STAT 110 - Section 5
Lecture 26
Professor Hao Wang
University of South Carolina
Spring 2012
Estimating p:
(1) Point estimator
pˆ
(2) Confidence interval estimator
• But, we do not know p! (If we did, we
wouldn’t need to make a confidence
interval…) So use in place of . This is an
approximate 95% confidence interval for p.
• CONFIDENCE INTERVAL FOR
• An approximate 95% confidence interval for p
is
confidence interval interpretation
• The "95%" in the confidence interval listed above represents
a level of certainty about our estimate.
• If we were to repeatedly make new estimates using exactly
the same procedure (by drawing a new sample, conducting
new interviews, calculating new estimates and new
confidence intervals), the confidence intervals would contain
the true value 95% of the time.
• (modified from
http://www.census.gov/did/www/saipe/methods/statecounty/ci
.html
A confidence interval for p
pˆ (1  pˆ )
pˆ  z value
n
z value
1
1.64
1.96
2.58
3
3.29
Percent Confidence
68.3%
90%
95%
99%
99.7%
99.9%
Assume your 99.7% confidence interval for p is
0.07  0.06  0.01 to 0.13
What will your 68% confidence interval be?
a)
0.07  0.01  0.06 to 0.08
b)
0.07  0.02  0.05 to 0.09
c)
0.07  0.03  0.04 to 0.10
Four more questions from last years Practice Exam 3
Hypothesis testing example 1
• James Bond insists that his martinis should
be shaken, not stirred. A skeptical bartender
tests Bond with 6 martinis (using six fair coin
flip to determine which drinks to shake and
which to stir). Bond errs on one and correctly
identifies the other 5 before passing out.
• Is this evidence sufficient to reject the null
hypothesis that Bond cannot tell the
difference and is merely guessing (i.e., his
probability of correct identification p=0.5). To
find out we test the null hypothesis
H0: Bond is merely guessing versus
alternative
H1: Bond is not merely guessing.
• Under the null hypothesis, what is the
probability that Bond correctly identifies 5 or
more out of the 6 martinis?
• Based on the answer to the above question,
what is your conclusion about Bond’s ability to
tell the difference at the significance level
5%?
Hypothesis testing example 2
• A company claims that it has only a 5%
complaint rate for its products. A consumer
protection group thinks the percent is higher.
A survey of a random sample of 400 product
owners shows that 33 had complaints is:
• Assume that the company’s claim is true and
that p is really 0.05 (5%). What is the
probability that a of 8% or more would be
observed?
• This is an example of a Hypothesis Test.
• The claim that p=0.05 is what is called the
“Null Hypothesis”.
• The probability 0.15% is called the “P-value”.
It is similar to the probability in a criminal trial
using DNA evidence… a small number is
evidence against the “Null Hypothesis”.
Chapter 22 – What is a Test of Significance
• A hypothesis test checks sample data against
a claim or assumption about the population.
The claim or assumption being tested is called
the null hypothesis such as:
H0: p = 0.50
The null hypothesis is the status quo. It is always
set up using an equal sign.
The statement we are looking for evidence of is called
the alternative hypothesis.
Three possible alternate hypotheses for seeing if a
coin was fair could be:
1) Ha: p < 0.50
or 2) Ha : p>0.50
or 3) Ha: p is not equal to 0.50
The alternative hypothesis is also called the
experimental hypothesis. It is opposed to the null
hypothesis.
The p-value is used to summarize the amount of
evidence we have against the null hypothesis.
The p-value is the probability that we would see a
statistic at least as extreme as the one observed if
the null hypothesis was true.
The smaller the p-value, the more evidence against
the null hypothesis.
The Level of significance ( a ) determines how
much evidence against H0 we require to reject
H0 and find in favor of the alternative hypothesis,
Ha.
If the p-value is less than a then we have
statistically significant evidence against the null
hypothesis.
If the null hypothesis was really true, then the
probability that you reject the null hypothesis is
equal to a. So you should pick a to be the
chance you are willing to take of rejecting H0
when it is really true.
