350 STAT Assignment 8 Solutions (60 points)
1.
(5 points = 0.5 each)
(a) Yes, > 100 is a statement about a population standard deviation, i.e., a statement about a population
parameter.
(b) No, this is a statement about the statistic x , not a statement about a population parameter.
~
(c) Yes, this is a statement about the population median .
(d) No, this is a statement about the statistic s (s is the sample, not population, standard deviation.
(e) Yes, the parameter here is the ratio of two other parameters; i.e, 1/2 describes some aspect of the
populations being sampled, so it is a parameter, not a statistic.
(f) No, saying that the difference between two samples means is -5.0 is a statement about sample results, not
about population parameters.
(g) Yes, this is a statement about the parameter of an exponential population.
(h) Yes, this is a statement about the proportion of successes in a population.
(i) Yes, this is a legitimate hypothesis because we can make a hypothesis about the population distribution [see
(4) at the beginning of this section].
(j) Yes, this is a legitimate hypothesis. We can make a hypothesis about the population parameters [see (3) at the
beginning of this section].
10.
(3 points = 0.5 each)
(a) H0 should not be rejected, since p-value = .084 > = .05.
(b) H0 should not be rejected, since p-value = .003 > = .001.
(c) H0 should be rejected, since p-value = .048 < = .05.
(d) H0 should be rejected, since p-value = .084 < = .10.
(e) H0 should not be rejected, since p-value = .039 > = .01.
(f) H0 should be rejected, since p-value = .017 < = .10.
12.
(4 points = 1 each)
34.43 34
(a) z 1.06
2.87
50
So, p-value = 2P(z > |2.87|) = 2(.0021) = .0042
[Note: Since Ha is a two-tailed test, the p-value is the sum of the area in the two tails. That is, p-value = P(z
< -2.87) + P(z > 2.87) = 2P(z > 2.87)]
33.57 34
(b) z 1.06
2.87 So, p-value = 2P(z >| -2.87|) = 2(.0021) = .0042
50
33.25 34
(c) z 1.89
2.24
32
So, p-value = 2P(z >| -2.24|) = 2(.0125) = .025
33.66 34
(d) z 2.53
1.57
36
So, p-value = 2P(z > |1.57|) = 2(.0582) = .1164
16.
(5) Let denote the true average penetration. Since we are concerned about the specifications not being met, the
relevant hypotheses are:
H0: the specifications are met ( = 50) , versus
Ha: the specifications are not met ( > 50).
52.750
The test statistic is: z
3.33 The corresponding p-value = P(z > 3.33) = .0004
4. 8
35
Since p-value = .0004 < = .05, we reject H0 and conclude that the true average penetration exceeds 50 mils.
Thus, we conclude the specifications have not been met.
18.
(3 points = 1 each)
(a) p-value = P(t > 3.2) = .003
(note: df = (n 1) = 14)
Since p-value = .003 < = .05, we reject H0. We claim > 20.
(b) p-value = P(t > 1.8) = .055
(note: df = (n 1) = 8)
Since p-value = .0055 > = .01, we fail to reject H0. We have insufficient evidence to claim > 20.
(c) p-value = P(t >-.2) = .578
(note: df = (n 1) = 23)
Since p-value = .578 >any sensible choice of , we fail to reject H0. We have insufficient evidence to claim
> 20.
22.
(5) Let denote the true average escape time. The relevant hypotheses are H0: = 360 versus Ha: > 360.
We choose these hypotheses because the wording in this exercise indicates that the investigators believed, a
priori, that would be at most 6 minutes (i.e., 360). This belief should represent the null hypothesis.
370.69360
The test statistic is: t
2.24
24.36
26
The df = (n 1) = 25.
The corresponding p-value = P(t > 2.24) .017
Since the p-value = .017 < = .05, we reject H0 and conclude that the data contradicts the investigators a priori
belief. That is, we have sufficient evidence to conclude that the true average escape time exceeds 6 minutes.
