QSCI 482 HW 3 DUE
1. Refer back to the example from Topic 1, "The Language of Singles
Bars". We already showed that the data do NOT conform to the null
hypothesis of uniformity (we rejected that null hypothesis at the .05
level of significance). Now use statistical partitioning (subdividing)
to show that while two of the categories "look like each other"
(uniformity between those two categories), they are different from the
third category in terms of frequency of counts. Be sure to include the
final step--showing that your chisquare test statistics from the
subdivisions add up [approximately] to the original, overall test
statistic; and that the df add up also.
Type Observed Expected
Compliments 85 106.666
Declarations 122 106.666
Questions 113 106.666
Partitioning fobs fexp (fobs-fexp)2/fobs
D 122 117.5 0.172
Q 113 117.5 0.172
Total 235 235
c2obs = 0.344
c2(0.05,1 df) = 3.841
…There
isn’t a statistically significant difference between declarations and
compliments at the 0.05 level of significance with 1 degree of freedom.
D+Q 235 213.333 2.201
C 85 106.666 4.401
Total 320 320
c2obs = 6.602
c2(0.05,1 df) = 3.841
…There is
a significant difference between compliments and the other two types at the
0.05 level of significance with 1 degree of freedom.
Note: The two observed c2 values above, 0.344 and 6.602 add up to 6.946, which is
pretty close to the original observed statistic from topic 1, 6.98, and the
degrees of freedom above add up to 2, as in the original problem.
2. A scientist has run a bioassay experiment, where each organism (say,
an amphipod) is exposed to a toxicant for a specified period of time
(96-hour bioassays are quite common). At the end of the 96 hours, the
status (alive or dead) of each organism is noted. There are 50
organisms in each group. The data are as follows:
Expected
TREATMENT: A B C D Total A B C D Total
Dead 01 10 15 18 44 11 11 11 11 44
Alive 49 40 35 32 156 39 39 39 39 156
Total 50 50 50 50 200
At the .05 level of significance, test the null hypothesis of "no effect
of treatment type on mortality" for the 4 treatments. Then, go through
the process of partitioning (subdivision) to get more information from
the original test result and to see which of the treatment groups can
be combined and which seem different.
H0: There is no difference in the effects of the
four treatment types on mortality.
Ha: There is some difference in the effects of
the treatment types.
Ho: p11
= p12 = p13 = p14
Ha: There is some inequality involved
a = 0.05 df =
(r-1)(c-1) = (2-1)(4-1) = 3
c2obs = (1-11)2/11 + (10-11)2/11 + (15-11)2/11
+ (18-11)2/11 + (49-39)2/39
+ (40-39)2/39 + (35-39)2/39 +
(32–39)2/39
= 9.091 + .091 + 1.455 + 4.455 + 2.564
+ 0.026 + 0.410 + 1.256
= 19.348
c2(0.05,3) = 7.815
Reject Ho: Conclude that there is a statistical
difference in the effects of the treatment types
Deviations Stdz
Treatment: A B C D A B C D
Dead -10 -1 +4 +7 -3.02 -0.30 1.21 2.11
Alive +10 +1 -4 -7 1.60 0.16 -0.64 -1.12
Let’s test to see
whether B, C and D might be combined
Obs: B C D Total Exp B C D Total
Dead 10 15 18 43 Dead 14.3 14.3 14.3 43
Alive 40 35 32 107 Alive 35.67 35.67 35.67
107
Total 50 50 50 150 Total 50 50 50 150
c2obs = 1.293+0.0343+0.09573+0.5256+0.01258+0.3776
= 3.2004
c2(0.05,2) = 5.991
The statistically
similar results of B, C and D indicate that they can be combined as one group
So, let’s compare A as a single group against
B,C and D as a pooled group.
Obs: A B,C&D Total Exp: A B,C&D Total
Dead 01 43 44 11 33 44
Alive 49 107 156 39 117 156
Total 50 150 200 50 150 200
c2obs = 3.0303+9.091+0.8547+2.5641
= 15.5401
c2(0.05,1) = 3.84
We do reject Ho here,
and conclude that A is distinct from B,C,D.
And, c2obs of 15.54 + c2obs of 3.20 = 18.74, which is roughly equal to our original
19.348. The df add up as well (1+2 = 3).
3. The population of body weights for a small mammal is normally
distributed with a mean of 64.0 g [grams] and a standard deviation of 12.2 g.
a. What is the probability that an individual drawn at random from this
population has a weight of at least 60 g?
m = 64.0 s =
12.2 x = 60
Z = x-m / s = 60-64 / 12.2 =
-0.328
Zval for 0.33 = 0.3707,
therefore 0.3707 probability that an individual drawn at random would have a
weight less than 60 grams, so there
is an 0.6293 probability of an
individual weighing 60 or more grams being drawn randomly.
b. What proportion of this population has a weight between 58 and 68 g?
Z58 = 58-64 /
12.2 =
-0.492 From table B.2 this has a
prob. val. of 0.3121 (prob. of value falling in the left tail of the
distribution)
Z68 = 68-64 /
12.2 = 0.329 From table B.2 this has a
prob. value of 0.3707 (prob. of value falling in the right tail of the
distribution)
Therefore,
1.0000-0.3121-0.3707 = the probability of a value falling between these two
amounts in the distribution.
Pr (58<x<68) = 0.3172 31.72% of the population weighs between
58 and 68g.
c. If a random sample of size 12 is drawn from this population, what is
the probability that the sample average will be between 59 and 65 g?
sXbar = sqrt(s/n) = sqrt
(12.2/sqrt 12) = 3.52
Pr (Xbar≤59) = Pr
[ Z ≤ (59-64.0)/3.52 ] = Pr(Z≤-1.42) = 0.0778
Pr (Xbar≥65) = Pr
[Z ≥ (65-64.0)/3.52 ] = Pr(Z≥.2841) = 0.3897
Pr
(59≤Xbar≤65) = 1.000-0.0778-0.3897 = 0.5325
d. How large a sample size would one have to take to end up with a
standard error of the mean no greater than 1.0 g?
sxbar = sqrt(s2/n) sxbar2 = s2/n 1.0 =
148.84/n n must be at least
149