Transcript Slide 1

Introduction to Statistical Inferences
Inference means making a statement about a population based
on an analysis of a random sample taken from the population.
Types of Inferences:
Estimation of a parameter, such a the mean. We make an
estimate and calculate a margin of error for the estimate. For
example, the mean age of shoreline students is 28.5 years with
a Margin of Error of ± 3 years.
Hypothesis Testing. We test the truth of a statement about a
population. We test the statement that the water quality meets
quality standards.
Both types of inference rely on the use of Sampling
Distributions.
Section 8.1, Page 152
1
Confidence Interval for Mean, μ
with known σ
A random sample of 36 rivets is selected and
each is tested for shearing strength. The sample
mean x = 924 lbs, σ = 18.

Our point estimate for the mean shearing
strength for the entire population would be
μ =924 lbs. Because this is just one sample, it is
unlikely that the sample mean of 924 lbs. exactly
equals the true mean of the population.
How close is the sample mean to the true mean
of the population?
We will use the sample mean to develop and
confidence interval or range of numbers for the
plausible value of the true mean.
Section 8.1, Page 154
2
Confidence Interval for Mean, μ
with known σ
x 
18
3
36

Because the sampling is normal, we know that
the area between μ – 6 and μ+6 contains 95%
of all the sample means.
If I pick a sample mean at random x and
construct an interval ( x -6, x +6), there is a
95% chance that this x will be within 6 units of
the true mean, and the interval will therefore
contain the true mean.

 x = 924, our 95%
For our example,
confidence 
interval is (918, 930)

Section 8.1, Page 154
3
Confidence Interval for Mean, μ
with known σ
When a sampling distribution for a sample
mean, x is normal, then a confidence interval for
μ, the true mean as follows:

Confidence Interval = x ± Margin of Error
Margin of Error = Critical Value × Standard
Error.
The critical value sometimes referred to as

confidence coefficient
is the number of standard
error units in the Margin of Error for a given
confidence level. We will use the notation z(α)
to refer to to the critical value for the confidence
level α.
The equation for the confidence interval for
confidence α: x ± Margin of Error =
x
± z(α) × x



Section 8.2, Page 156
4
Constructing An Interval
TI-83 Add-in Programs
A random sample of 100 commuting students was
obtained. The resulting sample mean was 10.22
miles. (σ = 6 miles) Find the 95% confidence interval
for the true mean.
Check conditions: Since the sample size is ≥ 30, the
sampling distribution will be normal, even if the
population is not normal.
C. I. = sample mean ± margin of error
= sample mean ± critical value * standard error.
Find the critical value for 95% confidence.
PRGM – 1:CRITVAL – ENTER
1 ENTER - .95 - ANSWER: CR VALUE = 1.96
Find the Standard Error of the Sampling Distribution
PRGM – STDERROR-ENTER
4:1 MEAN : 100 : 6 : ANSWER: SE = .60
C.I. = 10.22 ± 1.96 * 0.60 = 10.22 ± 1.176
C.I. = (10.22 – 1.176, 10.22 + 1.176 = (9.04, 11.40)
Section 8.2, Page 157
5
Constructing An Interval
Black Box Program
A random sample of 100 commuting students was
obtained. The resulting sample mean was 10.22 miles.
(σ = 6 miles) Find the 95% confidence interval for the
true mean.
STAT - TESTS – 7:Zinterval – ENTER
STATS : σ = 6 ; x = 10.22 ; n=100 ; C-Level = .95
Answer: (9.04, 11.40)
Using this output we can say:
1. We are 95% confident that the true mean commute

distance is in the the interval.
2. If we were to take 100 different samples, and
construct 100 different confidence intervals,
approximately 95 of them would contain the true mean
commute distance.
Find the Margin of Error for the confidence interval.
ME = .5(width of interval) = .5( 11.40 - 9.04) = 1.18.
Section 8.2, Page 157
6
Problems
Problems, Page 50
7
Problems
a. What is the variable being studied?
b. Find the 90% confidence interval estimate
for the mean speed.
c. Find the 95% confidence interval estimate
for the mean speed.
d. Which interval is larger. Why?
Problems, Page 178
8
Sample Size
TI-83 Add-in Program
To solve this problem, we need a relationship
between sample size and the variables given. The
margin of error (ME) is such a relationship.
ME  z( ) *

n
Solving forn :
z( ) *  2
n  

 ME 
We solve using the TI-83:
PGRM – SAMPLSIZ – ENTER –
3: KNOWN
σx ; CONF LEVEL = .99; ME = 75;

