Download MTH 4452 Test #2 - Solutions

MTH 4452 Test #2 - Solutions
Fall 2007
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers
1. In a study, 31 patients with an inflammation and swelling of the Achilles tendon
(Achilles Tendinopathy) were examined and found to have Achilles tendons with a
mean diameter of 9.8 mm and a standard deviation of 1.95 mm. The mean diameter
for Achilles tendons in the general population is 5.97 mm. Is there sufficient evidence
to indicate that the average diameter of Achilles tendons of patients with Achilles
Tendinopathy is greater than 5.97 mm? Test at the 5% level of significance.
First of all, we suspect that the average diameter of Achilles tendons of patients with
Achilles Tendinopathy is greater than 5.97 mm. If we consider only those people with
Achilles Tendinopathy, and let µ be the mean of the population of people with Achilles
Tendinopathy, then our hypotheses are:
H0 : µ = µ0 = 5.97
Ha : µ > µ0 = 5.97
Next, since the size of the sample, n = 31, is greater than 30, we can assume that
the sampling distribution for the sample means is normal with mean µ and standard
0
.
deviation (standard error) σ x̄ = √σn ≈ √sn . Our test statistic is z = x̄−µ
σ x̄
Also, since we are testing at the 5% level of significance, α = 0.05.
Our test is a one-tailed test, so our rejection region is as shown below:
Rejection Region
Area = 0.05
z" = 1.645
Note that zα = 1.645.
Observe: z =
x̄−µ0
σ x̄
³
´ = 10.936
= 9.8−5.97
1.95
√
31
Since z > zα = 1.645, z is in the rejection region, and therefore, we reject H0 and
accept Ha . There is sufficient evidence to indicate that the average diameter of Achilles
tendons of patients with Achilles Tendinopathy is greater than 5.97 mm.
2. The addition of MMT to gasoline has caused concern about human exposure to Mn. In
a study of ambient air concentrations of fine Mn, samples of air were taken in California
and at National Park sites. The information is given below.
Mean
Standard Deviation
Number of Sites
National Parks California
0.94
2.8
1.2
2.8
36
26
(a) At the 5% significance level, test to see if there is sufficient evidence that the mean
concentrations of Mn differ at the two types of sites. (Even though n2 < 30, use
the large sample z-test.)
Since we are testing to see if the mean concentrations of Mn differ, this will be a
two-tailed test.
H0 : (µ1 − µ2 ) = D0 = 0.
Ha : (µ1 − µ2 ) 6= D0 = 0.
We are testing at the 5% significance level, so α = 0.05.
Since this is a two-tailed test, we are interested in
α
2
= 0.025.
Our critical values are ±z α2 = ±1.96.
Our rejection region is shown below:
Area = 0.025
Area = 0.025
!z"/2 = !1.96
z"/2 = 1.96
Rejection Region
2
Our test statistic is z =
0.58441
Observe: z =
(µ1 −µ2 )−D0
σ x̄1 −x̄2
(µ1 −µ2 )−D0
, where
σ x̄1 −x̄2
=
(0.94−2.8)−0
0.58441
σ x̄1 −x̄2 =
r
s21
n1
+
s22
n2
=
r
(1.2)2
36
+
(2.8)2
26
=
= −3.1827
Since z = −3.1827 is in the rejection region, we reject H0 and accept Ha . We
conclude that there is sufficient evidence that the mean concentrations of Mn
differ at the two types of sites.
(b) What is the p-value of this test?
Since this is a two-tailed test, the p-value is P (|z| > 3.1827) . (i.e. The probability
of getting a z-value that is farther from the mean than −3.1827.)
Note: the area of the tails corresponding to z = ±3.1827 is 2 · (0.0007) = 0.0014.
The p-value is 0.0014.
3. A random sample of 1400 observations from a binomial population produced x = 529
successes.
(a) If your research hypothesis is that p differs from 0.4, then what hypotheses should
you test?
H0 : p = p0 = 0.4
Ha : p 6= p0 = 0.4
Note: This is a two-tailed test.
(b) Calculate the test statistic and its p-value. At the 1% level of significance, use the
p-value to determine if there is sufficient evidence to indicate that p is different
from 0.4.
First, note that np0 = (1400) (0.4) = 560 > 5, and nq0 = (1400) (0.6) = 840 > 5.
Thus, we have the right to assume that the sampling distribution for pˆ is normal.
pˆ =
x
n
=
529
1400
= 0.378
The test statistic is z =
z=
p̂−p0
σ p̂
=
0.378−0.4
0.0131
p̂−p0
,
σ p̂
where σ p̂ =
q
p0 q0
n
=
q
(0.4)(0.6)
1400
= 0.0131
= −1.6794
Since this is a two-tailed test, the p-value is P (|z| > 1.6794) . (i.e., The probability
of getting a z-value that is farther from the mean than −1.6794.)
Note: the area of the tails corresponding to z = ±1.6794 is 2 · (0.0465) = 0.093
The p-value is 0.093.
At the 1% level of significance, α = 0.01
3
Since the p-value is greater than α = 0.01, we do not reject H0 .
There is not sufficient evidence to indicate that p is different from 0.4.
4. An experiment was conducted to test the effect of a new drug on a viral infection. The
infection was induced in 100 mice, and the mice were randomly split into two groups
of 50. The first group, the control group, received no treatment for the infection. The
second received the drug. After a 30 day period, the proportions of survivors, p1 and
p2 , in the two groups were found to be 0.36 and 0.60, respectively.
(a) Is there sufficient evidence, at the 5% significance level, to indicate that the drug
is effective in treating the viral infection?
H0 : (p1 − p2 ) = D0 = 0
Ha : (p1 − p2 ) < D0 = 0
(Our test is a one-tailed test.)
At the 5% significance level, α = 0.05. The rejection region is shown below:
Area = 0.05
!z" = !1.645
Rejection Region
For the sake of being thorough, note that n1 p1 = 18; n1 q1 = 32; n2 p2 = 30; n2 q2 =
20. Since all of these are greater than or equal to 5, we can assume normality.
First note, that since the hypothesis is centered about zero, we will use a pooled
estimator for p̂ , when computing sp̂1 −p̂2 .
p̂ =
x1 +x2
n1 +n2
=
18+30
50+50
σ p̂1 −p̂2 ≈ sp̂1 −p̂2 =
= 0.48
q
p̂q̂
n1
+
Our test statistic is z =
p̂q̂
n2
=
q
(0.48)(0.52)
50
(p1 −p2 )−D0
σp̂1 −p̂2
=
+
(0.48)(0.52)
50
(0.36−0.60)−0
0.09992
= 0.09992
= −2.4019
Since z = −2.4019 is in the rejection region, we reject H0 and accept Ha .
4
There is sufficient evidence to indicate that the drug is effective in treating the
viral infection.
5