Answer:
There is sufficient evidence to conclude that the fund has moderate risk at the 0.05 level of significance
Explanation:
Given:
Sample size = 27months
n = 27
Let sigma = population standard deviation
Sigma = 3% = 0.03
s = sample standard deviation
s =2.19% = 0.0219
Alpha = 0.05
The mutual fund manager claims that his fund has moderate risk.
Hypothesis:
H0: Population standard deviation is equal to sample standard deviation.
H0 = sigma = 0.03
H1: Population standard deviation is not equal to sample standard deviation.
Alternative hypothesis:
H0< sigma = H0 <0.03
Degree of freedom = n-1 = 27-1 = 26
We would use the chi-square test to check if the standard deviation of a population is equal to a particular value
The test statistic:
x^2 = [(n-1)s^2]/ sigma^2
x^2 = [(27-1)*0.0219^2]/0.03^2
x^2 = [(26)*0.00047961]/0.03^2
x^2 = (0.01246986/0.03^2)
x^2 = 0.01246986/ 0.0009
x^2 = 13.8554
You can compare the value of your x^2 to a (degrees of freedom, alpha) chi squared critical value from a table (you can find such table online) or use an EXCEL formula
Using EXCEL formula to find p-value:
CHISQ.DIST(13.8554, 26, TRUE)
Type in the value of chi square as x,
Degree of freedom as = 26 and
TRUE (cumulative distribution function)
p-value = 0.025143
The level of significance = Alpha = 0.05
The decision rule states:
If p-value is less than or equal to level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.
Since p-value is less than level of significance, we would reject the null hypothesis.
0.025143 < 0.05
Conclusion:
There is sufficient evidence to conclude that the fund has moderate risk at the 0.05 level of significance.
