There is a new drug that is used to treat leukemia. The following data represents the remission time in weeks for a random sample of 21 patients using the drug.
1. There is a new drug
that is used to treat leukemia. The following data represents the remission
time in weeks for a random sample of 21 patients using the drug.
|
10 |
7 |
32 |
23 |
22 |
6 |
16 |
|
11 |
20 |
19 |
6 |
17 |
35 |
6 |
|
10 |
34 |
32 |
25 |
13 |
9 |
6 |
Let X be a random variable representing
the remission time in weeks for all patients using the new drug. Assume that
the distribution of x is normal. A previously used drug treatment has a mean
remission time of 12.5 weeks. Does the data indicate that the mean remission
time using the new drug is different from 12.5 week at a level of significance
of 0.01?
State the null hypothesis:
A. µ=12.5
B. µ?12.5
C. µ<12.5
D. µ>12.5
Answer: Choose an item.
State the alternative hypothesis:
A. µ=12.5
B. µ?12.5
C. µ<12.5
D. µ>12.5
Answer: Choose an item.
|
Let X be a random variable |
State the level of
significance:
A. 0.001
B. 0.01
C. 0.05
D. 0.10
Answer: Choose an item.
|
Let X be a random variable representing |
State the test
statistic:
A. 0.058
B. 0.552
C. 1.058
D. 2.106
Answer: Choose an item.
Perform calculations
Please write down your solutions or
copy and paste your Excel output here:
Then answer the following two
questions:
Critical value:
A. 0.050
B. 1.960
C. 2.086
D. 2.845
Answer: Choose an item.
P-value:
A. p <0.001
B. 0.001 = p <0.01
C. 0.01 = p <0.05
D. 0.05 = p
Answer: Choose an item.
Statistical Conclusion
A. Reject the null
hypothesis
B. Do not reject the null
hypothesis
Answer: Choose an item.
Experimental Conclusion
A. There is sufficient
evidence to conclude that the mean remission time using the new drug is
different from 12.5 week at a level of significance of 0.01.
B. There is no sufficient
evidence to conclude that the mean remission time using the new drug is
different from 12.5 week at a level of significance of 0.01.
Answer: Choose an item.
2. We wish to test the
claim that the mean body mass index (BMI) of men is equal to the mean BMI of
women. Use the data below to test this claim.
|
Men |
Women |
|
20 |
29 |
|
37 |
28 |
|
46 |
20 |
|
23 |
28 |
|
20 |
42 |
|
23 |
45 |
|
21 |
19 |
|
15 |
45 |
|
20 |
16 |
|
28 |
32 |
|
27 |
38 |
|
20 |
45 |
|
30 |
41 |
|
22 |
34 |
|
27 |
28 |
|
38 |
21 |
|
29 |
42 |
|
20 |
21 |
|
16 |
30 |
|
27 |
28 |
|
42 |
30 |
|
37 |
43 |
|
39 |
40 |
|
39 |
16 |
|
32 |
44 |
|
16 |
15 |
|
21 |
16 |
|
26 |
20 |
|
17 |
41 |
|
39 |
16 |
State the Null Hypothesis
A. µ1 = µ2
B. µ1 ? µ2
C. µ1 > µ2
D. µ1 < µ2
Where µ1 and µ2 are the mean body mass index
for men and women, respectively.
Answer: Choose an item.
State the alternative hypothesis:
A. µ1 = µ2
B. µ1 ? µ2
C. µ1 > µ2
D. µ1 < µ2
Answer: Choose an item.
State the Level of significance
State the level of significance:
A. 0.001
B. 0.01
C. 0.05
D. 0.10
Answer: Choose an item.
State the test statistic (its absolute
value, for example the absolute value of -1.5 is 1.5):
A. 0.058
B. 0.515
C. 1.273
D. 2.108
Answer: Choose an item.
Perform calculations
Please write down your solutions or
copy and paste your Excel output here:
Then answer the following two questions:
Critical value:
A. 0.050
B. 1.960
C. 2.002
D. 2.045
Answer: Choose an item.
P-value:
A. p <0.001
B. 0.001 = p <0.01
C. 0.01 = p <0.05
D. 0.05 = p
Answer: Choose an item.
Statistical Conclusion
A. Reject the null
hypothesis
B. Do not reject the null
hypothesis
Answer: Choose an item.
Experimental Conclusion
There is sufficient evidence to conclude that the
mean body mass index (BMI) of mean is
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