(a) (4) The quantile plots for both the round and square pads appear to be quite linear. So yes, it is plausible that
both force distributions are normal. Also note that the normal probability also plots suggest that normality is
plausible for both groups.
5500
5500
4500
Square
4500
Round
32.
3500
3500
2500
2500
1500
0.0
0.1
0.2
0.3
Quantile
0.4
0.5
0.0
0.1
0.2
0.3
Quantile
0.4
0.5
(b) (5) Our hypotheses are H 0: R S 0 versus H a : R S 0 . Both sample sizes are small, and since
normality is plausible, we use a t test statistic.
t
( xR xS )
2
sR
nR
2
sS
nS
(3878 3597) 0
2832 390 2
281
.583
481.86
Since this is such a small t-value (close to the center, 0), we would clearly fail to reject H 0 . Thus, it does not
appear that the true average force for round pads differs from that of square pads.
36.
Since the experimental design involved growing each of two types of wheat in different locations, the
locations serve as a blocking variable and the appropriate analysis is a paired t-test. The analysis requires we
compute the difference between Sundance winter and Manitou spring wheat yields for each of the 9 plot
locations.
(a) (3) In order to proceed, we must check the assumption that the 9 sample differences have been randomly
sampled from a population that is normally distributed. A normal probability plot of the 9 differences
suggests that the normality assumption is reasonable, plot not shown.
(b) (5) Let d denote the true average difference in yield for the winter wheat versus the spring wheat. Since we
are interested in determining if the average yield for winter wheat is more than 500 kg/ha higher than for
spring wheat, the relevant hypotheses are:
H 0: d 500 versus H a : d 500 The descriptive statistic for the nine differences are:
n 9 d 782.2 s d 236.7
The corresponding test statistic is:
782.2 500
t
3.58
236.7
9
With df = (n - 1) = 8, the p-value = P(t > 3.6) = .004
Since, p-value = .004 < most choices for (e.g ., .05 or .01) , we reject H0. That is, we have sufficient
evidence to claim that the true average yield for the winter wheat is more than 500 kg/ha higher than for the
spring wheat.
55.
(a) (4 points = 1each)
Sample size, n
100
z value
( x 0 )
=
n
(101100 )
15
=
n /15 =
100 /15 .67
P-value (upper)
.2514
n
400
1600
n /15 =
1600 /15 2.67
.0038
2500
62.
n /15 = 400 /15 1.33
n /15 = 2500 /15 3.33
.0004
.0918
(a) (3) No, it does not appear plausible that the distribution is normal. Notice that the mean value, x = 215, is not
nearly in the middle of the range of values, 5 to 1176. The midrange would be about 585. Since the mean is
so much lower than this, one would suspect the distribution is skewed.
However, it is not necessary to assume normality if the sample size is large enough, due to the central limit
theorem. Since this problem has a sample size which is large enough (i.e., 47 > 30), we can proceed with a test of
hypothesis about the true mean consumption.
(b) (5) Let denote the true mean consumption. Since we are interested in determining if there is evidence to
contradict the prior belief that was at most 200 mg, the following hypotheses should be tested.
H0: = 200 versus Ha: > 200. The value of the test statistic is:
x 200 215 200
z
.44
s
235
n
47
The corresponding p-value = P(z > .44) = .33.
Since p-value = .33 > most any choice of , we fail to reject H0. There is insufficient evidence to suggest
that the true mean caffeine consumption of adult women exceeds 200 mg per day.
66.
(6) Let p denote the true proportion of front-seat occupants involved in head-on collisions, in a certain region,
who sustain no injuries. Given the wording of the exercise, the relevant hypotheses are:
H0 :
1
p
3
versus Ha:
1
3
1
p
3
2
3
[Note: 319 106.3 and 319 212.7 are each 5 and 319 > 30].
So, the test statistic is:
z
95 1
p p0 319 3
1.35
1 2
p0 1 p0
n
3 3
319
The corresponding p-value = P(z < -1.35) = .0885
Since p-value = .0885 > =.05, we fail to reject H0. We have insufficient evidence to claim that less than onethird of all such accidents result in no injuries.