1.
DDT is a pesticide banned in the United States for its danger to humans and animals. In an experiment on the impact of DDT, six rats were exposed to DDT poisoning and six rats were not exposed. For each rat in the experiment, a measurement of nerve sensitivity was recorded. The researchers suspected that the mean nerve sensitivity for rats exposed to DDT is greater than that for rats not poisoned. The data follow:
Poisoned rats: 
12.207 
16.869 
25.050 
22.429 
8.456 
20.589 
Unpoisoned rats: 
11.074 
9.686 
12.064 
9.351 
8.182 
6.642 
Let μ _{1} be the mean nerve sensitivity for rats poisoned
with DDT. Let μ _{2} be the mean nerve sensitivity for rats
not poisoned with DDT.
The (Option 1) less conservative computed degrees of freedom
is:
A. 
7 

B. 
4. 

C. 
5. 

D. 
6. 
2,
An SRS of 100 flights of a large airline (airline 1) showed that 64 were on time. An SRS of 100 flights of another large airline (airline 2) showed that 80 were on time. Let p _{1} and p _{2} be the proportion of all flights that are on time for these two airlines. Using a 0.01 level of significance, what conclusion would you draw from this test?
A. 
accept the null hypothesis 

B. 
fail to reject the null hypothesis; there is not strong enough evidence 

C. 
reject the null hypothesis; the values for p_{1} and p_{2} indicate that there is strong enough evidence 

D. 
reject the null hypothesis; the Pvalue indicates that there is strong enough evidence 
3.
Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24hour period were recorded. The data follow.
Treatment 
Minutes of rest 
Control 
450 
413 
418 
Low dose 
466 
422 
435 
Medium dose 
421 
453 
419 
High dose 
364 
330 
389 
Assume the data are four independent SRSs, one from each of the
four populations of caffeine levels, and that the distribution of
the yields is Normal.
A partial ANOVA table produced by Minitab follows, along with the
means and standard deviation of the yields for the four
groups.
The null hypothesis for the ANOVA Ftest is that the
population mean rest is:
A. 
the same for all four levels of caffeine. 

B. 
increasing as the caffeine level gets larger. 

C. 
decreasing as the caffeine level gets larger. 

D. 
largest for the high level of caffeine. 
4.
In the United States, there has historically been a strong relationship between smoking and education, with welleducated people less likely to smoke. To examine whether this pattern has changed, a sample of 459 men was selected at random from those who had visited a health center for a routine checkup over the course of the past year. Education is classified into three categories corresponding to the highest level of education achieved, and smoking status is classified into four categories.
Smoking Status 
Education 
Nonsmoker 
Former 
Moderate 
Heavy 
Total 
High School 
56 
54 
41 
36 
187 
College 
37 
43 
27 
32 
139 
Graduate 
53 
28 
36 
16 
133 
Total 
146 
125 
104 
84 
459 
This is an r × c table. The number c has
value:
A. 
6. 

B. 
2. 

C. 
4. 

D. 
3 
1. The degrees of freedom is given by the formula is given by:
, where 1 denotes the sample estimates for poisoned rats and 2 denote the sample estimates for nonpoisoned rats. Here n1n2=6, s1=6.34 and s2= 1.95. Using these values we get df= 5.94. The nearest integer to this is 6.
Thus the correct option is D. 6
2. The pvalue of the test is 0.012. Thus the answer is A. accept the null hypothesis
3. A. the same for all four levels of caffeine
4. c is the number of columns. Thus the answer is D. 3
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