PSY 310: Statistical Inference

Given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 85\\ \sigma & =& 14\\ \stackrel{\_}{x\text{'}}& =& 70\\ \sigma \text{'}& =& 12\\ {x\text{'}}_i& =& m{x}_i+c\end{array}$$

Find:

$$\left(m,c\right)$$

The three pertinent characteristics for a set of numbers are: location, spread, and shape. These characteristics are affected only be certain linear transformations:

• Location — the position in the number line of the set — affected by additive and multiplicative operations
• Spread — the amount of the number line covered by the set — affected only by multiplicative operations
• Shape — the relative distances between elements of the set — unafected by linear transformations

Standard deviation is a measure of spread, so:

$$\begin{array}{ccc}\sigma \text{'}& =& m\sigma \\ m& =& \frac{\sigma \text{'}}{\sigma }\end{array}$$

Mean is a measure of location, so:

$$\begin{array}{ccc}\stackrel{\_}{x\text{'}}& =& m\stackrel{\_}{x}+c\\ c& =& \stackrel{\_}{x\text{'}}-m\stackrel{\_}{x}\end{array}$$

Substituting in from the original values:

$$\begin{array}{ccc}m& =& \frac{\sigma \text{'}}{\sigma }\\ & =& \frac{12}{14}\\ & =& \frac{6}{7}\end{array}$$ $$\begin{array}{ccc}c& =& \stackrel{\_}{x\text{'}}-m\stackrel{\_}{x}\\ & =& 70-\frac{6}{7}\left(85\right)\\ & =& \frac{-20}{7}\\ & \approxeq & -2.9\end{array}$$

Given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 70\\ \sigma & =& 12\\ z_i& =& 1.25\end{array}$$

Find:

$$\left(x_i\right)$$

The Z-score is the score standardized to a mean of 0 and standard deviation of 1.

$$\begin{array}{ccc}{z}_i& =& \frac{x_i-\stackrel{\_}{x}}{\sigma }\\ x_i& =& z_i\sigma +\stackrel{\_}{x}\end{array}$$

Substituting in the given values this is:

$$x_i=z_i\sigma +\stackrel{\_}{x}=2.5\left(12\right)+70=100$$

Given:

$$\begin{array}{ccc}{\sigma }_{x,y}& =& 10\\ {\sigma }_x& =& {\sigma }_y\\ & =& 12\end{array}$$

Find:

$$\left({\rho }_{x,y}\right)$$

This solution follows pretty simply from the definition:

$$\begin{array}{ccc}{\rho }_{x,y}& =& \frac{{\sigma }_{x,y}}{{\sigma }_x{\sigma }_y}\\ & =& \frac{10}{{12}^2}\\ & =& \frac{5}{72}\\ & \approxeq & 0.069\end{array}$$

Given:

$$\begin{array}{ccc}x& =& 78\\ \stackrel{\_}{x}& =& 70\\ \sigma & =& 10\end{array}$$

Find the percentile rank, $r\left(z\right)$.

The first step is to normalize the numeric score as a z-score:

$$\begin{array}{ccc}z& =& \frac{x-\stackrel{\_}{x}}{\sigma }\\ & =& \frac{78-70}{10}\\ & =& 0.8\end{array}$$

Then using a z-table, it is possible to look up $r\left(0.8\right)\approxeq \mathrm{78.81\%}$.

Given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 500\\ \sigma & =& 100\\ x_a& =& 550\\ x_b& =& 650\end{array}$$

Find:

$$\left(\underset{x_a}{\overset{x_b}{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}{\left(\frac{x-\stackrel{\_}{x}}{\sigma }\right)}^2}\right)$$

It is possible to solve that integral, but not necessary to get an approximation. An alternative is to convert the scores to z-scores and use a table lookup to find the areas:

$$\begin{array}{ccc}{z}_a& =& 0.5\\ z_b& =& 1.5\end{array}$$

Then using a z-table, one can find:

$$\begin{array}{ccccc}r\left(z_a\right)& =& r\left(0.5\right)& \approxeq & 0.6915\\ r\left(z_b\right)& =& r\left(1.5\right)& \approxeq & 0.9332\end{array}$$