σx = 900; Answer: n = 956
Section 8.2, Page 160
9
Problems
the standard deviation is 5 seconds.
Problems, Page 179
10
Hypothesis Testing
The evidence for Ha is the sample mean
Ho: The average GPA of students who take statistics
is 3.30 (or more)
Ha: The average GPA of students who take statistics
is less than 3.30.
Sample evidence in the form of the sample mean of a
sample of students will try to prove Ha is true. If Ha is
true, then Ho is false.
Section 8.3, Page 161
11
Writing Hypotheses
State authorities suspect the the
manager of investment fund is
guilty of embezzling money for his
own use.
In our system of justice, a
presumption of innocence is
essential to a trial procedure.
Ho: Manager is innocent
Ha: Manager is not innocent
The state will present evidence in
trial to try to prove Ha.
Section 8.3, Page 162
12
Problems
Problems, Page 179
13
Problems
Problems, Page 179
14
Hypothesis Test of Mean μ (σ Known)
Illustrative Problem
Problem: An aircraft manufacturer must
demonstrate that its rivets meet the required
specifications. One of the specs is: “The mean
shearing strength of all such rivets, μ, is at least
925 lbs. (σ=18). Each time the manufacturer
buys rivets, it is concerned that the mean strength
might be less than the 925-lb pound specification.
A random sample of 50 rivets is selected. The
sample mean is 921.18 and n = 50.
STEP 1: The set up
a. Describe the parameter of interest.
The parameter of interest is μ, the population
mean.
b.Write the Hypotheses.
Ho: μ = 925 (≥) (The mean is at least 925)
Ha: μ < 925 (The mean is less than 925)
Section 8.4, Page 167
15
Illustrative Problem (2)
STEP 2: Check assumptions for Normal Sampling
Distribution
Since σ is known, we will need a normal sampling
distribution. The sampling will be normal if the population
is normal, or if the sample size is ≥ 30. Since the sample
size is 50, the Central Limit Theorem insures that that the
sampling distribution is normal
STEP 3: The evidence for Ha
The evidence for Ha is that the sample mean is 921.18
lbs. This is less than the Ho value of 925 lbs. The are
two possible explanations for the difference between the
sample mean and the Ho mean:
1. Samples are subject to sampling variation. Ho is true
and the sample mean difference is explained by natural
sampling variation.
2. The difference is too great to be reasonably explained
by sampling variation. The difference is explained by the
fact that Ho is not true.
Section 8.4, PLage 169
16
Illustrative Problem (3)
STEP 4. The probability distribution.
We will use the probability distribution to calculate
the probability that if Ho, is true, the difference
between the evidence, x  921.18and x  925 is due
to sampling variation.
p-value = area
=0.0667.


x  925
x  921.18
Sampling Distribution
x 

n

18
50
p  value  p(x
  921.18, given x  925) 


PRGM – NORMDIST -1
LOWER BOUND = -2ND EE 99
UPPER BOUND = 921.18
MEAN = 925
SE(x)  18 / 50
ANSWER: AREA = p-value = 0.0667.



