Download Exercises 6.7

Solutions to Statistical Methods in the Biological and Health Sciences, by J. Susan Milton
Exercises 6.7
1. Since we have a right-tailed test, the P-value = P(T19 > 3).
Table VI tells us that P(T19 2.861) = 0.995 and P(T19 3.883) = 0.9995.
Therefore, P(T19 > 2.861)=1-0.995 = 0.005 and P(T19 > 3.883)=1-0.9995 = 0.0005.
Since our test statistic (to=3) falls between 2.861 and 3.883, our P-value must fall between
0.0005 and 0.005. That is, 0.0005 < P < 0.005. Based on the value of our P-value, we would
reject Ho only if were greater than or equal to 0.005.
2. Since we have a left-tailed test, the P-value = P(T23 < -2). By symmetry of the T-distribution,
P(T23 < -2) = P(T23 > 2).
Table VI tells us that P(T23 1.714) = 0.95 and P(T23 2.069) = 0.975.
Therefore, P(T23 > 1.714)=1-0.95 = 0.05 and P(T23 > 2.069)=1-0.975 = 0.025.
Since 2 falls between 1.714 and 2.069, our P-value must fall between 0.025 and 0.05. That
is, 0.025 < P < 0.05. Based on the value of our P-value, we would reject Ho only if were
greater than or equal to 0.05.
3. Since we have a two-tailed test, the P-value = 2P(T15 > 1.5).
Table VI tells us that P(T15 1.341) = 0.90 and P(T15 1.753) = 0.95.
Therefore, P(T15 > 1.341)=1-0.90 = 0.10 and P(T15 > 1.753)=1-0.95 = 0.05.
Since our test statistic (to=1.5) falls between 1.341 and 1.753, P(T15 > 1.5) must fall between
0.05 and 0.10, and thus our P-value=2P(T15 > 1.5) must fall between 2(0.05) and 2(0.10).
That is, 0.10 < P < 0.20. Based on the value of our P-value, we would only reject Ho if were greater than or equal to 0.20 (which is a pretty high tolerance for a Type I error).
5a. H0: µ = 0.035%
H1: µ > 0.035%
5b. n = 144
X̄ = 0.09%
s = 0.25%
t0 x̄ µ 0
s/ n
0.09 0.035
2.64
0.25/ 144
P-value = P(T143 > 2.64). Since n is large, use the last row in the T-table denoted by infinity
() sign. That is, find P(T > 2.64).
Table VI tells us that P(T 2.576) = 0.995 and P(T 3.291) = 0.9995.
Therefore, P(T > 2.576)=1-0.995 = 0.005 and P(T > 3.291)=1-0.9995 = 0.0005.
Since our test statistic (to=2.64) falls between 2.576 and 3.291, our P-value must fall between
0.0005 and 0.005. That is, 0.0005 < P < 0.005. Based on the value of our P-value, we would
reject Ho if were greater than or equal to 0.005.
6. H0: µ = 6
H1: µ > 6
n = 16
d.f. = n1 = 161 = 15
X̄ = 6.15
s = 0.266
x̄ µ 0
6.15 6
Test statistic = t0 2.26
s/ n
0.266/ 16
So, our p-value is “how likely is it that we would observe a sample average more extreme
than 6.15 if in fact the true average µ were 6?” Then translating the sample average to the
test statistic, our p-value = P(T15 > 2.26)
Using Table VI to calculate the P-value:
Table VI tells us that P(T15 2.131) = 0.975 and P(T15 2.602) = 0.99.
Therefore, P(T15 > 2.131)=1-0.975 = 0.025 and P(T15 > 2.602)=1-0.99 = 0.01.
Since our test statistic (to=2.26) falls between 2.131 and 2.602, our P-value must fall
between 0.01 and 0.025. That is, 0.01 < P < 0.025.
Using Minitab to calculate the P-value:
Cumulative Distribution Function
Student's t distribution with 15 DF
x
2.2600
P( X <= x)
0.9804
Minitab tells us P(T15 < 2.26) = 0.9804. Therefore, P(T15 > 2.26) must be 10.9804 =
0.0196. That is, our p-value is 0.0196.
Our P-value tells us that it is very unlikely (P=0.0196 =0.05) that we would observe a sample
average as large as we did if the true mean were 6, so we must reject Ho. There is sufficient
evidence at the =0.05 level to conclude that the mean maximum effective range for the bat’s
echolocation system is more than 6 meters.
Minitab will also perform the hypothesis test directly for us by selecting “1-sample t” under
“Basic Statistics.” Doing so, we get the following output:
T-Test of the Mean
Test of mu = 6.0000 vs mu > 6.0000
Variable
Distance
N
16
Mean
6.1500
StDev
0.2658
SE Mean
0.0665
T
2.26
P
0.020
7b. Test statistic: t0 x̄ µ 0
)/ n
5.3 5
1.37
1.2/ 30
Rejection region: Reject H0 if t0 1.311.
Decision: Reject H0, since 1.37 1.311.
7d. Test statistic: t0 x̄ µ 0
s/ n
9.3 10
1.75
2/ 25
Rejection region: Reject H0 if t0 -2.492
Decision: Do not reject H0, since -1.75 ¾ -2.492.
7e. Test statistic: t0 x̄ µ 0
s/ n
5.1 6
1.50
3/ 25
Rejection region: Reject H0 if t0 -2.064 or if t0 2.064.
Decision: Do not reject H0, since -1.50 does not fall in the rejection region.