The quantity between them then is:

$$\begin{array}{ccccccccc}{\Delta }_{a,b}& =& r\left(z_b\right)-r\left(z_a\right)& \approxeq & 0.9332-\mathrm{0.6915}& =& 0.2417& \approxeq & \mathrm{24\%}\end{array}$$

Given:

$$\begin{array}{ccc}{x}_a& =& 88\\ r\left(z_a\right)& =& \mathrm{79.8\%}\\ x_b& =& 75\\ r\left(z_b\right)& =& \mathrm{40.1\%}\end{array}$$

Find:

$$\left(\stackrel{\_}{x},\sigma \right)$$

First, use the z-table to get the z-scores for the given correlations:

$$\begin{array}{ccccccccccc}r\left(0.83\right)& \approxeq & 0.7967& <& r\left(z_a\right)& =& 0.798& <& r\left(0.84\right)& \approxeq & 0.7995\\ & & & z_a& \approxeq & 0.83& & & \\ r\left(-0.26\right)& \approxeq & 0.3974& <& r\left(z_b\right)& =& 0.401& <& r\left(-0.25\right)& \approxeq & 0.4013\\ & & & z_b& \approxeq & -0.25& & & \end{array}$$

Given these, you have the equations for the z-scores which combine the score, mean and deviation. Since there are two of them that makes two equations in two unknowns and they just need to be subtracted from each other:

$$\begin{array}{ccccc}{z}_a-z_b& =& \frac{x_a-\stackrel{\_}{x}}{\sigma }-\frac{x_b-\stackrel{\_}{x}}{\sigma }& =& \frac{x_a-x_b}{\sigma }\end{array}$$ $$\begin{array}{ccccccc}\sigma & =& \frac{x_a-x_b}{z_a-z_b}& \approxeq & \frac{88-75}{0.83-\left(-0.25\right)}& \approxeq & 12\end{array}$$ $$\begin{array}{ccccccc}\stackrel{\_}{x}& =& -\left(\sigma z_a-x_a\right)& \approxeq & -\left(12\left(0.83\right)-88\right)& \approxeq & 78\end{array}$$

Given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 10\\ {\sigma }_x& =& 100\\ \left|x\right|& =& 9\\ y_n& =& 15\\ y& =& x\bigcup \left\{y_n\right\}\end{array}$$

Find:

$$\left(\stackrel{\_}{y},{\sigma }_y\right)$$

Finding the new mean is straightforward:

$$\begin{array}{ccccccccccc}\stackrel{\_}{y}& =& \frac{\sum _{i=1}^{\left|y\right|}{y}_i}{\left|y\right|}& =& \frac{\sum _{i=1}^{\left|x\right|}{x}_i+y_n}{\left|x\right|+1}& =& \frac{\left|x\right|\stackrel{\_}{x}+y_n}{\left|x\right|+1}& =& \frac{\left(\right)\left(\right)+15}{\left(\right)+1}& =& 10.5\end{array}$$

The variance is a little trickier. First a transformation of ${\sigma }_x^2$ is needed:

$$\begin{array}{ccc}{\sigma }_x^2& =& \frac{\sum {\left(x_i-\stackrel{\_}{x}\right)}^2}{\left|x\right|-1}\\ \sum {\left(x_i-\stackrel{\_}{x}\right)}^2& =& {\sigma }_x^2\left(\left|x\right|-1\right)\end{array}$$

This equation can then be used to find the new deviation:

$$\begin{array}{ccc}{\sigma }_y^2& =& \frac{\sum _{i=1}^{\left|y\right|}{\left(y_i-\stackrel{\_}{y}\right)}^2}{\left|y\right|-1}\\ & =& \frac{\sum _{i=1}^{\left|x\right|}{\left(x_i-\stackrel{\_}{x}\right)}^2+{\left(y_n-\stackrel{\_}{y}\right)}^2}{\left|x\right|}\\ & =& \frac{{\sigma }_x^2\left(\left|x\right|-1\right)+{\left(y_n-\stackrel{\_}{y}\right)}^2}{\left|x\right|}\\ & =& \frac{\left(100\right)\left(9-1\right)+{\left(15-10.5\right)}^2}{9}\\ & =& \frac{3281}{36}\\ & \approxeq & 91\end{array}$$

