new drug that is used to treat leukemia accounting homework help

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new drug that is used to treat leukemia accounting homework help

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. Âµ=12.5
B. Âµâ‰ 12.5
C. Âµ<12.5
D. Âµ>12.5
Answer: Choose an item.
State the alternative hypothesis:
A. Âµ=12.5
B. Âµâ‰ 12.5
C. Âµ<12.5
D. Âµ>12.5
Answer: Choose an item.

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 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 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 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. Î¼1 = μ2
B. Î¼1 â‰  μ2
C. Î¼1 > μ2
D. Î¼1 < 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. Î¼1 = μ2
B. Î¼1 â‰  μ2
C. Î¼1 > μ2
D. Î¼1 < 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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