Download Assignment 18 9.4.4 Let WD denotes week day (M

Assignment 18
9.4.4
Let WD denotes week day (M-F), WE denotes week end (Sa-Su.
Assumptions:
1. All expected frequencies are 1 or greater
2. At most 20% of the expected frequencies are less than 5.
3. Simple Random Sample
STEP 1:
HO: The proportion of child births occurring during the week is 5/7
HA: The proportion of child births occurring during the week is not 5/7.
STEP 2:
Births
Observed
Proportion Expected
o-e
WD
WE
Total
716
216
932
5/7
2/7
50.2857
-50.2857
665.7143
266.2857
(o-e)^2
Chi-Square subtotal
(o-e)^2/e
2528.6516 3.7984
2528.6516 9.4960
13.2944
STEP 3: All of the expected frequencies are greater than one and none of the expected
frequencies are less than 5
STEP 4:   0.05
STEP 5: From the Table in Step 2  s 2  13.2944
STEP 6: The distribution has 2-1=1 degrees of freedom, from Table 9 the p-value <0.005 since
 0.005  7.879
STEP 7: The p-value is less than the significance level since   0.05 so we do not reject the
null hypothesis
STEP 8: Since the p-value is less than 0.005, the null hypothesis is rejected. From this study,
there is sufficient evidence to support that the proportion of births during the week is not 5/7.
9.4.5
Let WF denote white feathers, DF denote dark feathers, SC denote small comb, and LC denote
large comb.
Assumptions:
4. All expected frequencies are 1 or greater
5. At most 20% of the expected frequencies are less than 5.
6. Simple Random Sample
STEP 1:
HO: The distribution of the white feathers & small comb, white feathers & large comb, dark
feathers & small comb, and dark feathers & large comb for the chick offspring are consistent
with the Mendelian expected ratios of 9:3:3:1
HA: The distribution of the chick offspring is not consistent with the Mendelian expectations.
STEP 2:
Chicks
WF & SC
WF &
LC
DF & SC
DF & LC
Total
Observed
Proportion Expected
o-e
(o-e)^2
111
37
9/16
3/16
106.875
35.625
4.125
1.375
17.01563
1.890625
34
8
190
3/16
1/16
35.625
11.875
-1.625
-3.875
2.640625
15.01563
Chi-Square subtotal
(o-e)^2/e
0.159211
0.05307
0.074123
1.264474
1.550877
STEP 3: All of the expected frequencies are greater than one and none of the expected
frequencies are less than 5
STEP 4:   0.10
STEP 5: From the Table in Step 2  s 2  1.550877
STEP 6: The distribution has 4-1=3 degrees of freedom, from Table 9 the p-value >0.2 since
 0.2  4.64
STEP 7: The p-value is greater than the significance level since   0.10 so we do not reject the
null hypothesis
STEP 8: Since the p-value is greater than 0.2, there do not reject the null hypothesis. From this
study, there is not sufficient evidence to support that the chick offspring are not consistent with
the Mendelian distribution for feather color and comb height.
9.4.8
Assumptions:
1. All expected frequencies are 1 or greater
2. At most 20% of the expected frequencies are less than 5.
3. Simple Random Sample
STEP 1:
HO: All of the rats in each triplet are equally likely to develop cancer so the drug will not cause
the treated rat to develop cancer prior to the other two control rats in each triplet.
HA: The treated rat in each triplet is more likely to develop cancer from the drug then the two
control rats.
STEP 2:
Rats
Observed
Tumor
12
first in
the
treated rat
Tumor
8
first in
one of the
two
control
rats
Total
20
Proportion Expected
o-e
(o-e)^2
1/3
6.6667
5.3333
28.4441
Chi-Square subtotal
(o-e)^2/e
4.2666
2/3
13.3333
-5.333
28.4441
2.1333
6.3999
STEP 3: All of the expected frequencies are greater than one and none of the expected
frequencies are less than 5
STEP 4:   0.01
STEP 5: From the Table in Step 2  s 2  6.3999
STEP 6: The distribution has 2-1=1 degrees of freedom. From Table 9  0.02  5.41 and
 0.01  6.635 so
1
1
 0.010  p  value   0.02
2
2
0.005  p  value  0.01
STEP 7: The p-value is less than the significance level since   0.01 so we reject the null
hypothesis
STEP 8: Since the p-value is between 0.005 and 0.01, the null hypothesis is rejected. From this
study, there is sufficient evidence to support that the rats that are treated with the drug are more
likely to develop cancer then the rats that are not treated with the drug.
