test of significant without establishing the hypothesis

0

0

0

00

:

:

or

orH

H

a

Step 3: Decide what test we have to run, in case of proportion, we use t-test

nSx

t 0

Step 1: Identify and define the parameter.

Topic 20 – Test of Significant: MeanStep 4: Run the test from calculator

Step 5: From the calculator write down the p-value T-test

Step 6: Compare your p-value with α – alpha – Significant Level

If p-value is smaller than α we “reject” the null hypothesis, then it is statistically significant based on data.

If p-value is greater than the α we “Fail to reject” the null hypothesis, then it is not statistically significant based on data.

Last step: we write conclusion based on step 6 at significant level α

• p- value > 0.1: little or no evidence against H0 • 0.05 < p- value <= 0.10: some evidence against H0 • 0.01 < p- value <= 0.05: moderate evidence against H0 • 0.001 < p- value <= 0.01: strong evidence against H0 • p- value <= 0.001: very strong evidence against H0

Ho: µ = 115Ha: µ 115

Ho: µ = 115Ha: µ < 115

Ho: µ = 115Ha: µ > 115

Few Possible cases to look at:

A teacher suspects that the mean for older students is higher than 115

A teacher suspects that the mean for older students is same or more than 115

A teacher suspects that the mean for older students is also 115

Higher than means (> 115)The opposite of higher than is less than or equal to 115 ( 115)Comparing the two, null hypothesis is the comparison that includes equality (=)

Same or more than means (> 115)The opposite of same or more than is less than 115 (< 115)

Same means (= 115)The opposite of same is not equal to 115 ( 115)

One-sided alternative

One-sided alternative

Two-sided alternative

Fuel economy.

According to the Environmental Protection Agency (EPA), the Honda Civic hybrid car gets 51 miles per gallon (mpg) on the highway. The EPA ratings often overstate true fuel economy. Larry keeps careful records of the gas mileage of his new Civic hybrid for 3000 miles of highway driving. His result is x-bar= 47.2 mpg. Larry wonders whether the data show that his true long-term average highway mileage is less than 51 mpg. What are his null and alternative hypotheses?

AnswerLarry wonders whether the data show that his true long-term average highway mileage is less than 51 mpg.

H0: µ = 51 mpg; Ha: µ < 51 mpg.

Problem

If a researcher is interested in testing whether the mean is different from some claimed value, 55, then the null and alternative are

test the hypotheses H0: μ = 55, Ha: μ ≠ 55

Stating hypotheses.

In planning a study of the birth weights of babies whose mothers did not see a doctor before delivery, a researcher states the hypotheses asH0 : x-bar = 1000 grams

Ha : x-bar < 1000 grams

What’s wrong with this?

Hypotheses should be stated in terms of µ, not x-bar  .

Topic 20 – Test of Significant: Mean, σ is unknown

Use if the sample data is given, use the Stat, Edit and enter data in the calculator before running the T-test

µ0 is mean–value in questionList: L1 where the raw data is entered by you

µ: is the alternative hypothesis

Use if the information about sample data is given.

µ0 is mean–value in questionX-bar is sample meanSx is Sample Standard deviationn is sample sizeµ: is the alternative hypothesis

Improving your SAT score. We suspect that on the average students will score higher on their second attempt at the SAT mathematics exam than on their first attempt. Suppose we know that the changes in score (second try minus first try) follow a Normal distribution. Here are the results for 46 randomly chosen high school students:

−30 24 47 70 −62 55 −41 −32 128 −11

−43 122 −10 56 32 −30 −28 −19 1 17

57 −14 −58 77 27 −33 51 17 −67 29

94 −11 2 12 −53 −49 49 8 −24 96

120 2 −33 −2 −39 99

Do these data give good evidence that the mean change in the population is greater than zero?

Activity 20- 2: Sleeping Times

The null hypothesis is that the mean sleep time of the population is 7 hours. In symbols, the null hypothesis is H0 : µ = 7.0 hours. The alternative hypothesis is that the mean sleep time of the population is not 7 hours. In symbols, the alternative hypothesis is Ha : µ ≠ 7.0 hours.

Sample Number

Sample Size

Sample Mean

Sample SD

Test Statistic

p- value

1 10 6.6 0.825 2 10 6.6 1.5973 30 6.6 0.825 4 30 6.6 1.597

Exercise 20-8: UFO Sighters’ Personality – Page 432Exercise 20-10: Credit Card Usage - Page 433Exercise 20-21: Pet Ownership - Page 436Exercise 20-14: Age Guesses – Page 434

EXTRA PROBLEMS

Problem

Assume that you are conducting a test of significance using a significance level of α = 0.10. If your test yields a P-value of 0.08, what is the appropriate conclusion?

P-value = 0.08 < 0.10 Reject Null, It is statistically significant

Problem

The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but measurements on a random sample of 400 cigarettes of this brand gave a mean of x = 1.52. Is this evidence that the mean nicotine content is actually higher than advertised? at significance level α = 0.01. You conclude

test the hypotheses H0: μ = 1.5, Ha: μ > 1.5

Is this evidence that the mean nicotine content is actually higher than advertised? State the hypothesis

ProblemA researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 more recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation of sample is 3 years, what is the p-value of the test? at significance level α = 0.05. You conclude