Chapter 3.1, Problem 20E

To determine

Identify the mutual fund that performed better.

Answer to Problem 20E

The Mutual fund T performed better than Stivers.

Explanation of Solution

Calculation:

In the Mutual fund S, $10,000 is invested at the beginning of 2004 and in Mutual fund T, $5,000 is invested at the beginning of 2004. The given table represents the value of each investment at the end of each subsequent year.

The geometric mean is often used for finding the mean growth rate.

The mean growth factor over these eight periods is as follows:

${\overline{x}}_g=\sqrt[8]{\left(x_1\right)\left(x_2\right)\left(x_3\right)\left(x_4\right)\left(x_5\right)\left(x_6\right)\left(x_7\right)\left(x_8\right)}$

Here, $x_i$’s are the growth rate for the i^th year.

The growing factor of each year can be obtained by dividing the value of end of year by the previous year value.

The growth factor of Mutual fund S for the 2004 can be obtained as follows:

$\begin{array}{c}\text{Growth rate of }{x}_1=\frac{11,000}{10,000}\\ =1.100\end{array}$

The growth factor of Mutual fund T for the 2004 can be obtained as follows:

$\begin{array}{c}\text{Growth rate of }{x}_1=\frac{5,600}{5,000}\\ =1.12\end{array}$

Similarly, the growth actor for remaining years can be obtained as shown in the table given below:

Years	Mutual fund S		Mutual fund T
	End of Year Value	Growth Factor	End of Year Value	Growth Factor
2004	$11,000	1.100	$5,600	1.120
2005	$12,000	1.091	$6,300	1.125
2006	$13,000	1.083	$6,900	1.095
2007	$14,000	1.077	$7,600	1.101
2008	$15,000	1.071	$8,500	1.118
2009	$16,000	1.067	$9,200	1.082
2010	$17,000	1.063	$9,900	1.076
2011	$18,000	1.059	$10,600	1.071

For the Mutual fund S:

The balance at the end of the 2011 is the initial investment times the product of all 8 growth factors.

$18,000=10,000\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]$

That is,

$\begin{array}{c}\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]=\frac{18,000}{10,000}\\ =1.8\end{array}$

Substitute $\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]=1.8$ in the formula of geometric mean.

$\begin{array}{c}{\overline{x}}_g=\sqrt[8]{1.8}\\ =1.07624\end{array}$

Thus, the mean annual return for the Mutual fund S is $\left(1.07624-1\right)100\%=\underset{\_}{7.624\%}$

For the Mutual fund T:

The balance at the end of the 2011 is the initial investment times the product of all 8 growth factors.

$10,600=5,000\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]$

That is,

$\begin{array}{c}\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]=\frac{10,600}{5,000}\\ =2.12\end{array}$

Substitute $\left[\left(x_1\right)\left(x_2\right)\mathrm{...}\left(x_8\right)\right]=2.12$ in the formula of the geometric mean.

$\begin{array}{c}{\overline{x}}_g=\sqrt[8]{2.12}\\ =1.09848\end{array}$

Thus, the mean annual return for the Mutual fund T is $\left(1.09848-1\right)100\%=\underset{\_}{9.848\%}$

The annual return earned by the Mutual fund T is 9.8% and by the Mutual fund S is 7.6%. Hence, the annual return earned by the Mutual fund T is more than that of Mutual fund S.

Thus, it can be concluded that the Mutual fund T is performed better than Mutual fund S.