Save - Invest - Prosper
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Income |
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Income |
Employment |
WEalth |
Consumption |
Investment |
Published
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Exported: 26 Nov 2011
Original URL: http://knol.google.com/k/-/-/2utb2lsm2k7a/498
a) Population and a sample;
A Population is defined as all members of a specified group.
A Sample is a subset of a defined population.
b) Parameter;
A Parameter is a descriptive measure or characteristic of a defined population. It is thus the actual true measure of the population’s characteristic(s).
c) Types of measurement scales;
Nominal Scale: categorizes each member of the population or sample using an integer for each category. It is the weakest level of measurement with no implied ranking or intensity.
Ordinal Scale: each member of the population or sample is placed into a category and these categories are ordered with respect to some characteristic. It is a stronger level of measurement because it allows ordering across members. Think Letter Grades.
Interval Scale: each member is assigned a number from a scale. This scale provides a ranking across members and assurance that differences between scale values are equal. Scale values can thus be added or subtracted in a meaningful way. Think temperature either Celsius or Fahrenheit.
Ratio Scale: All the characteristics of interval scales plus a true zero point. Allows computation of meaningful ratios, as well as addition and subtraction. Think rates of asset returns or height.
d) Frequency distribution;
Frequency Distribution: is a tabular display of data summarized into a relatively small number of intervals. The frequency distribution is the list of intervals together with the corresponding measures of frequency for the variable of interest.
e) Calculation of a holding period return;
Holding Period Return: is expressed in percent terms, i.e. independent of currency units, and is calculated over a period of time.
Rt = [Pt - Pt-1 + Dt ]/ Pt-1
Where
Holding Period Return = Rt
Share Price end of time t = Pt
Share Price end of time t-1 = Pt-1
Holding Period Return, Rt, consists of capital gains over the period plus distributions during the period divided by the beginning price (distribution yield).
f) Use of intervals to summarize data;
An interval is a set of values in which observations on a random variable’s outcomes may fall. The set of intervals must be mutually exclusive and exhaustive, that is each observation must fall into one and only one interval.
The frequency with which observations fall into each interval is used to construct the frequency distribution for a random variable’s outcomes.
g) Relative frequencies, given a frequency distribution;
A frequency distribution shows the absolute number of observations in each interval. A relative frequency divides each corresponding absolute frequency by the total number of observations. Thus a relative frequency distribution shows the percentage of total observations in each interval.
h) Histogram and frequency polygon;
A histogram is the graphical equivalent of a frequency distribution; it is a bar chart where continuous data on a random variable’s observations have been grouped into a frequency distribution.
A frequency polygon is the line graph equivalent of a frequency distribution; it is a line graph that joins the frequency for each interval, plotted at the midpoint of that interval.
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A randomly selected sample of XYZ Investement Advisors' portfolios show the following:
Sample size 25
Average Sharpe Ratio 2.6
Standard Deviation 1.6
The firm claims its methodology produces portfolios with an average Sharpe ratio of 3.0. Verify.
Null hypothesis: H0: µ0 = 3.0
Alternative Hypothesis H1: µ0 is not equal to 3.0
This is a two tailed test.
Degrees of freedom = n –1 = sample size –1 = 25-1 = 24
Probability = alpha/2 = .05/2 = .025
From students t-Distribution table, the critical t is found to be 2.064. Hence acceptance range of t-statistic is –2.064 to 2.064.
T-calculated for the sample
Standard error = Sx/Square root of n = 1.6/5 = 0.32
T calculated = (X -µ0)/Sx = (2.6 – 3.0)/0.32 = -1.25
Since the –1.25 falls within the acceptance range, we cannot reject the null hypothesis.
Acceptance range around the hypothesized value =
1.0 + (2.064*0.32) = 2.34 to 3.66
The sample average falls within this range, so we cannot reject the null hypothesis.
One-Tailed Hypothesis Tests
We want to do the above test with the idea of the firm that their Sharpe ratio are greater than or equal to 3.0. Hence it becomes a one-tail test. Also the population variance is known as 2.44.
Null hypo = H0 = Mu > 3.0
Alt. Hypo = H1 = Mu < 3.0
Now p = alpha/1 = 0.050
From standard normal tables critical value is –1.645. The acceptance range for the test statistic is –1.645 to +infinity.
The test statistic is population standard deviation/Square root of the sample size. = S.R(2.44)/S.R.(25) = 1.56/5 = 0.31
Test Statistic calc = Z calc = (X - )/ Sx = (2.6 – 3.0)/0.31
= -1.29
The –1.29 is within the range, thus we cannot reject the null hypothesis.
