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What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?Mean(to decimals)Standard deviation(to decimals)b. Develop 9...

Question

What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?Mean(to decimals)Standard deviation(to decimals)b. Develop 95% confidence interval for the population mean years to maturity Round the answer to four decimal places_yearsWhat is the sample mean yield on corporate bonds and what is the sample standard deviation?Mean(to decimals)Standard deviation(to decimals)Develop a 95% confidence interval for the population mean yield on corporate bonds Roun

What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation? Mean (to decimals) Standard deviation (to decimals) b. Develop 95% confidence interval for the population mean years to maturity Round the answer to four decimal places_ years What is the sample mean yield on corporate bonds and what is the sample standard deviation? Mean (to decimals) Standard deviation (to decimals) Develop a 95% confidence interval for the population mean yield on corporate bonds Round the answer to four decima places percent



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The average annual total return for U.S. Diversified Equity mutual fiom 1999 to 2003 was 4.1$\%$ (Business Week, January $26,2004$ ). A researcher would like to conduct a hypothesis test to see whether the returns for mid-cap growth funds over the same period are significantly different from the average for U.S. Diversified Equity funds.
a. Formulate the hypotheses that can be used to determine whether the mean annual return for mid-cap growth funds differ from the mean for U.S. Diversified Equity funds.
b. A sample of 40 mid-cap growth funds provides a mean return of $\overline{x}=3.4 \%$ . Assume the population standard deviation for mid-cap growth funds is known from previous studies to be $\sigma=2 \%$ . Use the sample results to compute the test statistic and p-value for the hypothesis test.
c. At $\alpha=.05,$ what is your conclusion?

I'll be discussing sort of estimators, unbiased estimators and biased and um normal distributions. So the problem gives us that given a population of this data set. So it's like some um presidents age that they were assassinated simply. So one of them is 56, 49, 58 and 46. It says the part a find find the average of these four items. So I just use Excel um you know from from now on, I'll just be using cell. If you'd like to grab a brush up on on on how to find averages, feel free to look at previous formulas. So the average of these four items is 52.25. Now it asks us to find the distribution of sample means given that the samples are unequal to. So what this means is that if I were to take this for this four item population and I take samples of two. So either 56 49 Maybe 49, 56, 58, uh 56 etc etcetera. And find each of those um means what is the meaning of the entire thing. So um there are 16 different possibilities that I can choose from given and equals two. And I'm selecting from a four group of four. Have listed the 16 possibilities right here. So in this um box region right here. So 56, 56, 47 and etc. Now Each of these is a sample. So this is sample one, sample to example three. And so um let me just quickly go ahead and bring it down and there are 16 different samples just as a set. Now I find the mean or the average of each of these samples. Right? And so all these are just um averages. I get this thing, this data set. Now I find the probability distribution of this particular data set and it is this right here. So all these are just unique values of this right here and these one and twos and ones refer to the number of counts. So for example 46 out of this data set, there's only 1 46 47.5, there's 2 47.5 in this data set, one right here and one right here and I kept doing this for the rest of the data set. Now I find the proportion Of this number one divided by the total number of values here. So this right here would just be one which is the C25 and count. This is just 16. So um you know Like so one divided by 16 sorry One divided by 16 gives me .065 which is the same here. So I just did two divided by um .16 to develop .62 divided by 16, 1 daughter was 16 etc. So now I have these two columns right here let me just make them different color and I'll do this is basically a probability distribution. Now I have to find the meat of this probability distribution to do that. I'll just find a weighted average. So I just multiply this value and this value and I take the entire some of this. And what I what I find is that this equal to 52.25. Notice how this is the exact same thing as this. 52.5. So what this shows is that if I find the mean of all the population datasets, population items, I get 52.5. And also if I find the means of the Sample. So sorry, let me rephrase this mean of these sample distribution given that the samples are an equals two, I also get 52.25. Because these two values are the exact same. We can confirm that the getting the means that the sample distribution is a unbiased estimator of the mean of the population.

In this question, we want a margin of error of two years with a 95% confidence, a standard deviation of 13.36 and we want to figure out the sample size that will do that. So I plugged in all of our numbers into this formula for margin of error. And now I'm going to solve for N. By using some algebra, I'm going to divide both sides by 1.96 which will leave 1.204 on the left, and then I'm gonna move that square root of an out of the denominator by multiplying both sides by the square root of an and that will cancel out the square root event over there on the right. Now, I'm going to get rid of this 1.2 away from the square root of end by dividing both sides by 1.204 Mhm. And that's gonna give me the square root of n is equal to 13.9 to 8. And then you get rid of the square root, I'll square both sides. And that will give me sample size of 171 42 Which will we will round up to the nearest whole number, 172 people as our sample size, or 172 million is the same. Millionaires are the sample size. And the reason that we use the 13.36 the sample standard deviation here instead of sigma. We're allowed to do that because the sample side, the original sample size was more than 30. It was actually 36.

The following solution for number 19. And we look at the ages of the death row inmates. So there's someone that thinks that it's not equal to the 40.7 that was earlier. So she sampled 300 sorry 32 inmates on death row found the mean age to be 38.9 with the standard deviation of 9.6. So we're testing that it's not equal to the 40.7 at the five percent level of significance. Um So I'm gonna use the P value method on this because this way I can you know use the calculator. So t equals and then the p value equals. So what I'm gonna do with that PVS I'm gonna explicitly compared to the alpha And that will tell me whether to reject or not reject. So I'm going to go into the T. 84. And if I go to stat and the air over to test is that second option? T test the mu not. Is the hypothesized value 40.7? The X. Bar 38.9 the s. Is 9.6. And the sample size is 32. And then just change that alternative hypothesis to not equal to. And then I calculate and I get negative 1.607. And the P value is pretty big. 297 there. Okay so let's go and write those down. So negative one point 0607 and the p value is 0 to 97 which is in fact greater than alpha. So I will fail to reject. Mhm. Each nuts. Okay so what that means is there's not enough evidence to suggest that Um the mean age of death row inmates is is not equal to 40.7. So then we do a 95% confidence interval, 95% confidence interval, and then we'll interpret that. Um So if we go to stat and air over to test and we're gonna go down to the eighth option, the t interval interval. Unfortunately for us, everything is all set up ready to go And we go and calculate and that's 35 4 To 42.4. So let's go and write those down. Mhm. He's 35 0.4 42.4. Which means if we're going to interpret this for 95% confident that the mean age for all death row inmates in the us is between 35.4 and 42.4 years.

Problem. 22. The critical values using table six four degrees of freedom equal to end minus one, which is 30 minus one, is equal 29. So the chi square off one minus Alfa over two is equal to 14.257 and the chi square or all five people to 49.588 So the boundaries for the standard deviation is n minus one over Chi Square of Alfa over to DR as is equal to 6.256 and the other boundary in minus one over Chi square, off one minus Alfa over to dot ESC is equal to 11.667 The other boundaries for the variants are the square value off these values, which is 39.14 and 136.12


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