Homework
#6
Answers
4.1(a). Null hypothesis: Last night’s game was actually
an NHL
hockey game.
4.1(b). On the basis of the null hypothesis, I would expect
between 0 and 6 points maximum per team.
The number of points scored was substantially larger than 6 points per
team so it would be highly unlikely that scoring this many points would occur
were this a hockey game (as I assumed in the null hypothesis). Therefore, I would reject the null hypothesis
and conclude that the scores reported must have been for something other than a
hockey game, which explains why I believe the person was mistaken about reading
those scores for a hockey game.
4.2(b). No, $4.25 is a common observation, so it is
unlikely that
you were overcharged for lunch.
4.2(c). I set up the null hypothesis that I was
charged
correctly. Therefore, I would expect to receive about
$1.00 in change, give or take a quarter or so.
The change that I received is in line with that expectation (there is a
31.7% probability that I was not overcharged) and, therefore, I have no basis
for rejecting H_{0}.
4.3.
A type I error would be concluding that
I had been shortchanged when in fact I had not.
4.4.
A type II error would be concluding
that I had not been shortchanged when in fact I had.
4.5.
The critical value would be that amount
of change below which I would decide that I had been shortchanged (or the price
above which I would decide that I had been overcharged). The rejection region would be all amounts
less than the critical value in the case of getting too little change (or above
the critical value in the case of getting overcharged)—i.e., all amounts that
would lead to rejection of H_{0}.
1. The sample
with the mean of 40 would have a mean
that more
closely approximates µ The
larger the n, the less of a problem error (or random noise) will be. Since error is assumed to be random, the
larger your n, the more likely you will have an extreme observation on one end
and an extreme observation on the other end of a distribution—thus, the noise
and error are assumed to balance out (equal zero) with a great enough n and the
greater then, the closer to 0 error will be.
2. µ=600
σ=100
n=65
SE=_{}
The
standard error you could expect with a sample of 65 scores would be 12.40.
3(a). µ=250
σ=20
95%→1.96
X=250+(1.96)(20)=289.2
X=250-(1.96)(20)=210.8
0
The 95% confidence interval for the
distribution would be 210.8 to 289.2.
3(b). µ=250
σ=20
80%→1.28
X=250+(1.28)(20)=275.6
X=250-(1.28)(20)=224.4
The 80% confidence interval for this
distribution would be 224.4 and 275.6.
4. µ=80
99%→2.58
Upper
bound=97
97=80+(2.58)(σ)
17=2.58 σ
σ=_{}
X=80-(2.58)(6.589)=63
The standard deviation is 6.589 and the
lower bound of the 99% confidence interval is 63.
5.
X_{A} |
X_{freq}_{ A} |
X_{weighted}_{} |
_{}(X_{A}-_{})^{2} |
20 |
2 |
40 |
2(28.09) |
19 |
1 |
19 |
18.49 |
18 |
1 |
18 |
10.89 |
17 |
5 |
85 |
5(5.29) |
16 |
1 |
16 |
1.69 |
15 |
1 |
15 |
.09 |
14 |
1 |
14 |
.49 |
13 |
1 |
13 |
2.89 |
12 |
3 |
36 |
3(7.29) |
10 |
2 |
20 |
2(22.09) |
9 |
2 |
18 |
2(32.49) |
N=20 _{} _{}
_{}
σ_{A}_{=}
_{ }
Mode=17
STEM | LEAF
0 | 99
1*| 00
1t| 2223
1f| 45
1s| 677777
1.| 89
2*| 00
Figure
5. Stem and leaf plot of the population data.
If you did
histograms or frequency distributions, this is good, but I was looking more for
stem-and-leaf and boxplots. [I did not display them because there are so
many different ways to group them/graph them appropriately].
Median location=_{}
Median=_{}
Hinge location=_{}
Lower hinge=12
Upper hinge=17
H-spread=17-12=5
H-spread*1.5=5(1.5)=7.5
Upper fence=17+7.5=24.5
Lower fence=12-7.5=4.5
Lower adjacent value=9
Upper adjacent value=20
4 6 8 10 12 14
16 18 20
22 24 26
---------------------------------------------
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| |---------| |
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| |---------| |
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Figure 5(b). Boxplot of population data.
5(b).
_{}
_{}
z=_{}
z=1.22→2(.1112)=.2224
There
is a 22.24% likelihood of getting a score as extreme as 19 in this population.
Z=_{}
z=-3.32→2(.0006)=.0012
There
is a .12% likelihood of getting a score as extreme as 3 in this population.
Z=_{}
Z=3.77→2(.0001)=.0026
There
is a .02% likelihood of getting a score as extreme as 28 in this population.