Download 11-2

Lesson 11 - 2
Carrying Out
Significance Tests
Vocabulary
• Hypothesis – a statement or claim regarding a characteristic of
one or more populations
• Hypothesis Testing – procedure, base on sample evidence and
probability, used to test hypotheses
• Null Hypothesis – H0, is a statement to be tested; assumed to
be true until evidence indicates otherwise
• Alternative Hypothesis – H1, is a claim to be tested.(what we will
test to see if evidence supports the possibility)
• Level of Significance – probability of making a Type I error, α
Inference Toolbox
To test a claim about an unknown population parameter
• Step 1: Hypotheses
– Identify population of interest and parameter
– State null and alternative hypotheses
• Step 2: Conditions
– Check 3 conditions (SRS, Normality, Independence); if met,
then continue (or proceed under caution if not)
• Step 3: Calculations
– State test or test statistic
– Use calculator to calculate test statistic and p-value
• Step 4: Interpretation
– Conclusion, connection and context about P-value and the
hypotheses
Z Test for a Population Mean
• Note: not usually seen in AP world (t-test more common)
Hypothesis Testing Approaches
• Classical
– Logic: If the sample mean is too many standard deviations
from the mean stated in the null hypothesis, then we reject the
null hypothesis (accept the alternative)
• P-Value
– Logic: Assuming H0 is true, if the probability of getting a
sample mean as extreme or more extreme than the one
obtained is small, then we reject the null hypothesis (accept
the alternative).
• Confidence Intervals
– Logic: If the sample mean lies in the confidence interval about
the status quo, then we fail to reject the null hypothesis
Classical Approach
-zα/2
-zα
zα/2
zα
Critical Regions
Test Statistic:
x – μ0
z0 = ------------σ/√n
Reject null hypothesis, if
Left-Tailed
Two-Tailed
Right-Tailed
z0 < - zα
z0 < - zα/2
or
z0 > z α/2
z 0 > zα
P-Value Approach
z0
-|z0|
|z0|
P-Value is the
area highlighted
Test Statistic:
x – μ0
z0 = ------------σ/√n
Reject null hypothesis, if
P-Value < α
z0
Confidence Interval Approach
Confidence Interval:
x – zα/2 · σ/√n
Lower
Bound
x + zα/2 · σ/√n
Upper
Bound
μ0
Reject null hypothesis, if
μ0 is not in the confidence interval
Example 1
Assume that cell phone bills are normally distributed. A
simple random sample of 12 cell phone bills finds x-bar
= $65.014. The mean in 2004 was $50.64. Assume σ =
$18.49. Test if the average bill is different today at the α
= 0.05 level. Use each approach.
Step 1: Hypothesis
H0:  = $50.64 (mean cell phone bill is unchanged)
Ha:  ≠ $50.64 (mean cell phone bill has changed)
Step 2: Conditions
SRS: stated in problem
Normality: population is normally distributed
Independence: far more than 120 cell phones
Example 1: Classical Approach
A simple random sample of 12 cell phone bills finds x-bar = $65.014. The
mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is
different today at the α = 0.05 level. Use the classical approach.
not equal  two-tailed
X-bar – μ
65.014 – 50.64
14.374
Z0 = --------------- = ---------------------- = ------------- = 2.69
σ / √n
18.49/√12
5.3376
Zc = 1.96
Using alpha, α = 0.05 the shaded region are the
rejection regions. The sample mean would be too
many standard deviations away from the population
mean. Since z0 lies in the rejection region, we would
reject H0.
Zc (α/2 = 0.025) = 1.96
Example 1: P-Value
A simple random sample of 12 cell phone bills finds x-bar = $65.014. The
mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is
different today at the α = 0.05 level. Use the P-value approach.
not equal  two-tailed
X-bar – μ
65.014 – 50.64
14.374
Z0 = --------------- = ---------------------- = ------------- = 2.69
σ / √n
18.49/√12
5.3376
-Z0 = -2.69
The shaded region is the probability of obtaining a
sample mean that is greater than $65.014; which is
equal to 2(0.0036) = 0.0072. Using alpha, α = 0.05,
we would reject H0 because the p-value is less than α.
P( z < Z0 = -2.69) = 0.0036 (double this to get p-value because its two-sided!)
Using Your Calculator: Z-Test
• For classical or p-value approaches
• Press STAT
– Tab over to TESTS
– Select Z-Test and ENTER
•
•
•
•
Highlight Stats
Entry μ0, σ, x-bar, and n from summary stats
Highlight test type (two-sided, left, or right)
Highlight Calculate and ENTER
• Read z-critical and/or p-value off screen
From previous problem:
z0 = 2.693 and p-value = 0.0071
Example 1: Confidence Interval
A simple random sample of 12 cell phone bills finds x-bar = $65.014. The
mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is
different today at the α = 0.05 level. Use confidence intervals.
Confidence Interval = Point Estimate ± Margin of Error
= x-bar ± Zα/2 σ / √n
= 65.014 ± 1.96 (18.49) / √12
Zc (α/2) = 1.96
50.64
= 65.014 ± 10.4617
x-bar
54.55
75.48
The shaded region is the region outside the 1- α, or 95%
confidence interval. Since the old population mean lies
outside the confidence interval, then we would reject H0.
