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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?

a) $H_{0} : \mu_{1}-\mu_{2} \leq 0 ; H_{a} : \mu_{1}-\mu_{2}>0$b) $p>0.20 ;$ Do not reject $H_{0}$

Intro Stats / AP Statistics

Chapter 10

Comparisons Involving Means, Experimental Design, and Analysis of Variance

Descriptive Statistics

The Chi-Square Distribution

Experiment

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The first thing we have to do in this problem is come up with the hypotheses so that when the nor reject no hypothesis is rejected, it leads to the conclusion that the load mutual funds have a higher mean annual return over the five year period. 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 mutual funds. So you cannot read it this way, or you can write it as a difference of means and say that the mood the mute for the load mutual funds minus the mu for the non load mutual funds is greater than zero. And now, then all hypotheses no hypothesis is just gonna be the opposite of whatever we wrote down before. So the mute 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. So this is our null hypothesis and alternative hypothesis, and now we're going to use our data set to construct um, the hypothesis test with an Alfa of 0.5 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 and the first sample mean is simply in excel the average of the load The load column, which is 16.23 So this is the sample mean for the load. This is the sample mean for the non bowed, which is equal to 15.7, 15.23 and 16 or 16.23 and 15.7. And now the standard deviation for the load is equal to you, 3.5 to approximately equal to 3.52 And the standard deviation for the non load is approximately equal two 3.31 Now, in order to find a P value, we will need to come up with a test statistic. And because we're not given, um, a population standard deviation, we have to come up with a sample standard deviation. So we will come up with a T test statistic and the T te statistic follows the following formula. Our point estimate, which is the difference of our means minus the hypothesis difference The no difference which is zero over here over a square it of the first samples standard deviation squared over for a sample size plus the second sample standard deviation square over the second sample size. And remember that a standard deviation squared is simply the variance so that if you hear that, um, that just refers to the samples or the standard deviation squares. So the difference in our means is going to be, uh, 0.5 to 0.52 My over or minus zero over square root of 3.52 squared over our first sample size, or just 30 plus the second variance, which is 3.31 squared over 30. So ultimately, we get a T test statistic of 0.6, so this leads to 82 statistic of 0.6 And now, in order to use our tea table to come up with a teacher statistic, we have to come up with the degrees of freedom. Degrees of freedom formula is complicated, so stick with me. It is the first sample variance over the first sample size, plus the second sample variance over the second sample size, and some of that is squared over one divided by the first sample size minus one times the first sample variance squared over the first sample size squared, plus one over the second sample size minus one times the variance of the second sample. Over the second sample size squared, so this is equal to 3.52 squared 3.52 squared over 30. Close three point 32 squared over 30 squared over one over 29 times 3.5 two squared over 30 squared plus won over 29 times, 3.32 squared over 30 square. And finally we get a degrees of freedom of when we round down 57 there are degrees of freedom is equal to 57. Now, when we look at her tea table, we have a test a tous tick of 0.6004 and a degrees of freedom A 57. So we get that our that the area of the right end of Artie distribution. This is where T equals zero and this is where T is equal to 0.6 and our degrees of freedom is equal to 57 this area over here. Oh, sorry. This area down here is equal to 0.8 for eight. So probability that tea is less than or equal to 0.6 is equal to 0.848 But we're not interested in this. We're interested in this area. And this area is simply one minus the probability that tea is less than or equal to zero point six, which is equal to point one. Where we get that our that this is so that this area over here is greater than 0.2. So you get that this. Okay, we got that. This is greater then point to. So now that we have an Alfa equaling 0.5 and we have a P value greater than 0.2 because our P value is much greater than our Alfa 0.2 is greater than 0.5 We reject the knoll, so Rp is greater than point to, which is greater than 0.0.5 We failed to reject the knoll. We failed to reject the no. So what that means is that we cannot conclude that, um, the mean for the load mutual funds has a higher annual return over the five year period than the mean for the non lewd mutual funds

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