We can either:
• Reject the null hypothesis and conclude the
alternate is true.
• Fail to reject the null hypothesis and not have
enough evidence to say it is false.
We never get to accept the null hypothesis and
say that it is true!
Example
• A new drug is advertised as being 80%
effective. A consumer advocacy group thinks
that it isn’t that effective and is looking for
evidence that it doesn’t work well.
• What is the null hypothesis?
• What is the alternate hypothesis?
• Historically, 76% of students in an introductory
psychology course have correctly answered the
professor’s favorite test question on Freud. This
semester the professor gave a randomly selected
group of students an extra lecture on the subject
and wants to see if it will help them do better on the
question.
• The null hypothesis (H0) should be:
A) p < 0.76
D) p=0.76
B) p ≠ 0.76
E) p<0.24
C) p > 0.76
• Historically, 76% of students in an introductory
psychology course have correctly answered the
professor’s favorite test question on Freud. This
semester the professor gave a randomly selected
group of students an extra lecture on the subject
and wants to see if it will help them do better on the
question.
• The alternate hypothesis (HA) should be:
A) p < 0.76
D) p=0.76
B) p ≠ 0.76
E) p<0.24
C) p > 0.76
McDonalds claims their quarter-pounders weigh 0.25
pounds. A disgruntled customer thinks they’re being
swindled and that they don’t weigh that much and
wants to find out.
The null hypothesis (H0) for testing this would be:
A) m < 0.25 lbs.
B)
x = 0.25 lbs.
x ≥ 0.25 lbs.
E) x ≠ 0.25 lbs.
D)
C) m = 0.25 lbs.
Recall that
x is the mean of the sample.
m (pronounced mu) is the symbol used for the mean of
the population.
McDonalds claims their quarter-pounders weigh
0.25 pounds. A disgruntled customer thinks
they’re being swindled and that they don’t weigh
that much and wants to find out.
The alternate hypothesis (HA) for testing this would
be:
A) m < 0.25 lbs.
D) m ≥ 0.25 lbs.
B) m = 0.25 lbs.
E) m ≠ 0.25 lbs.
C) m > 0.25 lbs.
A random sample of 100 quarter-pounders was
weighed in order to test the null hypothesis
H0: m=0.25 against the alternate hypothesis
HA: m<0.25 .
The average weight of the 100 burgers was 0.249
lbs. and the p-value was 0.1752.
If a=0.05 we would
A) Reject the null hypothesis and conclude the
burgers were too light.
B) Accept the null hypothesis and conclude the
burgers weighed 0.25 lbs. on average.
C) Conclude that we do not have enough evidence
to say the burgers are too light.
A random sample of 100 quarter-pounders was
weighed in order to test the null hypothesis
H0: m=0.25 against the alternate hypothesis
HA: m<0.25 .
The average weight of the 100 burgers was 0.231
lbs. and the p-value was 0.03851.
If a=0.05 we would
A) Reject the null hypothesis and conclude the
burgers were too light.
B) Accept the null hypothesis and conclude the
burgers weighed 0.25 lbs. on average.
C) Conclude that we do not have enough evidence
to say the burgers are too light.
McDonalds claims their quarter-pounders weigh
0.25 pounds. The manager thinks that the
employees are wasting meat and making them
too heavy.
The alternative hypothesis (HA) for testing this
would be that average weight (called m) is:
A) m < 0.25 lbs.
D) m ≤ 0.25 lbs.
B) m = 0.25 lbs.
E) m ≠ 0.25 lbs.
C) m > 0.25 lbs.
Identify the null hypothesis (HO) :
a. No change in VOD
b. Reduction in VOD
c. Increase in VOD
Identify the alternative hypothesis (HA) :
a. No change in VOD
b. Reduction in VOD
Increase in VOD
c.
At what α-level would you be able to reject the null hypothesis
using the circled p-value?
A) ® =0.01
B) ® =0.05
C) ® =0.10