Given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 670\\ {\sigma }_x& =& 70\\ {\rho }_{x,y}& =& 0.57\\ \stackrel{\_}{y}& =& 3.3\\ {\sigma }_y& =& 0.7\\ x_i& =& 595\\ y_i& =& 3.62\end{array}$$

Find:

$$\left(r\left({x_i|}_{y_i}\right)\right)$$

Finding the answer to this question requires finding the conditional mean and deviation. Those equations are:

$$\begin{array}{ccc}{\mu }_{{y|}_{x=a}}& =& mx+b\\ m& =& \rho \left(\frac{{\sigma }_y}{{\sigma }_x}\right)\\ b& =& \stackrel{\_}{y}-b\stackrel{\_}{x}\\ {\sigma }_{{y|}_{x=a}}^2& =& \left(1-{\rho }_{x,y}^2\right){\sigma }_y^2\end{array}$$

These equations reduce significantly when dealing with z-scores where $\stackrel{\_}{x}=\stackrel{\_}{y}=0$ and ${\sigma }_x={\sigma }_y=1$.

$$\begin{array}{ccc}{\mu }_{{z_y|}_{z_x=a}}& =& {\rho }_{x,y}{z}_x\\ {\sigma }_{{z_y|}_{z_x=a}}& =& \sqrt{1-{\rho }_{x,y}^2}\end{array}$$

So the question becomes, what is the rank of $z_y$ given ${\mu }_{{z_y|}_{z_x=a}}$ and ${\sigma }_{{z_y|}_{z_x=a}}$? This requires one more normalization to a standard mean and then it's a simple table lookup.

$$\begin{array}{ccccccc}{z}_{x_i}& =& \frac{x_i-\stackrel{\_}{x}}{{\sigma }_y}& =& -\frac{15}{14}& \approxeq & -1.07\\ {\mu }_{{z_y|}_{z_x}}& =& {\rho }_{x,y}{z}_x& \approxeq & .57\left(-1.07\right)& \approxeq & -0.610\\ {\sigma }_{{z_y|}_{z_x}}& =& \sqrt{1-{\rho }_{x,y}^2}& =& \sqrt{1-{.57}^2}& \approxeq & 0.823\\ z_{y_i}& =& \frac{y_i-\stackrel{\_}{y}}{{\sigma }_y}& =& \frac{16}{35}& \approxeq & 0.457\\ {z_y|}_{z_x}& =& \frac{z_y-{\mu }_{{z_y|}_{z_x}}}{{\sigma }_{{z_y|}_{z_x}}}& \approxeq & \frac{0.457-\left(-0.610\right)}{0.823}& \approxeq & 1.296\\ r\left({x_i|}_{y_i}\right)& =& r\left({z_y|}_{z_x}\right)& \approxeq & r\left(1.296\right)& \approxeq & \mathrm{90.3\%}\end{array}$$

The method of figuring this out is beyond the scope of this class. To derive an answer, one simply uses table 5.1 on pg. 74 of Glass and Hopkins. It gives a value of 5 for the expected range with a sample size of 100. This 5 is for a deviation of 1, so the range where σ = 15 is (5)(15) = 75. This range extends on either side of the mean, so the maximum is:

$$\begin{array}{ccccc}\mu +\frac{\sigma R_{100}}{2}& =& 120+\frac{\left(15\right)\left(5\right)}{2}& =& 157.5\end{array}$$

The solution is not what I did initially, which is, given:

$$\begin{array}{ccc}\stackrel{\_}{x}& =& 120\\ \sigma & =& 15\\ \left|x\right|& =& 100\end{array}$$

Find:

$$\left(max\left(x_i\right)\right)$$

Using the equation from #13 for the maximum z-score for a given sample size, we find:

$$\begin{array}{ccccccc}max\left(z_i\right)& =& \frac{n-1}{\sqrt{n}}& =& \frac{99}{\sqrt{100}}& =& 9.9\end{array}$$