Section 8.4, Page 169
17
Illustrative Problem (4)
Using Black Box Program to Calculate p-value
Problem: An aircraft manufacturer must demonstrate
that its rivets meet the required specifications. One of
the specs is: “The mean shearing strength of all such
rivets, μ, is at least 925 lbs (σ=18) . Each time the
manufacturer buys rivets, it is concerned that the
mean strength might be less than the 925-lb pound
specification. A random sample of 50 rivets is
selected. The sample mean is 921.18 and n = 50.
STAT-TESTS-1:Ztest
Input: Stats
μo: 925 (This is the Ho parameter value, μ=925)
σ: 18
x : 921.18
n: 50
μ: <μ0 (The is the alternate Hypotheses)
Calculate
Answer: P =0.0667 = p-value
Section 8.4, Page 169
18
Illustrative Problem (5)
STEP 5. Decision
We only have two choices for a decision:
1. We reject Ho
2. We fail to reject Ho
Recall that there were only two possibilities that could
explain the difference between the Ho mean and the
sample mean:
a. Ho is true and the difference is due to sampling
variation.
b. Ho is not true.
The p-value tells us how likely it is that a. is the correct
explanation for the evidence. If a. is unlikely – it has a
very small probability of occurring- then we conclude
b. must be the correct explanation for the evidence,
the sample mean.
Decision Criteria (Significance Level for p-value)
If the p-value falls below the significance level, α, then
a. is considered too unlikely, and we reject Ho, and
conclude Ha is true. If α is not specifically stated in
the problem, it is assumed to be 0.05.
Since our problem has a p-value of 0.0667 > 0.05, we
Section 8.4, Page 177
fail to reject Ho.
19
What does the P-value really mean?
The p-value is a probability! It is the probability that, if H0 is
true, the difference between the H0 value and the sample
statistic is due to sampling variation.
If the p-value is very small, then the difference between
the H0 value and the sample statistic is unlikely due to
sampling variation, so we must conclude that sampling
variation is an unlikely explanation for the difference. We
therefore conclude that HA must be true.
Sometimes, in the press, we see that a study was inclusive
because the study results are likely caused by chance. Or,
that the study results are conclusive because the results
are unlikely due to chance. In this case, “chance” means
normal sampling variation. We also say that the results of
the study are not “statistically significant.” There is nothing
really “statistically significant” when the null hypothesis is
not proved.
On the other hand, when null hypothesis is proved, it is a
“big deal” and we say the results are “statistically
significant.”
Section 8.4
20
Problems
a.
b.
c.
d.
e.
Write the appropriate hypotheses.
What condition must be met? Is it met? Explain.
What is its mean and standard error of the sampling
distribution?
Find the p-value.
What is your decision? Explain.
Problems, Page 180
21
Problems
a.
b.
c.
d.
e.
Write the appropriate hypotheses.
What condition must be met? Is it met? Explain.
Sketch the sampling distribution and show its mean
and standard deviation?
Find the p-value.
What is your decision? Explain.
Problems, Page 181
22
Two Tailed Test
In this problem, sample evidence larger than the
mean or evidence smaller than the mean can
cause us to reject the null hypothesis.
The appropriate hypotheses are:
Ho: μ = 82 (The new test mean test value is 82)
Ha: μ ≠ 82 (The new test mean is either larger
than 82 or smaller than 82)
Section 8.4, Page 171
23
Two Tailed Test Continued
P-value = sum of the two
symmetrical areas
μ=82
x  79 x  85
p  value  p(x  79 or x  85, given   82)



Left Tail:
PRGM - NORMDIST 1
LOWER BOUND = -2ND EE99
UPPER BOUND = 79
MEAN = 82
SE(x) 
8 / 36
ANSWER: 0.0122
p-value = Left tail area + right tail area = 2*Left tail area


=0.0244.
Since the p-value is less than 0.05, we reject the null
hypothesis and conclude the alternative hypothesis is true –
the mean of the new test is different than the mean of the
old test.
Section 8.4, Page 171
24
Problems
Test the claim that the BMI of the cardiovascular
technologists is different than the BMI of the general
population. Use α = .05. Assume the population of
the BMI of the cardiovascular technologists is normal.
a.State the necessary hypotheses.
b.Is the sampling distribution normal. Why?
c.Find the p-value.
d.State your conclusion.
e.If you made an error, what type of error did you
make?
Problems Page 181
25
Types of Errors
Type I Error: Reject a true H0.
Type II Error: Failure to reject a false H0.
The probability of a Type I error is the α level or
significance level. Recall that we reject Ho if the pvalue is 5% or less. If the p-value =5%, there is a 5%
chance that the scenario of Ho true and the evidence is
due to sampling variation is the correct scenario. In the
long run, we will make an error rejecting Ho 5% of the
time.
We can reduce the probability of a type I error by
reducing the α level to 1%. If the p-value =1%, there is
a 1% chance that the scenario of Ho true and the
evidence is due to sampling variation is the correct
scenario. In the long run, we will make a type I error
only 1% of the time.
Reducing the α level will reduce the probability of a
type I error, but it will increase the probability of a type
II error, fail to reject a false Ho.
Section 8.3, Page 162
26
Problems
Problems, Page 179
27
Problems
Problems, Page 179
28