Calculating statistical values from given data

In summary, The analyst collected data on the performance of the $X$ stock for $10$ different years and calculated the arithmetic mean, median, mode, standard deviation, coefficient of variability, and coefficient of asymmetry. The data does not have any extreme values according to the cut-off limits $z=\pm 2$ or $z=\pm 3$. The analyst claims that the average yield next year will be at least $7.5\%$, and assuming a normal distribution, a t-test can be used to verify this claim.
  • #1
mathmari
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Hey! :giggle:

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

1631077089551.png


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?
 
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  • #2
mathmari said:
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 $b_1 = \frac 1{n}\sum\left(\frac{x_i-\bar x}{s}\right)^3$. 🤔

mathmari said:
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. 🤔

mathmari said:
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. 🤔
 

Related to Calculating statistical values from given data

1. How do I calculate the mean from a set of data?

The mean, also known as the average, can be calculated by adding all the values in the data set and dividing by the total number of values.

2. What is the difference between median and mode?

The median is the middle value in a data set when the values are arranged in ascending or descending order. The mode is the most frequently occurring value in the data set.

3. How do I calculate the standard deviation?

The standard deviation measures the spread of data around the mean. It can be calculated by finding the difference between each value and the mean, squaring those differences, finding the average of those squared differences, and then taking the square root of that average.

4. What is the significance of the p-value in statistical analysis?

The p-value is a measure of the probability of obtaining a result at least as extreme as the one observed, assuming that the null hypothesis is true. It is used to determine the significance of a result and whether it can be considered statistically significant.

5. How do I choose the appropriate statistical test for my data?

The choice of statistical test depends on the type of data and the research question being asked. Factors to consider include the type of variables (categorical or continuous), the number of groups being compared, and the assumptions of the test. Consulting with a statistician or using online resources can help in selecting the appropriate test.

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