Given that, it is striaghtforward to find the score:

$$\begin{array}{ccccccc}x& =& \sigma z+\stackrel{\_}{x}& =& 15\left(9.9\right)+120& =& 268.5\end{array}$$

Given:

$$\begin{array}{ccc}{\sigma }_x& =& 10\\ {\sigma }_y& =& 5\end{array}$$

Find:

$$\left(max\left({\sigma }_{x,y}\right)\right)$$

σ increases as ρ increases and ρ is -1 ≤ ρ ≤ 1, so:

$$\begin{array}{ccc}max\left({\rho }_{x,y}\right)& =& \frac{max\left({\sigma }_{x,y}\right)}{{\sigma }_x{\sigma }_y}\\ max\left({\sigma }_{x,y}\right)& =& {\sigma }_x{\sigma }_{y}max\left({\rho }_{x,y}\right)\\ & =& \left(1\right)\left(10\right)\left(5\right)\\ & =& 50\end{array}$$

Given:

$$\begin{array}{ccc}x& =& \left\{\right\}\end{array}$$

Find:

1. $$\begin{array}{ccccc}\sum _{i=2}^5\frac{x_i}{x_{i-1}}& =& \frac{5}{1}+\frac{8}{5}+\frac{9}{8}+\frac{11}{9}& =& \frac{3221}{360}\end{array}$$
2. $$\begin{array}{ccccc}\sum _{i=1}^5\sqrt{x_i}& =& \sqrt{1}+\sqrt{5}+\sqrt{8}+\sqrt{9}+\sqrt{11}& \approxeq & 12.38\end{array}$$
3. $$\begin{array}{ccccc}{\left(\sum _{i=1}^5{x}_i\right)}^2& =& {\left(1+5+8+9+11\right)}^2& =& 1156\end{array}$$

It would not change. The correlation is a combination of scores which have been scaled to a uniform mean and standard deviation. Changing between inches and centimeters is a linear transform that will be canceled out in computing the correlation.

Given:

$$\begin{array}{ccc}\left|x\right|& =& 127\end{array}$$

Find:

$$\left(\underset{\sigma \to \mathrm{\infty }}{lim}z\right)$$

This derivation relies on a transformation of the variance when dealing with z-scores: (Recall that ${\sigma }_z=1$ and ${\mu }_z=0$.)

$$\begin{array}{ccc}{\sigma }_z^2& =& \frac{\sum {\left(z_i-\stackrel{\_}{z}\right)}^2}{n-1}\\ 1& =& \frac{\sum {\left(z_i-0\right)}^2}{n-1}\\ \sum {z_i}^2& =& n-1\end{array}$$ $$\begin{array}{ccc}\sum {z_i}^2& =& n-1\end{array}$$

Assume then that to take $\sigma \to \mathrm{\infty }$ for a set $x$, you take the element $x_n\to \mathrm{\infty }$. As you do this the z-scores for the elements $x_1$ to $x_{n-1}$ will approach some uniform value, $-k$. Since the mean for the set of z-scores is by definition 0, the z-score for $x_n$ has to be $\left(n-1\right)k$. From the previous equation then:

$$\begin{array}{ccc}n-1& =& \sum {z_i}^2\\ & =& \sum _{i=1}^{n-1}{z_i}^2+{z_n}^2\\ & =& \sum _{i=1}^{n-1}{\left(-k\right)}^2+{\left(\left(n-1\right)k\right)}^2\\ & =& \left(n-1\right)k^2+{\left(n-1\right)}^2{k}^2& =& n-1\\ k& =& \sqrt{\frac{n-1}{\left(n-1\right)+{\left(n-1\right)}^2}}\\ & =& \frac{1}{\sqrt{n}}\end{array}$$

So, substituting back into the original equation:

$$\begin{array}{ccccc}\underset{\sigma \to \mathrm{\infty }}{lim}z& =& \left(n-1\right)k& =& \frac{n-1}{\sqrt{n}}\end{array}$$

The original values can then be substituted in to get:

$$\begin{array}{ccccc}\underset{\sigma \to \mathrm{\infty }}{lim}z& =& \frac{126}{\sqrt{\mathrm{127}}}& \approxeq & 11.18\end{array}$$