Using Your Calculator: Z-Interval
• Press STAT
– Tab over to TESTS
– Select Z-Interval and ENTER
•
•
•
•
Highlight Stats
Entry σ, x-bar, and n from summary stats
Entry your confidence level (1- α)
Highlight Calculate and ENTER
• Read confidence interval off of screen
– If μ0 is in the interval, then FTR
– If μ0 is outside the interval, then REJ
From previous problem:
u0 = 50.64 and interval (54.552, 75.476)
Therefore Reject
Example 2
National Center for Health has that the mean systolic
blood pressure for males 35 to 44 years of age is 128 and
σ=15. The medical director for a company examines the
medical records of 72 male executives in the age group
and finds that their mean blood pressure is 129.93. Is
there evidence to support that their blood pressure is
different?
Step 1: Hypothesis
H0:  = 128 (younger male executives’ mean blood pressure is 128)
Ha:  ≠ 128 (their blood pressure is different than 128)
Step 2: Conditions
SRS: possible issue, but selected from free annual exams
Normality: sample size large enough for CLT to apply
Independence: have to assume more than 720 young male
executives in the company (large company!!)
Example 2: P-Value
Step 3: Calculations:
X-bar – μ
129.93 – 128
1.93
Z0 = --------------- = ---------------------- = ------------- = 1.092
σ / √n
15/√72
1.7678
From calculator:
z = 1.0918 p-value = 0.2749
Step 4: Interpretation:
More than 27% of the time with a sample size of 72 from the general
population of males in the 35-44 age group, we would get blood pressure
values this extreme or more  Fail to reject H0; not enough evidence to say
that this companies executives differ from the general population.
Example 3
Medical director for a large company institutes a health
promotion campaign to encourage employees to
exercise more and eat a healthier diet. One measure of
the effectiveness of such a program is a drop in blood
pressure. The director chooses a random sample of 50
employees and compares their blood pressures from
physical exams given before the campaign and again a
year later. The mean change in systolic blood pressure
for these n=50 employees is -6. We take the population
standard deviation to be σ=20. The director decides to
use an α=0.05 significance level.
Example 3 cont
Hypothesis:
H0: μ = 0 blood pressure is same
Ha: μ < 0 Regime lowers blood pressure
Conditions:
1: SRS -- stated in the problem statement
2: Normality -- unknown underlying distribution, but large
sample size of 50 says x-bar will be Normally distributed (CLT)
3: Independence -- since sampling is w/o replacement; assume
company has over 500 employees
Example 3 cont
Calculations:
x – μ0
-6 – 0
Z0 = ----------- = -----------------σ/√n
20/√50
= -2.12
Interpretation:
P-value = 0.0170  so only 1.7% of
the time could we get a more
extreme value. Since this is less
than α = 0.05, we reject H0 and
conclude that the mean difference
in blood pressure is negative (so
the regime may have worked!)
P-value = P(z < Z0) = P(z < -2.12)
= 0.0170 (unusual !)
Example 4
The Deely lab analyzes specimens of a drug to
determine the concentration of the active ingredient.
The results are not precise and repeated measurements
follow a Normal distribution quite closely. The analysis
procedure has no bias, so the mean of the population of
all measurements is the true concentration of the
specimen. The standard deviation of this distribution
was found to be σ=0.0068 grams per liter.
A client sends a specimen for which the concentration
of active ingredients is supposed to be 0.86%. Deely’s
three analyses give concentrations of 0.8403, 0.8363,
and 0.8447. Is there significant evidence at the 1% level
that the concentration is not 0.86%? Use a confidence
interval approach as well as z-test.
Example 4 cont
Hypothesis:
H0: μ = 0.86 grams per liter
Ha: μ  0.86 grams per liter
Conditions:
1: SRS -- assume that each analyses represents an observation
in a simple random sample
2: Normality -- stated in the problem that distribution is Normal
3: Independence -- assume each test is independent from the
others (more than 30 specimens)
Example 4 cont
Calculations:
x – μ0
0.8404 – 0.86
Z0 = ----------- = -----------------σ/√n
0.0068/√3
= -4.99
Interpretation:
P-value = 0.0004  so only .04% of
the time could we get a more extreme
value. Since this is less than α =
0.01, we reject H0 and conclude that
the mean concentration of active
ingredients is not 0.86
P-value = 2P(z < Z0) = 2P(z < -4.99)
= 0.0004 (unusual !)
x-bar  z* σ / √n
0.8404  2.576 (0.0068) / √3
0.8404  0.0101
(0.8303, 0.8505)
Summary and Homework
• Summary
– A hypothesis test of means compares whether the true
mean is either
• Equal to, or not equal to, μ0
• Equal to, or less than, μ0
• Equal to, or more than, μ0
X-bar – μ
Z0 = --------------σ / √n
– There are three equivalent methods of performing the
hypothesis test
• The classical approach
• The P-value approach
• The confidence interval approach
• Homework
– pg 714 ; 11.36-37, 38 a&c, 39