8a. H0: µ = 10
H1: µ > 10
8b. n = 9
X̄ = 11
s = 1.94
d.f. = n1 = 91 = 8
Test statistic: t0 x̄ µ 0
s/ n
11 10
1.55
1.94/ 9
So, our p-value is “how likely is it that we would observe a sample average more extreme
than 11 if in fact the true average µ were 10?” Then translating the sample average to the test
statistic, our p-value = P(T8 > 1.55)
Using Table VI to calculate the P-value:
Table VI tells us that P(T8 1.397) = 0.90 and P(T8 1.86) = 0.95.
Therefore, P(T8 > 1.397)=1-0.90 = 0.10 and P(T8 > 1.86)=1-0.95 = 0.05.
Since our test statistic (to=1.55) falls between 1.397 and 1.86, our P-value must fall
between 0.05 and 0.10. That is, 0.05 < P < 0.10.
Using Minitab to calculate the P-value:
Cumulative Distribution Function
Student's t distribution with 8 DF
x
P( X <= x)
1.5500
0.9201
Minitab tells us P(T8 < 1.55) = 0.9201. Therefore, P(T8 > 1.55) must be 1 0.9201=
0.0799. That is our p-value is 0.0799.
Our P-value tells us that the probability of observing the sample we did, if the true mean were 10,
is 0.08. Therefore, if we compare 0.08 to =0.10, we must reject Ho, since 0.08 < 0.10. There is
sufficient evidence at the =0.05 level to conclude that the mean maximum effective range for
the bat’s echolocation system is more than 6 meters. (Note: that by setting = 0.10, it means we
are willing to commit a Type I error about 10% of the time in this situation.)
Minitab will also perform the hypothesis test directly for us by selecting “1-sample t” under
“Basic Statistics.” Doing so, we get the following output:
T-Test of the Mean
Test of mu = 10.000 vs mu > 10.000
Variable
Radiate
N
9
Mean
11.000
StDev
1.936
SE Mean
0.645
T
1.55
P
0.080
Alternatively, we could use the rejection region approach:
Rejection region: Reject H0 if t0 1.397.
Decision: Reject H0, since 1.55 1.397.
There is sufficient evidence to conclude that the radiation has increased above the safe limit.
Because we rejected H0, there is a possibility that we committed a Type I error.
9a. H0: µ = 2.5
H1: µ g 2.5
9b. n = 10
d.f. = n1 = 101 = 9
X̄ = 2.66
s = 0.20
x̄ µ 0
2.66 2.5
Test statistic: t0 2.53
s/ n
0.20/ 10
Rejection Region Approach:
Rejection region: Reject H0 if t0 -1.833 or if t0 1.833. Decision: Reject H0, since 2.53 1.833.
There is sufficient evidence to conclude that the time to closure is different from 2.5 seconds.
Because we rejected H0, there is a possibility that we committed a Type I error.
Using Minitab to calculate the P-value:
Cumulative Distribution Function
Student's t distribution with 9 DF
x
P( X <= x)
2.5300
0.9839
Minitab tells us P(T9 < 2.53) = 0.9839. Therefore, P(T9 > 2.53) must be 1 0.9839= 0.0161. So,
since we have a two-tailed test, our p-value is 2 × 0.0161= 0.0322.
Our P-value tells us that the probability of observing the sample we did, if the true mean were 2.5
is 0.0322. Therefore, if we compare 0.0322 to =0.10, we must reject Ho, since 0.0322 < 0.10.
There is sufficient evidence at the =0.10 level to conclude that the mean time from touch to
complete closure is not equal to 2.5 seconds. (Note: that by setting = 0.10, it means we are
willing to commit a Type I error about 10% of the time in this situation.)
Minitab will also perform the hypothesis test directly for us by selecting “1-sample t” under
“Basic Statistics.” Doing so, we get the following output:
T-Test of the Mean
Test of mu = 2.5000 vs mu not = 2.5000
Variable
Time
N
10
Mean
2.6600
StDev
0.2011
SE Mean
0.0636
T
2.52
P
0.033
10. H0: µ = 17
H1: µ < 17
n = 16 d.f. = n1 = 161 = 15
X̄ = 15.4375
s = 13.24
x̄ µ 0
15.4375 17
Test statistic: t0 0.47
s/ n
13.24/ 16
Rejection region approach:
Rejection region: Reject H0 if t0 -1.753.
Decision: Do not reject H0, since -0.47 does not fall in the rejection region.
There is insufficient evidence to conclude that the number of days has been reduced.
(Because we did not reject H0, there is a possibility that we committed a Type II error.)
Using Minitab to calculate the P-value:
Cumulative Distribution Function
Student's t distribution with 15 DF
x
P( X <= x)
-0.4700
0.3226
Minitab tells us P(T15 < 0.47) = 0.3226. Since our test is left-tailed, our p-value is 0.3226.
Our P-value tells us that the probability of observing the sample we did, if the true mean were 17,
is 0.3226. Therefore, if we compare 0.3266 to =0.05, we do not reject Ho, since 0.3266 > 0.05.
That is, it is not surprising to observe the sample we did, so there is not enough evidence at the
=0.05 level to conclude that the mean number of days is less than 17. Since we failed to reject
the null hypothesis, there is a possibility that we committed a Type II error.
Minitab will also perform the hypothesis test directly for us by selecting “1-sample t” under
“Basic Statistics.” Doing so, we get the following output:
T-Test of the Mean
Test of mu = 17.00 vs mu < 17.00
Variable
Days
N
16
Mean
15.44
StDev
13.24
SE Mean
3.31
T
-0.47
P
0.32
Solutions prepared by Laura J. Simon, Pennsylvania State University