Normal Distr Practice

Major League baseball attendance in 2011 averaged 30,000 with a standard deviation of 6,000.

i. What percentage of teams had between 25,000 and 32,000 in attendance?

ii. What percentage of teams had between 32,000 and 40,000 in attendance?

Iii. How many fans would a team have to draw to be in top 10% of teams in terms of attendance?

Q3 Raw Scores n=24

2│4445668 (7) 3│013568 (6) 4│01123444456 (11)

Mean—35.58; Median=37; Mode=44; s=8.01;

Part IVSignificantly DifferentUsing Inferential Statistics

Chapter 9 Significantly Significant

What it Means for You and Me

What you will learn in Chapter 9

What significance is and why it is important Significance vs. Meaningfulness

Type I Error Type II Error How inferential statistics works How to determine the right statistical test for

your purposes

The Concept of Significance

Any difference between groups that is due to a systematic influence rather than chance Must assume that all other factors that might

contribute to differences are controlled

If Only We Were Perfect…

Significance level The risk associated with not being 100% positive

that what occurred in the experiment is a result of what you did or what is being tested

The goal is to eliminate competing reasons for differences as much as possible.

Statistical Significance The degree of risk you are willing to take that you

will reject a null hypothesis when it is actually true.

The World’s Most Important Table

Type I Errors (Level of Significance)

The probability of rejecting a null hypothesis when it is true

Conventional levels are set between .01 and .05

Usually represented in a report as

p < .05 In other words, we reject the null hypothesis

with a 5% chance of making a Type 1 error;

Type II Errors

The probability of accepting a null hypothesis when it is false

As your sample characteristics become closer to the population, the probability that you will accept a false null hypothesis decreases

You Try!! Label following as a Correct Decision, either Type 1 or Type 11 error

i. Researcher concludes children’s reading scores in 5th and 6th grades are not significantly different, when in fact, they are significantly different;

ii.Researcher concludes children’s reading scores in 5th and 6th grades are significantly different, when in fact, they are not significantly different;

iii. Researcher concludes children’s reading scores in 5th and 6th grades are significantly different, when in fact, they are significantly different;

Type 1 and 2 Errors and Correct Decisions Null hypothesis (Ho): No sig. difference between men

and women concerning empathy   Claim that there is NO diff between men and women concerning empathy, when in fact, there is NO

difference (Correct decision) Claim that there is a diff between men and women concerning empathy, when in fact, there is NO

difference (incorrect decision) TYPE 1 ERROR

Type 1 and 2 Errors and Correct Decisions Claim that there is NO diff between men and

women concerning empathy, when in fact, there is a difference (incorrect decision) TYPE 2 ERROR

Claim that there is a diff between men and women concerning empathy, when in fact, there is a difference (Correct decision)

Confidence Intervals

A recent survey asked college students about their video game playing habits. From a population of 30,000 students, 1303 were randomly sampled. By their own estimation, sample members averaged 3.65 hours of playing per weekend day, with a standard deviation of 2.64 hours per day. What is the 95% confidence interval estimate for the population’s average number of video playing hours per weekend day?

Steps to Finding Confidence Interval 1) Calculate standard error of the mean(SEM)

σ/ sqrt n 2) Calculate margin of error/confidence limit ( z score * SEM) 95% =1.96 standard

deviations 99%= 2.58 standard deviations 3) Add and subtract margin of

error/confidence limit from the sample mean to get lower and upper limits.

Calculate the standard error

ó/√n OR s/√n i. σ= 15 n=45 ii. σ= 15 n=75

Margin of Error/Confidence Limit

Construct margin of error/confidence limit for 95% and 99%

Construct Confidence Interval with lower and upper bound

You Try!!

I. An international dormitory wants to see how long, on average, its members have been in the United States. Out of 105 people in the dorm, 45 were sampled and averaged 4.5 years in country with a standard deviation of 2.12 years. Construct a 95 and 99 percent confidence interval estimate for the dorms member’s average time in the United States.

Confidence Interval for Proportion

Out of 50 SJSU students, 35 believed that undocumented students should be able to get in-state tuition. Find 95% Confidence Interval for the true population estimate who is for granting in-state tuition to undocumented students.

You Try!!

In a recent student election at SJSU, a candidate running for President garnered 325 votes out of 800 total. Find 99 % Confidence Interval for the true population estimate who would vote for the candidate for President.

How Inference Works

A representative sample of the population is chosen.

A test is given, means are computed and compared

A conclusion is reached as to whether the scores are statistically significant

Based on the results of the sample, an inference is made about the population.

The Picture Worth a Thousand Words

Deciding What Test to Use

Test of Significance (pp.175-176)

1. A statement of null hypothesis.2. Set the level of risk associated with the null

hypothesis.3. Select the appropriate test.4. Compute the test statistic (obtained) value5. Determine the value needed to reject the null

hypothesis using appropriate table of critical values6. Compare the obtained value to the critical value7. If obtained value is more extreme, reject null

hypothesis8. If obtained value is not more extreme, accept null

hypothesis

Hypothesis Testing: Z Test

A local school district recently implemented an experimental program for science education. After 1 year, the 36 children in this program obtained an average score of 61.8 on a national science achievement test. This national test is standardized so that the national average is 60 with a standard deviation of 6. Did the students in the special program score above the national average? Test at α = .05.

You Try!!

There is widely held belief that men are at a greater advantage in terms of salary than women in technological fields. The average of engineering salaries of men is known with men making, on average, an introductory salary of $50,000 with a population standard deviation of $5,000. Then a sample of 50 female engineers making and introductory salary, on average, of $45,000 was taken. Using the Z test at 99% significance level, test null hypothesis that there is no difference between engineering salaries of both men and woman.

Glossary Terms to Know

Significance level Statistical significance Type I error Type II error Obtained value

Test statistic value Critical value