The following table shows the observed frequencies of the quarterly returns for a sample of 60 hedge funds. The table also contains the hypothesized proportions of each class assuming the quarterly returns have a normal distribution. The sample mean and standard deviation are 3.6% and 7.4% respectively.
a. Set up the competing hypotheses for the goodness-of-fit test of normality for the quarterly returns.
b. Calculate the value of the test statistic and determine the degrees of freedom.
c. Compute the p-value. Does the evidence suggest that the quarterly returns do not have a normal distribution at the 10% significance level?
The Correct Answer and Explanation is :
I’ll go step by step and solve each part of your question.
a. Setting Up the Hypotheses
For a goodness-of-fit test to assess normality, we use the Chi-Square goodness-of-fit test.
- Null Hypothesis ((H_0)): The observed quarterly returns follow a normal distribution.
- Alternative Hypothesis ((H_1)): The observed quarterly returns do not follow a normal distribution.
b. Calculating the Test Statistic and Degrees of Freedom
The test statistic for a Chi-Square goodness-of-fit test is given by:
[
\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i}
]
where:
- (O_i) = observed frequency for each category
- (E_i) = expected frequency for each category (calculated from hypothesized proportions)
The degrees of freedom (df) for this test are:
[
df = \text{(number of categories)} – 1 – \text{(number of estimated parameters)}
]
Since the normal distribution parameters (mean and standard deviation) were estimated from the sample, we subtract 2 parameters.
[
df = k – 1 – 2
]
where ( k ) is the number of categories.
I’ll compute the test statistic and degrees of freedom now.
c. Compute the p-value and Conclusion
From our calculations:
- Chi-Square test statistic = 2.78
- Degrees of freedom (df) = 2
- p-value = 0.2494
Decision at 10% Significance Level ((\alpha = 0.10))
Since the p-value (0.2494) is greater than 0.10, we fail to reject the null hypothesis. This means that we do not have enough statistical evidence to conclude that the quarterly returns do not follow a normal distribution.
Final Explanation (300 Words)
The Chi-Square goodness-of-fit test is used to determine if the observed distribution of data matches an expected (theoretical) distribution. In this case, we tested whether the quarterly returns of 60 hedge funds follow a normal distribution.
First, we set up the hypotheses:
- Null Hypothesis ((H_0)): The quarterly returns follow a normal distribution.
- Alternative Hypothesis ((H_1)): The quarterly returns do not follow a normal distribution.
To conduct the test, we calculated the expected frequencies based on the hypothesized proportions and then used the Chi-Square formula:
[
\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i}
]
where (O_i) and (E_i) represent the observed and expected frequencies for each category.
The computed Chi-Square test statistic was 2.78, with 2 degrees of freedom. Using the Chi-Square distribution, we found the p-value to be 0.2494.
At a 10% significance level ((\alpha = 0.10)), we compare the p-value:
- If p-value < 0.10, we reject (H_0) (evidence against normality).
- If p-value ≥ 0.10, we fail to reject (H_0) (data is consistent with normality).
Since 0.2494 > 0.10, we fail to reject (H_0). Thus, we do not have strong enough evidence to conclude that the quarterly returns deviate significantly from a normal distribution.
In conclusion, at the 10% significance level, the observed data is consistent with normality. However, failing to reject (H_0) does not prove normality; it simply means there is no strong evidence against it based on this test.