5

Muluab Tund iIS professionally managed investment scheme that pools money from many investors and invests variety ol securities. Growth funds focus primarily on inc...

Question

Muluab Tund iIS professionally managed investment scheme that pools money from many investors and invests variety ol securities. Growth funds focus primarily on increasing the value of investments_ wnereas blended funds seek balance between current income and growth Here data on the expense ratio (expenses as % of assets) for samples or 20 large-cap balanced funds (BI) and 20 large-cap growth funds (Gr) ("large-cap refers t0 the sizes companies which the funds invest; the population sizes a

muluab Tund iIS professionally managed investment scheme that pools money from many investors and invests variety ol securities. Growth funds focus primarily on increasing the value of investments_ wnereas blended funds seek balance between current income and growth Here data on the expense ratio (expenses as % of assets) for samples or 20 large-cap balanced funds (BI) and 20 large-cap growth funds (Gr) ("large-cap refers t0 the sizes companies which the funds invest; the population sizes are 825 and 762 respectively). 1,01 1.22 1.63 1,32 1.27 1.26 0.76 1,06 0.66 0.96 2.88 1.04 0.75 0.10 0.79 1.59 1.27 0.94 0.85 '54 1.06 1.25 2.16 1.56 0.99 1.11 1.08 1.82 2.07 0.91 0.81 1.38 0.64 1.50 1.00 1.12 1.78 1.03 1.16 Calculate and compare the values or %, > , and for the two types of funds_ (Round your answers three decima places:) For the sample balanced funds have % Jno for the sample of growth funds have Jnd The sample mean for the balanced funds ~Select-- than the sample mean for the growth funds_ The sample median for the balanced funds Select - than the sample median for the growth funds_ The sample standard deviation for the balanced funds Select _ than the sample standard deviation for the growth funds_ (b) Construct comparative boxplot for the two types of funds Expense Ratio Expense Ratio 3.Or



Answers

Annual holdings turnover for a mutual fund is the percentage of a fund's assets that are sold
during a particular year. Generally speaking, a fund with a low value of turnover is more stable
and risk averse, whereas a high value of turnover indicates a substantial amount of buying and
selling in an attempt to take advantage of short-term market fluctuations. Here are values of
turnover for a sample of 20 large-cap blended funds extracted from Morningstar.com:
$\begin{array}{llll}{1.03} & {1.23} & {1.10} & {1.64} & {1.30} \\ {0.94} & {2.86} & {1.05} & {0.75} & {0.09}\end{array}$
$\begin{array}{llll}{1.27} & {1.25} & {0.78} & {1.05} & {0.64} \\ {0.79} & {1.61} & {1.26} & {0.93} & {0.84}\end{array}$
(a) Would you use the one-sample $t$ test to decide whether there is compelling evidence for
concluding that the population mean turnover is less than 100$\% ?$ Explain.
(b) A normal probability plot of the 20 In(turnover) values shows a very pronounced linear
pattern, suggesting it is reasonable to assume that the turnover distribution is lognormal.
Recall that $X$ has a lognormal distribution if $\ln (X)$ is normally distributed with mean value
$\mu$ and standard deviation $\sigma .$ Because $\mu$ is also the median of the $\ln (X)$ distribution, $e^{\mu}$ is the
median of the $X$ distribution. Use this information to decide whether there is compelling
evidence for concluding that the median of the turnover population distribution is less than
100$\% .$

