An investor with a stock portfolio worth several hundred thousand dollars sued his broker and brokerage firm because he felt tha
t lack of diversification in his portfolio led to poor performance for many years in a row. In an effort to avoid close public scrutiny, the firm agreed to settle the conflict by an arbitration panel. The arbitration panel compared a sample of 39 months of the investor's returns with the average of the Standard & Poor's 500-stock index for the same period in order to determine whether there was a substantial decrease. Historically, the S&P has a mean return of 0.95. Suppose that you are a member of this arbitration panel. Conduct a hypothesis test to determine if the investor's portfolio performed significantly worse than the performance of the S&P 500. Use a level of significance of alpha=0.05 . Open the data file Rates of Return. Use this information to answer questions 1 through 5.
1. Give a 95% confidence interval for the true mean return of the investor's portfolio. Round your answer accurate to three decimal places in interval notation. Be sure to put the lower bound in the first box and the upper bound in the second. [Example: (42.335, 54.859)]
( , )
2. Is the alternative hypothesis for this test one-tailed or two-tailed?
a. One-tailed
b. Two-tailed
3. What is the t-score for this test? Give your answer accurate to three decimal places. (Example: -3.234)
4. What is the P-value for this test? Give your answer accurate to three decimal places. (Example: 0.034)
5. Based on the results of this test, is there enough evidence to say that the investor's portfolio performed significantly worse than the S&P 500?
a. Yes, because we rejected the null.
b. Yes, because we failed to reject the null.
c. No, because we rejected the null.
d. No, because we failed to reject the null.
1. Give a 95% confidence interval for the true mean return of the investor's portfolio. Round your answer accurate to three decimal places in interval notation. Be sure to put the lower bound in the first box and the upper bound in the second. [Example: (42.335, 54.859)]
( , )
2. Is the alternative hypothesis for this test one-tailed or two-tailed?
a. One-tailed
b. Two-tailed
3. What is the t-score for this test? Give your answer accurate to three decimal places. (Example: -3.234)
4. What is the P-value for this test? Give your answer accurate to three decimal places. (Example: 0.034)
5. Based on the results of this test, is there enough evidence to say that the investor's portfolio performed significantly worse than the S&P 500?
a. Yes, because we rejected the null.
b. Yes, because we failed to reject the null.
c. No, because we rejected the null.
d. No, because we failed to reject the null.
1 answer:
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Answer:
The expected value of the sample proportion is of 0.07.
1. P(p < .02) = 0.0823
2. P(p > .15) = 0.0132
3. P(.05 < p < .09) = 0.4176
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
7% of all registered voters belong to the Green party. 50 voters:
This means that
So, for the normal distribution:
The expected value of the sample proportion is of 0.07.
1. Determine P(p < .02).
This is the pvalue of Z when X = 0.02. So
By the Central Limit Theorem
has a pvalue of 0.0823
So
P(p < .02) = 0.0823
2. Determine P(p > .15).
This is 1 subtracted by the pvalue of Z when X = 0.15. So
has a pvalue of 0.9868
1 - 0.9868 = 0.0132
So
P(p > .15) = 0.0132
3. Determine P(.05 < p < .09).
This is the pvalue of Z when X = 0.09 subtracted by the pvalue of Z when X = 0.05. So
X = 0.09
has a pvalue of 0.7088
X = 0.05
has a pvalue of 0.2912
0.7088 - 0.2912 = 0.4176. So
P(.05 < p < .09) = 0.4176