Problem 9.4.10HO: The distribution of black, brown, and white Mongolian gerbil progeny are consistent with
the Mendelian expected ratios of 1:2:1
HA: The distribution of black, brown, and white Mongolian gerbil progeny are not consistent
with the Mendelian expected ratios of 1:2:1
Gerbil
Black
Brown
Observed
40
59
Proportion Expected
1/4
35.25
2/4
70.5
White
Total
42
141
1/4
35.25
 s 2  3.81 . The df=2 and from Table 9,  0.2  3.22 and  0.1  4.61 . With a p-value between
(0.1, 0.2), we do not reject HO there is not sufficient evidence to support that the gerbil progeny
are not consistent with the Mendelian expected ratio.
9.4.12
HO: The inheritance pattern for each of the leaf types of the cowpea plants are Type I=
II=
12
, Type
16
1
3
, and Type III=
16
16
HA: The inheritance pattern for each of the leaf types of the cowpea plants are not Type I=
Type II=
12
,
16
1
3
, and Type III=
16
16
Cowpea plant
Type I
Type II
Type III
Total
Observed
179
44
23
246
Expected
184.5
46.125
15.375
 s 2  4.04 . The df=2 and from Table 9,  0.2  3.22 and  0.1  4.61 . With a p-value between
(0.1, 0.2), we do not reject HO there is not sufficient evidence to support that the cowpea plant
leaf types are not consistent with the probabilities stated.
10.2.7
Condition
Seizure-free
Not Seizure-free
Total
Valproate
6
(6.49)
14
(13.51)
20
Phenytoin
6
(5.51)
11
(11.49)
17
Total
12
25
37
Assumptions
1. All expected frequencies are 1 or greater
2. At most 20% of the expected frequencies are less than 5.
3. Simple random sample
STEP 1:
The null and alternative hypotheses are
H O : The Pr(Seizur e - free | Valproate)  Pr(Seizure - free | Phenytoin )
Patients were seizure-free whether they took valproate or phenytoin
H A : The Pr(Seizur e - free | Valproate)  Pr(Seizure - free | Phenytoin )
Patients had different numbers of seizures whether they took valproate or phenytoin
STEP 2:
The expected frequencies are
12  20
Seizure - free & Valproate 
 6.49
37
12  17
Seizure - free & Phenytoin 
 5.51
37
25  20
Not Seizure - free & Valproate 
 13.51
37
25  17
Not Seizure - free & Phenytoin 
 11.49
37
STEP 3:
The assumptions are met. None of the expected frequencies are 1 or less and none of the
expected frequencies are less than five.
STEP 4:   0.10
STEP 5:
 s2 
 s2
 s2
6  6.49 2  6  5.512  14  13.512  11  11.49 2
6.49
5.51
13.51
11.49
 0.0401  0.0436  0.0178  0.0226
 0.1241
STEP 6:  2 has df  1. The nearest value to the chi - square test statistic is  02.20  1.64 so the
p-value>0.20
STEP 7
P-value
Significance level
p-value > 0.20
>   0.10
Do not reject Ho
STEP 8
With a p-value > 0.20, the null hypothesis is not rejected. The proportion of patients that are
seizure free given they have taken phenytoin or valproate are the same.
10.2.12
Let p denote the probability of having a cold. Let 1 denote that the person has five or fewer
social relationships. Let 2 denote that the person has six or more types of social relationships.
The null and alternative hypotheses are as follows:
H O : The number of social relationsh ips does not affect the cold incidence so p1  p 2
H A : The number of social relationsh ips does affect the cold incidence so p1  p 2
Condition
Cold
No cold
Total
Five or fewer
57
(48.58)
66
(74.42)
123
Six or more
52
(60.42)
101
(92.58)
153
Total
109
167
276
 s2  4.35 With the degrees of freedom being 1, from Table 9 we find that  02.05  3.84 and
 02.02  5.41 . This means that the 0.02 < p-value < 0.05. Since the significance level is   0.05 .
The p-value is less than the significance level so we reject the null hypothesis H O . In
conclusion, the more types of social relationships that a person changes the probability of getting
a cold.
Reject Ho
STEP 6:
With a p  value  0.001 , the null hypothesis is rejected. There is a linear relationship between
fat-free mass and energy expenditure.
b. This study was an observational study. The researchers simply measured both fat-free mass
and energy expenditure of each subject.
c. Since this is an observational study, a cause and effect relationship can not be stated. What can
be stated is that from the observational study there is a statistically significant correlation
between fat free mass and energy expenditure.