And this example will be dealing with the normal cumulative distribution function in regards to finance and funds. So the first thing we're asked is whether or not we can assume that our X distribution is normal and the answer is yes. And that's because we're dealing with a sample of 100 and as long as and is greater than or equal to 30 than the central limit theorem applies. Yeah. And we can say X is approximately normal now that we know that we can establish a Z scores and either use the table or calculators to calculate probability. So now we're looking at the X bar which is the average based on a certain number of months. And in the first situation were asked what's going to happen in nine months and what's the probability that our return will be between 1% and 2%. So the first thing I need to do is I need to convert these two Z scores. Now everything is in percentages. So that's nice. I don't have to change it to a decimal so my Z score will be one minus my me now remember my mean is going to be the same as the distribution because it is behaving normally. And my standard deviation for X bar will be the standard deviation Divided by the square root event. So this will be 0.8 Divided by the squared of nine, Which is 0.2667%. So it'll be 1 -1.4%. Technically both of these are percentages Divided by 0.2667%. And when you work that out We have a Z score of negative 1.50 And we'll calculate the Z score for two following the same steps. And when you do that We get 2.2 eight. No, I like to use the calculator on this. You can use your table and you can look up the values in your table and do a little subtraction. Or you can go to the calculator. Second, vars choose number two. In this case we've got negative 1.5, 2.25 Are mean and standard deviation or zero and one because we're using Z scores enter, enter and you can see the probability is approximately .9209. And when we round that, that becomes .9210. Now on the next one, Were asked to look at what happens in 18 months. So now and will be 18. Our men will still be 1.4 but the standard deviation Is now going to be 0.8 divided by the square root of 18. And that's going to give us approximately 0.1886%. And we're still asked about the probability that our average will fall Between one and 2%. So changing this to a Z score, my score minus the mean divided by the standard deviation. That's AZ score of negative 2.12. Then I'm going to change the 2% to its corresponding Z score And that gives us 3.18 using my calculator, then with the Z scores. So second bars too -2.12, three point 18 And this is .982 two or rounded .9823. Yes. Now we're asked to look at what happens to our probability when our standard deviation changes, So at 18 or excuse me, as the population changes and the standard deviation changes. So the square root of 18 gave us a standard deviation of about .1886%, which then gave us a probability of .9823. Here we had nine That gave us a standard deviation of .2, so the smaller population gave us a larger standard deviation, which actually ended up giving us a smaller percent. So the probability will increase as the standard deviation decreases, because dividing by a smaller value will give us a larger result. So we can also say this is because the standard deviation, I'm going to abbreviate decreases as the sample size or the population increases. And then finally, we need to do a little interpretation. So, after 18 months X bar turns out to be more than 2%. Would that shake your confidence in the statement about your mean being 1.4%? So what we're gonna do is we're going to go back and we're going to take the values from the previous problem, we're going to take the 2% result that Z score. And we're going to use this in our calculator to find the probability That Z is actually greater than 3.18. So I want to use the calculator for this. I have to do a little bit of a little bit of a trick number two CDF. So because I'm going greater, my lower bound is 3.18 And then to tell my calculator that I want infinity, I type one and then second comma because there's two little letters that say e up there 9 9. That's the calculators way of recognizing infinity. And this is 7.3. But it has this either the -4. This is like scientific notation. So that means I actually need to move my decimal .4 places to the left. So this is actually the probability that that will occur, and that is so small that it's very unlikely that I that X bar would be greater than or equal to 2%.

So here we have created, um, cross tabulation, a table for types, off fun and average annual return. Over five years, period Onda, we have calculated the row totals and column totals. Once we have done, we are done with cross tabulation. We can easily we can easily make Quincy distribution out off the cross tabulation. For example, we have created this frequency distribution out of this cross revelation table. And this frequency distribution consists off data on five years average return. So here are the classes. We have just destroyed them in rows. Here in the cross revolution, they are in column. But here, in frequency distribution, we have rested them in rows. And these are the sub particular frequencies. You can see that. And this is the somewhat product. So in the end, we have created African see distribution on, uh, on the data front type, So it is very simple. This is the first column off the cross regulation, and this reconstitution will be the same. And here we will pick this last column, and that's appropriate hair to the So there appears to be a relationship between fun type and average return. Over the past five years because the frequency is not off. Frequency is not roughly the same in each row and each column of the cross tabulation. So there has to be some relationship between the front type and average rate of return over the five years period.

So to find the know an alternative hypothesis for this question, we'll have to go to where it says a researcher would like to conduct a hypothesis, test dot, dot dot To see if the midcap Growth funds are significantly different from the average just different from the average for US diversified equity funds. So what we know is that this isn't a left or right side or left or right tailed Z test. This is, in fact, a, um a to tail T test. So for our alternative hypothesis, we're going to have a not equal to sign. And we're going to have that Our mu is not equal to this, um, average annual total return of 4.1%. And on the flip side, we know that our no hypothesis is that are amused equal to 4.1. Now, also given in the problem, we know that we have a sample of 40 a sample of 40 a mean return of 3.4% eh? Population standard deviation of 2%. And we're testing our P value against an Alfa value of 0.5 So the first thing we have to do before we can do anything is find a Z value. Okay, So to find a Z value, we're going to take the sample mean and subtract from it. The population mean divided by the population, standard deviation divided by the sample size. So for us, that is going to be Ah, 3.4 minus 4.1, divided by two, divided by the square root of 40. And once we do that basic math, we're going to get a value of two negative. 2.21 Okay, so our Z score is negative. 2.21 All right, So if our if we draw a normal distribution with a Z of zero in the middle, we get that our Z score over here is negative. 2.21 But we want to do is find the area to the left of the Z score. And underneath this curve, what this area represents is the probability that our Z score is less than are equal to negative 2.21 No, the way we can compute this is using our ze table. We find that our p value or this probability is equal to 0.1 36. Now we take that value okay of 0.136 and to find our p value, we multiply that by two because we're using a two tailed T test. Okay, Z test. Sorry. A two tailed Z test two times 0.136 which is equal to 0.27 two. And now we compare this P value This P value toe are Alfa value of 0.5 So, um, 0.27 to is less than 0.5 Because this is less than our RPI value is less than our alpha value. We have sufficient evidence. Two reject the no and conclude that, um, our sample our population conclude that the sorry. Let me read this question to conclude that the mean annual clued that the mean annual return rate is not four point 1% with a key value of what was our P value 0.272 at an Alfa of 0.5

