The expense ratio (in percentage) for large-cap growth mutual funds have the following sample data0.52, 1.06, 1.26, 2.17, 1.55, 0.99, 1.10, 1.07, 1.81, 2.05 0.91, 0.79, 1.39, 0.62, 1.52, 1.02, 1.10, 1.78, 1.01, 1.15The data is believed to be normally distributed. a. Is there compelling evidence for concluding that the population mean expense ratio exceeds 1%? Carry out a test of the relevant hypothesis using a significance level 0.01.b. Referring back to part (a), describe in context type I and II errors and say which error you might have made in reaching your conclusion. The source from which the data was obtained reported that ? = 1.33 for the population of all 762 such funds. So did you actually commit an error in reaching your conclusion?c. Suppose that ? = 0.5, determine and interpret the power of the test in part (a) for the actual value of ? stated in (b).

Part a:  Since the calculated value of t = 2.412 does not fall in the critical region t > ± 2.539 we conclude that population mean expense ratio does not exceed 1% at 0.01 significance level.

Part b:  Yes we accepted the null hypothesis when it is false so we made a type II error.

Part c: Power = 0.6591=0.66

Step-by-step explanation:

Part a:

Let the null and alternate hypotheses be

H0: u ≤ 1 against the claim Ha: u > 1

The significance level is ∝=0.01

As the standard deviation of the population is not given we use t- test which has n-1= 20-1= 19 degrees of freedom

This is one tailed test the critical region is t > ± 2.539

The mean is calculated to be 1.24 and standard deviation of the sample = s= 0.4486

We take population mean = u= 1

The test statistic is

t = x`-u/ s/ √n

t= 1.24-1 / 0.445/√20

t= 0.24/0.0995

t= 2.412

Since the calculated value of t = 2.412 does not fall in the critical region t > ± 2.539 we conclude that population mean expense ratio does not exceed 1% at 0.01 significance level.

Part b:

Type I errors are those errors when the null hypothesis is true and is rejected.

Type II errors are those in which the null hypothesis is false and is accepted.

If we had rejected the null hypothesis as it was true then we would have made a type I error.

Population mean = u = 1.33

And our null hypothesis is u= 1

Yes we accepted the null hypothesis when it is false so we made a type II error.

Part c:

When σ= 0.5

For Part a: μ= 1.33

Power = P (reject H0/ H0 is false)

= 1-β

P (reject H0/ H0 is false) = P (z > 2.539+ (u`-u0) / σ*√n)

= P (z > 2.539+ (1-1.33) /0.5 *√20)

=P (z > 2.539-2.9516)

= P (z > 0.4126)

= 0.5 - P (z= 0.2938)                                         from the area of table.

= 0.5- 0.1591

=0.3409

Power = 1-β = 1- 0.3409= 0.6591=0.66