Mutual funds are classified as load or n o-load funds. Load...

Mutual funds are classified as load or $n o-$load funds. Load funds require an investor to pay an initial fee based on a percentage of the amount invested in the fund. The no-load funds do not require this initial fee. Some financial advisors argue that the load mutual funds may be worth the extra fee because these funds provide a higher mean rate of return than the no-load mutual funds. A sample of 30 load mutual funds and a sample of 30 no-load mutual funds were selected. Data were collected on the annual return for the funds over a five-year period. The data are contained in the data set Mutual. The data for the first five load and first five no-load mutual funds are as follows. a. Formulate $H_{0}$ and $H_{\text { a such that rejection of } H_{0} \text { leads to the conclusion that the load mutual funds have a higher mean annual return over the five-year period. b. Use the 60 mutual funds in the data set Mutual to conduct the hypothesis test. What is the $p$ -value? At $\alpha=.05,$ what is your conclusion?

Mutual funds are classified as load or $n o-$load funds. Load funds require an investor to pay an initial fee based on a percentage of the amount invested in the fund. The no-load funds do not require this initial fee. Some financial advisors argue that the load mutual funds may be worth the extra fee because these funds provide a higher mean rate of return than the no-load mutual funds. A sample of 30 load mutual funds and a sample of 30 no-load mutual funds were selected. Data were collected on the annual return for the funds over a five-year period. The data are contained in the data set Mutual. The data for the first five load and first five no-load mutual funds are as follows.
a. Formulate $H_{0}$ and $H_{\text { a such that rejection of } H_{0} \text { leads to the conclusion that the load mutual funds have a higher mean annual return over the five-year period.
b. Use the 60 mutual funds in the data set Mutual to conduct the hypothesis test. What is the $p$ -value? At $\alpha=.05,$ what is your conclusion?

Key Concepts

Hypothesis Testing

Hypothesis testing is a systematic method used to decide whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. It involves the formulation of a null hypothesis (which represents no effect or status quo) and an alternative hypothesis (which represents a claim or effect), followed by the use of statistical tests to determine if the observed data are consistent with the null hypothesis.

Null and Alternative Hypotheses

The null hypothesis (H0) represents the default position that there is no difference or no effect, while the alternative hypothesis (Ha) represents the claim that there is a difference or effect. Correctly formulating these hypotheses is critical because they define the statistical test to be performed and the interpretation of the resulting p-value.

Two-Sample t-Test

The two-sample t-test is a statistical method used to compare the means of two independent groups. This test determines whether any observed difference in the sample means is statistically significant, given the variability in the data. It is especially useful when assessing differences between groups on a continuous outcome, assuming the data approximately follow a normal distribution.

p-Value Interpretation

The p-value is the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is true. A lower p-value indicates stronger evidence against the null hypothesis. When the p-value is less than the predetermined significance level, it suggests that the observed data are unlikely to occur if the null hypothesis were true, leading to its rejection.

Significance Level

The significance level, often denoted by ?, is the threshold set before testing to decide when to reject the null hypothesis. It represents the maximum probability of committing a Type I error, which is the error of rejecting a true null hypothesis. Common significance levels are 0.05, 0.01, or 0.10, and setting this level helps maintain consistency and rigor in statistical testing.

Transcript

00:01 The first thing we're asked to do in this problem is come up with the hypotheses so that when the null -reject, null hypothesis is rejected, it leads to the conclusion that the load mutual funds have a higher mean annual return over the five -year period.

00:15 So basically that means that our alternative hypothesis is going to be that the mean for the load mutual funds is higher than the mean for the no -load.

00:30 Mutual funds.

00:34 So you can not write it this way or you can write it as a difference of means and say that the mode the mu for the load mutual funds minus the mu for the non -load mutual funds is greater than zero.

00:49 And now the null hypothesis is just going to be the opposite of whatever we wrote down before.

00:54 So the mu for the load mutual funds is less than or equal to the mu for the non -load or the difference of means is less than or equal to zero.

01:07 This is our null hypothesis and alternative hypothesis.

01:11 And now we're going to use our data set to construct the hypothesis test with an alpha of 0 .05.

01:19 So the first thing we need to do is come up with the sample means and sample standard deviations so that we can find a test statistic.

01:28 And the first sample mean is simply in excel the average of the load column column, which is 16 .23.

01:41 So this is the sample mean for the load.

01:45 This is the sample mean for the non -load which is equal to 15 .7, 15 .23 and 16 or 16 .23 and 15 .7.

02:00 And now the standard deviation for the load is equal to 3 .52, approximately equal to 3 .52.

02:11 And the standard deviation for the non -load is approximately equal to 3 .31.

02:19 Now in order to find a p value, we will need to come up with a test statistic.

02:24 And because we're not given a population standard deviation, we have to come up with a sample standard deviation.

02:32 So we will come up with a t test statistic.

02:35 And the t test statistic follows the following formula.

02:38 Our point estimate, which is the difference of our means minus the hypothesis this difference, the null difference, which is 0 over here over the square root of the first samples standard deviation squared over the first sample size plus the second sample standard deviation square over the second sample size.

03:05 And remember that a standard deviation squared is simply the variance.

03:09 So that if you hear that, that just refers to the sample or the standard deviation squared.

03:15 So the difference in our means is going to be 0 .52, so 0 .52, minus 0 over, over square root of 3 .52 squared over our first sample size, which is 30, plus the second variance, which is 3 .31 squared over 30.

03:50 So ultimately we get a t test statistic of 0 .6.

03:54 So this leads to a t test statistic of 0 .06.

03:59 And now in order to use our t table to come up with a t test statistic, we have to come up with a degrees of freedom...

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