Download Bio-Statistics Test of Hypotheses Exercises

Bio-Statistics
(SBE 304)
Test of Hypotheses
Exercises
Dr. Ayman Eldeib
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Hypothesis Tests
Problem I
Medical researchers have developed a new artificial heart constructed primarily
of titanium and plastic. The heart will last and operate almost indefinitely once
it is implanted in the patient’s body, but the battery pack needs to be recharged
about every four hours. A random sample of 50 battery packs is selected and
subjected to a life test. The average life of these batteries is 4.05 hours. Assume
that battery life is normally distributed with standard deviation σ= 0.2 hour.
a.
b.
Is there evidence to support the claim that mean battery life exceeds 4
hours? Use α = 0.05.
What sample size would be required to detect a true mean battery life
of 4.5 hours if we wanted the power of the test to be at least 0.9?
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem I
State the hypotheses
Null hypothesis: µ = 4
Alternative hypothesis: µ > 4
Note that these hypotheses constitute a one-tailed test. The null hypothesis will
be rejected if the sample mean is too big .
Formulate an analysis plan
Analyze sample data
Fall 2010
For this analysis, the significance level is 0.05.
The test method is a z-test.
Using sample data, we compute the standard error (SE),
and the z-score test statistic.
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem I
Analyze sample data
Using sample data, we compute the standard error (SE),
and the z-score test statistic.
SE = _σ / sqrt(n) = 0.20 / sqrt(50) = 0.20/7.07 = 0.283
z0 = (x - µ) / SE = (4.05 - 4)/0.283 = 1.77
Reject H0 if z0 > zα where z0.05 = 1.65
Interpret results
Fall 2010
Since 1.77>1.65, we reject the null hypothesis and there is
sufficient evidence to conclude that the true average battery
life exceeds 4 hours at the significance level α = 0.05.
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem I
b) Sample Size for an Upper-Tail α-Level Test
α=0.05, β=0.01, it follows that zα = z0.05 = 1.645, and zβ = z0.01 = 1.29 Then,
2
n=
( zα + z β ) σ
( µa − µ 0) 2
Fall 2010
2
2
2
(1.645 + 1.29) (0.2)
=
= 34.7 ≅ 35
2
( 4.5 − 4)
Bio-Statistics
Hypotheses Test
SBE 304
Hypothesis Tests
Example II
Problem
A 1992 article in the Journal of the American Medical Association (“A
Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body
Temperature, and Other Legacies of Carl Reinhold August Wundrlich”)
reported body temperature, gender, and heart rate for a number of subjects.
The body temperatures for 25 female subjects follow: 97.8, 97.2, 97.4,
97.6, 97.8, 97.9, 98.0, 98.0, 98.0, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4,
98.5, 98.6, 98.6, 98.7, 98.8, 98.8, 98.9, 98.9, and 99.0.
a.
b.
Fall 2010
Test the hypotheses H0: µ= 98.6 versus H1: µ ≠ 98.6 , using α = 0.05
Find the P-value
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem II
State the hypotheses
Null hypothesis: µ = 98.6
Alternative hypothesis: µ ≠ 98.6
Note that these hypotheses constitute a two-tailed test. The null hypothesis will
be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan
Analyze sample data
Fall 2010
For this analysis, the significance level is 0.05.
The test method is a t-test.
Using sample data, we compute the standard error (SE),
degrees of freedom (DF), and the t-score test statistic (t).
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem II
Analyze sample data
Using sample data, we compute the standard error (SE),
degrees of freedom (DF), and the t-score test statistic (t).
SE = s / sqrt(n) = 0.4821/ sqrt(25)
DF =_ n - 1 = 25 - 1 = 24
t0 = (x - µ) / SE = (98.264 – 98.6)/SE = -3.48
Reject H0 if |t0| > tα/2,n-1 where t α/2,n-1 = 2.064
Since 3.48 > 2.064, reject the null hypothesis and there is
Interpret results sufficient evidence to conclude that the true mean female body
temperature is not equal to 98.6 °F at α = 0.05
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem II
b) P-Value
Since we have a two-tailed test, the P-value is the probability
that the t-score having 24 degrees of freedom is less than - 3.48
or greater than 3.48.
We use the t Distribution Table to find P(t < - 3.48) = 0.001,
and P(t > 3.48) = 0.001.
Thus, the P-value = 0.001 + 0.001 = 0.002.
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Hypothesis Tests
Problem III
The fraction of defective integrated circuits produced in a
photolithography process is being studied. A random sample of
300 circuits is tested, revealing 13 defectives.
a. Find a 95% two-sided CI on the fraction of defective
circuits produced by this particular tool
b. Do the data support the claim that the fraction of defective
units produced is less than 0.05, using α = 0.05?
c. Find the P-value
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem III
a) CI
A 95% two-sided confidence interval for p can be computed as follows:
13/300 – 1.96*sqrt((13/300) (287/300)/300) ≤ p ≤
13/300 + 1.96*sqrt((13/300) (287/300)/300)
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem III
b) State the hypotheses
Null hypothesis: p = 0.05
Alternative hypothesis: p < 0.05
Note that these hypotheses constitute a one-tailed test. The null hypothesis will
be rejected if the sample mean is too small.
Formulate an analysis plan
Analyze sample data
Fall 2010
For this analysis, the significance level is 0.05.
The test method is a z-test.
Using sample data, we compute the standard error (SE)
and the z-score test statistic (z).
Bio-Statistics
Hypotheses Test
SBE 304
Cont.
Hypothesis Tests
Problem III
Analyze sample data
Z0 =
Y − np 0
=
np 0(1 − p 0)
Using sample data, we compute the standard error (SE)
and the z-score test statistic (z).
)
p − p0
13 − 300(0.05)
=
= −0.53
p0 (1 − p0 ) / n
(300)(0.05)(0.95)
Reject H0 if z0 < -zα where -z0.05 = -1.65.
Interpret results
C) P-Value
Fall 2010
Since −0.53 > −1.65, do not reject null hypothesis and
conclude that the true fraction of defective integrated
circuits is not significantly less than 0.05, at α = 0.05
P − Value = P( Z ≤ −0.53) = 0.29806
Bio-Statistics
Hypotheses Test
SBE 304
Questions ?
Fall 2010
Bio-Statistics
Hypotheses Test
SBE 304