Alright in this question, were asked to create a cross tabulation report with fun type as the row variable and expense ratio as the column variable. Additionally we should group the columns beginning at 0.25 in intervals of .25. Next will create a frequency distribution for expense ratio. And finally will form some of the conclusions. Um Yeah or some conclusions about fun type and expense ratio. So we're giving the state a set which represents uh 46 different mutual funds. They're fun type there in asset value, five year average return, expense ratio and Morningstar inc. Um So what we're gonna do is create a cross tabulation um with the on tape that's a real variable and the expense ratio as the common variable. So to do this will insert a pivot table, select our data set here. So I'm pressing control, shift right arrow and then control shift down arrow to select all my data there. Um I'm brought up on the left side here, the pivot table menu, and I know I want to put on type as the row variable and then expense ratio as the column variable. You can see here if I put uh oops account of fun type in the values I get the pivot table filled out but I have a lot of columns. So what I want to do next is group these columns, so it's more easily understood. So to do this I can click on this first value here and right click select group. And then here I want to start it at 0.25. Go up to our maximum value. So our max we're going to go up to in this Cases uh 1.49 and we're gonna okay two intervals of .25. So I'm going to click Okay here I see my pivot tables updated and it is much easier to read. I can see the different groupings here. That answers our first question here. Next we want to create a frequency distribution for expense ratio. So to do this, we can insert another pivot table, do the same thing to select our data. Yeah. And here we want to put uh fun type as the row, but an expense ratio in values. Oh not sorry. We want to switch these around. So now we have yeah, switched them around. We have expense ratio as the row variable in the account of the fund names or fun types. You'll get the same result either way as the value. So, we here we're breaking it down by fun type easily seeing the distribution for each expense ratio. So this last question here asked us um what conclusions can we draw about the fun type and expense ratio. So by looking at our charts here, we can see that um where the PE fund type, A lot of the funds are uh center and expense ratio of 1 to 1.25 and then 1.25 and up. We see some clustering here for type F I. On. And we feel also a lot of i. E. Funds in this same uh expense ratio category. We can see this even better down here. We see a lot of funds in this range. They do not look equally distributed. So there does appear to be a relationship between the fun type and the expense ratio.


