Transcript PPT

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 IV
Significantly Different
Using 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) value
5. Determine the value needed to reject the null
hypothesis using appropriate table of critical values
6. Compare the obtained value to the critical value
7. If obtained value is more extreme, reject null
hypothesis
8. 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