Question

In a random sample of 32 large cap mutual funds, the average YTD return is 3.05% with a standard deviation of 9.77%. Conduct by hand an appropriate hypothesis test at the 5% level to test whether the average YTD return for a large cap mutual fund is positive. (Again, that means to write the hypotheses, compute the test statistic, and estimate the p-value.) Give your test conclusion in terms of the question being asked. Confirm your test statistic and result with Minitab. Also, interpret the associated confidence bound for this test.

Answer 1

Answer:

$t=\frac{3.05-0}{\frac{9.77}{\sqrt{32}}}=1.766$    

$p_v =P(t_{(31)}>1.766)=0.0436$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 0% at 5% of signficance.  

The output for Minitab is on the figure attached.

The confidence bound is given as 0.12 represent the value asociated to the one tailed confidence interval calculated:

$\bar X -t_{\alpha/2} \frac{s}{\sqrt{n}}$

$3.05 -1.695 \frac{9.77}{\sqrt{32}}= 0.123$

So the minimum value that we expect for the return at 95% of confidence is 0.12 %

Explanation:

Data given and notation  

$\bar X=3.05$ represent the sample mean in%

$s=9.77$ represent the sample standard deviation

$n=32$ sample size  

$\mu_o =0$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean for the YTD return for a large cap is positive, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 0$  

Alternative hypothesis:$\mu > 0$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{3.05-0}{\frac{9.77}{\sqrt{32}}}=1.766$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=32-1=31$  

Since is a one side test the p value would be:  

$p_v =P(t_{(31)}>1.766)=0.0436$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 0% at 5% of signficance.  

The output for Minitab is on the figure attached.

The confidence bound is given as 0.12 represent the value asociated to the one tailed confidence interval calculated:

$\bar X -t_{\alpha/2} \frac{s}{\sqrt{n}}$

$3.05 -1.695 \frac{9.77}{\sqrt{32}}= 0.123$

So the minimum value that we expect for the return at 95% of confidence is 0.12 %