2. Suppose we have a stock trader who is interested to see if a new trading strategy will earn her more money. In order to see if the new strategy works, she will collect data on her profit 60 days before the change and 60 days after the change. She’s been trading for a long time, so she knows the population standard deviation of profit: o1= $22 (after the new strategy), and o2=$31 (before the new strategy). Sample 1 (after the new strategy) Sample 2 (before the new strategy) n1=60 n2=60 x1=$315 x2=$299 What is the point estimate for the difference in profit before and after the new strategy? a. $16 b. $21 c. $8 d. None of the above. 3. Calculate a 95% interval estimate for the difference. What is the margin of error? a. 16 b. 4.91 c. 10.12 d. 9.62 4. What is the result for the 95% interval estimate? (Note that with rounding you may be off by a few decimals, just choose the closest one) a. None of the above. b. $11.09 to $20.91 c. $6.38 to $25.62 d. $10.88 to $31.12 5. Now let's suppose the trader wants to be sure her new strategy is better, i.e., she will make more profit by switching to the new strategy. What would the hypothesis test for this look like? When necessary, use a level of significance = 0.05. a. H0: u1> u2 ha: u1 =< u2 b. h0: u1 <= u2 ha: u1> u2 c. h0: u1-u2<=0 ha: u1-u2 >0 d. (b) and (c) 6. Using the same information above: What is the value of test statistic? a. 3.26 b. None of the above. c. 1.28 d. 1.96 7. What is the p-value? a. 0.05 b. 0.0006 c. 0.0003 d. 0.9994 e. 0.0005 f. None of the above 8. Assuming the test statistic is less than than the critical value, what is your conclusion? a. Reject the null hypothesis b. Reject the alternative hypothesis c. Fail to reject the null hypothesis d. Both b. and c. are correct 9. Assuming we reject the null hypothesis in the test above above, what is your interpretation of this result? a. The old strategy earns more profit than the new strategy. b. There is no difference in profit between strategies. c. The new strategy earns more profit than the old strategy. d. There is no change in profit by switching strategies.
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
2. Suppose we have a stock trader who is interested to see if a new trading strategy will earn her more money. In order to see if the new strategy works, she will collect data on her profit 60 days before the change and 60 days after the change. She’s been trading for a long time, so she knows the population standard deviation of profit:
o1= $22 (after the new strategy), and o2=$31 (before the new strategy).
|
Sample 1 (after the new strategy) |
Sample 2 (before the new strategy) |
|
n1=60
|
n2=60
|
|
x1=$315
|
x2=$299
|
What is the point estimate for the difference in profit before and after the new strategy?
| a. |
$16 |
|
| b. |
$21 |
|
| c. |
$8 |
|
| d. |
None of the above. |
3. Calculate a 95%
| a. |
16 |
|
| b. |
4.91 |
|
| c. |
10.12 |
|
| d. |
9.62 |
What is the result for the 95% interval estimate? (Note that with rounding you may be off by a few decimals, just choose the closest one)
| a. |
None of the above. |
|
| b. |
$11.09 to $20.91 |
|
| c. |
$6.38 to $25.62 |
|
| d. |
$10.88 to $31.12 |
5.
Now let's suppose the trader wants to be sure her new strategy is better, i.e., she will make more profit by switching to the new strategy. What would the hypothesis test for this look like? When necessary, use a level of significance = 0.05.
| a. |
H0: u1> u2 ha: u1 =< u2 |
|
| b. |
h0: u1 <= u2 ha: u1> u2 |
|
| c. |
h0: u1-u2<=0 ha: u1-u2 >0 |
|
| d. |
(b) and (c) |
6. Using the same information above: What is the value of test statistic?
| a. |
3.26 |
|
| b. |
None of the above. |
|
| c. |
1.28 |
|
| d. |
1.96 |
7. What is the p-value?
| a. |
0.05 |
|
| b. |
0.0006 |
|
| c. |
0.0003 |
|
| d. |
0.9994 |
|
| e. |
0.0005 |
|
| f. |
None of the above |
8. Assuming the test statistic is less than than the critical value, what is your conclusion?
| a. |
Reject the null hypothesis |
|
| b. |
Reject the alternative hypothesis |
|
| c. |
Fail to reject the null hypothesis |
|
| d. |
Both b. and c. are correct |
9. Assuming we reject the null hypothesis in the test above above, what is your interpretation of this result?
| a. |
The old strategy earns more profit than the new strategy. |
|
| b. |
There is no difference in profit between strategies. |
|
| c. |
The new strategy earns more profit than the old strategy. |
|
| d. |
There is no change in profit by switching strategies. |
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