Calculating statistical values from given data

[mathmari]

Hey! :giggle:

Analyst has collected the following data on the performance of the $X$ stock for $10$ different years.

a) Calculate the arithmetic mean, the median, the mode, the standard deviation, the coefficient of variability and of asymmetry. You interpreted your results.

b) Does the data have extreme values (consider the cut-off limit $z = \pm 2$ or $z = \pm 3$).

c) The analyst claims that the average yield next year will be at least $7.5\%$. Is the claim correct? Do you think you would have an increased risk if you bought the $X$ share?
I have done the follwoing :

a) The arithmetic mean is $$\frac{5\%+3\%+8\%+12\%+2\%+18\%+5\%+8\%+24\%+16\%}{10}=\frac{101\%}{10}=10.1\%$$
For the median, we write the percentages in an increasing order then we calculate the half of the sum of $5$-th and $6$-th percentage.

Therefore the median is equal to $$\frac{8\%+8\%}{2}=8\%$$

The modeis the value that appears most often, so $8\%$.

For the standard deviation wesubtract from eachgiven value the mean, square the result and take the sum, divide that by the number $10$ and take the square root, right?

So we get $$\sigma=\sqrt{\frac{1}{10}\left [(0.05-0.101)^2+(0.03-0.101)^2+(0.08-0.101)^2+(0.12-0.101)^2+(0.02-0.101)^2+(0.180-0.101)^2+(0.05-0.101)^2+(0.08-0.101)^2+(0.24-0.101)^2+(0.16-0.101)^2\right ]}=\sqrt{\frac{1}{10}\cdot 0.04709}\approx 0.0686$$

The coefficient of variability is $$CV=\frac{\sigma}{\mu}=\frac{0.0686}{0.101}\approx 67.92\%$$ The coefficient of asymmetry is \begin{align*}\beta_1&=\frac{1}{10}\cdot \sum_{i=1}^{10}\left (\frac{x_i-\mu}{\sigma}\right )^3\\ & =\frac{1}{10\cdot \sigma^3}\cdot \left [\left (x_i-\mu\right )^3\right ]\\ & =\frac{1}{10\cdot 0.0686^3}\cdot \left [(0.05-0.101)^3+(0.03-0.101)^3+(0.08-0.101)^3+(0.12-0.101)^3+(0.02-0.101)^3+(0.180-0.101)^3+(0.05-0.101)^3+(0.08-0.101)^3+(0.24-0.101)^3+(0.16-0.101)^3\right ]\\ & =\frac{1}{0.00322828856}\cdot 0.00221772 \\ & \approx 0.68696\end{align*}

Is everything correct so far? How can we interpret the results?

b) Could you give me a hint for that?

c) Do we have to check what distribution we have in this case? Or how can we know that?

 

[I like Serena]

For the standard deviation wesubtract from eachgiven value the mean, square the result and take the sum, divide that by the number $10$ and take the square root, right?

So we get $$\sigma=...$$

Is everything correct so far? How can we interpret the results?

First the symbols, we use the Greek symbols $\mu$ and $\sigma$ to denote the so called population mean and standard variation.
They only apply if we are measuring the whole population, and as such they are usually unknown.
We use the Latin symbols $\bar x$ and $s$ to denote the mean and standard deviation of a sample of the population.

More specifically, the formula for the standard deviation of a sample is different from the one of the entire population.
That is because we "lose a degree of freedom" since we don't know the population mean $\mu$ and can only estimate it as $\bar x$ through the same sample.

In this case it means we must divide by $n-1=10-1=9$ to find the standard deviation $s$.
So $s=\sqrt{\frac 1{10-1}\cdot0.04709}$.

It also means we have $CV=\frac{s}{\bar x}$ and $b_1 = \frac 1{n}\sum\left(\frac{x_i-\bar x}{s}\right)^3$.

b) Could you give me a hint for that?

The cut-off limit $z=\pm 2$ means that we look at 2 standard deviations from the mean ($\bar x\pm 2s$) and we check if we have values further away from the mean than that.

c) Do we have to check what distribution we have in this case? Or how can we know that?

To check the claim we first formulate a null hypothesis and an alternative hypothesis, which we then try to verify.
The usual way to verify it, is by assuming that the "entire population" has a normal distribution, and that we have an independent sample of that population.
Questions (a) and (b) seem to be intended to verify if it is "reasonable" to assume a normal distribution.
It's also why question (a) asks to "interpret the results".
Assuming it is reasonable, we can do a t-test.