Similar Solved Questions

5 answers
Hi Aonnguppy Question Submit Question Help: to 3 Tenswe which decimal the values alplacge are number score separating normal I the bottom mean decimal WOJ} %L9 108,8 place. 1 the top Land a standard obtained 338 . using deviation of 99.5. exact Z-scores Ue lnla
Hi Aonnguppy Question Submit Question Help: to 3 Tenswe which decimal the values alplacge are number score separating normal I the bottom mean decimal WOJ} %L9 108,8 place. 1 the top Land a standard obtained 338 . using deviation of 99.5. exact Z-scores Ue lnla...
5 answers
Consider the gas phase reaction Ozlg) 02ig) with rate (symbolized with r) (1/3) d[OzVdt: The proposed mechanism is M+02 0 + M K1,k1 0 + 03 ~ 2 02 where M is any molecule_ a) Use the steady state approximation t0 derive an expression for the concentration of the intermediate in terms of the concentrations of the other species in the mechanism: b) With your result in (a) and the equation k2 [O] [Oz) obtain the rate law for the reaction:
Consider the gas phase reaction Ozlg) 02ig) with rate (symbolized with r) (1/3) d[OzVdt: The proposed mechanism is M+02 0 + M K1,k1 0 + 03 ~ 2 02 where M is any molecule_ a) Use the steady state approximation t0 derive an expression for the concentration of the intermediate in terms of the concentra...
5 answers
Describe explicitly (as in exercises 1 and 2 ) the linear transformation $T$ from $F^{2}$ into $F^{2}$ such that $T epsilon_{1}=(a, b), T epsilon_{2}=(c, d)$.
Describe explicitly (as in exercises 1 and 2 ) the linear transformation $T$ from $F^{2}$ into $F^{2}$ such that $T epsilon_{1}=(a, b), T epsilon_{2}=(c, d)$....
5 answers
Why do salts of weak bases and strong acids give acidic aqueous solutions? Illustrate with $mathrm{NH}_{4} mathrm{NO}_{3}$, a common fertilizer.
Why do salts of weak bases and strong acids give acidic aqueous solutions? Illustrate with $mathrm{NH}_{4} mathrm{NO}_{3}$, a common fertilizer....
5 answers
2 Find the functions fog, gof , fof ,and gog and their domains flx) =x?,g(x) = Vx-3
2 Find the functions fog, gof , fof ,and gog and their domains flx) =x?,g(x) = Vx-3...
5 answers
Simplify each expression. Justify each step.$$(10 p) 11$$
Simplify each expression. Justify each step. $$ (10 p) 11 $$...
5 answers
How can ionic compounds be neutral if they consist of positive and negative ions?
How can ionic compounds be neutral if they consist of positive and negative ions?...
1 answers
Figure 10.58 shows a propeller blade that rotates at 2000 rev/min about a perpendicular axis at point $B .$ Point $A$ is at the outer tip of the blade, at radial distance 1.50 $\mathrm{m} .$ (a) What is the difference in the magnitudes $a$ of the centripetal acceleration of point $A$ and of a point at radial distance 0.150 $\mathrm{m} ?$ (b) Find the slope of a plot of $a$ versus radial distance along the blade.
Figure 10.58 shows a propeller blade that rotates at 2000 rev/min about a perpendicular axis at point $B .$ Point $A$ is at the outer tip of the blade, at radial distance 1.50 $\mathrm{m} .$ (a) What is the difference in the magnitudes $a$ of the centripetal acceleration of point $A$ and of a p...
1 answers
Relation R on R is define as follows : for all m and n in R,m R n <=> m <= nQuestion : Prove that R is a partial order relation on RComplete steps pls. Thank youu
Relation R on R is define as follows : for all m and n in R, m R n <=> m <= n Question : Prove that R is a partial order relation on R Complete steps pls. Thank youu...
5 answers
8 20 + 11 Ca(HcOua(s) 1 1l 8 CuClz V Which of the following equilibrium systems 2 6 1 1 1Slawnioa ujum
8 20 + 11 Ca(HcOua(s) 1 1l 8 CuClz V Which of the following equilibrium systems 2 6 1 1 1 Slawnioa ujum...
5 answers
10 of 21neTianeco a54 1 791 Hans of H 0 nere FroducedNmen 2 O51Tr ola Eiocaior 8 Jeenttieoccbustion an:hau #FHAnIEmnincr foruula a4ti Eokcula fomul one dracalEe molar mits oitbe cciFouad %15 jouEdio |+ 70.Dgmol Datenmin t OrprnelEoftt rlemcnbm tbe dtider pE ueuleulquation=tpincalonrunaIcleculu falch
10 of 21 ne Tian eco a54 1 791 Hans of H 0 nere Froduced Nmen 2 O51Tr ola Eiocaior 8 Jeenttieo ccbustion an:hau #FHAnI Emnincr foruula a4ti Eokcula fomul one dracal Ee molar mits oitbe cciFouad %15 jouEdio |+ 70.Dgmol Datenmin t Orprnel Eoftt rlemcnbm tbe dtider pE ueuleul quation =tpincalonruna Icl...
5 answers
Daie: 084ho Wva Kesty CMs OSPIO uLJEConagroup009Qualitative Analysis Flow ChartF" Fe" N" G NH;Pircinuon ottWmcnVm IonKAinL Buller SystcmE Septration olnit- from (4'CWrcatment wtth Strong ALdConirm_tan ofFe""-Conhrmation ofCat-Conhrmation af N?1
daie: 084ho Wva Kesty CMs OSPIO uL JEConagroup 009 Qualitative Analysis Flow Chart F" Fe" N" G NH; Pircinuon ott WmcnVm Ion KAinL Buller Systcm E Septration olnit- from (4' CWrcatment wtth Strong ALd Conirm_tan ofFe"" -Conhrmation ofCat- Conhrmation af N? 1...
5 answers
5 720 g 1 3 8 8 L 2 1 2 1 L
5 720 g 1 3 8 8 L 2 1 2 1 L...
5 answers
Knecolncns 10Servico Catalog | 54Prriciple Papet[-/6 Points]DETAILSSERCP1O 18.P,006 0/4 Submissions UsedConsider the combination of resistors shown in the Figure below: 12.0 0 O0 N6(nL6,U0 0AOALFind Ute equlvalent reslstance between point and(6) Ir 4 volrage 0/ 39,6 Is applied between polnts and B find the Current In each resistor; 12 05 ( 4 0Additional Materials0Bodk
Knecolncns 10 Servico Catalog | 54 Prriciple Papet [-/6 Points] DETAILS SERCP1O 18.P,006 0/4 Submissions Used Consider the combination of resistors shown in the Figure below: 12.0 0 O0 N 6(nL 6,U0 0 AOAL Find Ute equlvalent reslstance between point and (6) Ir 4 volrage 0/ 39,6 Is applied between pol...

-- 0.021290--