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Manager A Manager B
10-year Avg. Annual . Return 12% 17%
Std. Dev. Of returns 20% 22%
It is assumed that the distribution of returns is approximately normally distributed. Therefore, test statistic is t.
Null Hypothesis: H0: Mu of A – Mu of B = 0
Alt. Hypo: H1: Mu of A – Mu of B is not equal to zero.
The t critical value for a two tailed test with a 5% level of significance and 18 degrees of freedom is 2.101. The calculated value of t is -.5318, which is within acceptable range for the statistic and therefore, we cannot reject the null hypothesis.
http://knol.google.com/k/narayana-rao-k-v-s-s/-/2utb2lsm2k7a/3579
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We are trying to verify claims about the mean differences between the returns of the advisory firm’s stock selections and the corresponding returns on the S&P 500 index.
Step 1: Term the difference as d.
Null Hypo. H0 = Mu of d = 3.0%
Alt. Hypo H1 = Mu of d is not equal to 3.0%
Standard error of d = Sd/S.R.(n) = 4%/S.R.(100)
= 4/100 = .4%
T calculated = (d – Mu of d)/Sd = (2.4% - 3.0%) / 0.4%
= -1.5
degrees of freedom = n – 1 = 100-1 = 99
t critical = t (.025, 99) = 1.984
Acceptance range for t is –1.984 to 1.984
T calculated is –1.5 and it is within acceptance range and hence null hypothesis is not rejected.
http://knol.google.com/k/narayana-rao-k-v-s-s/-/2utb2lsm2k7a/3579#
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The test statistic that is used to decide whether to reject the null or alternative hypothesis is the chi-square (χ2) statistic, which is defined as:
= (n-1)Sx2/
The chi-square distribution is a family of distribution each of which is defined by its degrees of freedom. The number of degrees of freedom is n-1.
It is to be noted that the chi square statistic is an adequate test statistic for testing hypotheses concerning population variances for normally distributed populations; it is not a good test statistic for testing hypotheses about the variances of populations that are not normally distributed.
Example: An investment management firm submits its investment record for the last 10 years. The std. Deviation of returns if 19%. But the firm claims that its targeted std. Deviation is 15%. Test at 5% significance level whether it is plausible.
Null Hypo: H0 = Sigma square of x is < 225
Alt. Hypo: H1 = Sigma square of x is > 225.
The degrees of freedom are 9, which is n-1 or 10-1. The calculated chi square value is
= (n-1)Sx2/sigma square = (1—1)(19)2/152
= 14.44
Null hypothesis is stated as one tailed test. Hence chi-square (.05, and 9) is 16.919. The calculated value of Chi square falls within the acceptance range of 0 to 16.919, thus the null hypothesis cannot be rejected.
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To test if the variances of two populations are equal, or if the variance of one population is greater than (or less than) the variance of another population.
The test statistic that is used to decide whether to reject these types of null or alternative hypotheses is the F-Statistic, which is defined as:
F = S12/ S22
The F-distribution is a family of distributions that are defined by two separate degrees of freedom: the degrees of freedom in the numerator (n-1) and the d of f in the denominator (n-1).
Example: Perform a hypothesis test to determine at a 5% level of significance whether the manager’s standard deviations are equal.
Manager A Manager B
!0-year Avg. Annual Return 12% 25%
Std. Deviation of Returns 20% 30%
Variance 400 900
Null H0: Sigma square of A = Sigma square of B
Alt H1: Sigma square of A ≠ Sigma square of B
F calculated = SA2/ SB2 = 900/400 = 2.25
Note: By convention, the higher variance is put in the numerator when calculating the F-statistic.
The d of f of both numerator and denominator are 10-1 = 9.
This is a two tailed test and hence F(.025,9,9) is 4.03.
The acceptance range is 0 to 4.03. Since the calculated value of F is 2.25, it is inside the acceptance range. We cannot reject the null hypothesis.
http://knol.google.com/k/narayana-rao-k-v-s-s/-/2utb2lsm2k7a/3579
Narayana Rao - 15 Dec 2010Published
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Original URL: http://knol.google.com/k/-/-/2utb2lsm2k7a/481
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Included in Research methodology Knol Book Chapter
http://knol.google.com/k/narayana-rao-k-v-s-s/-/2utb2lsm2k7a/3579
Narayana Rao - 15 Dec 2010