Chapter 18.6, Problem 11RE

To determine

Fill in the blank with suitable answer for the given statements.

Answer to Problem 11RE

i) For the number of 2’s, the observed value is 0.8 SEs above the expected value.

ii) For the sum of draws, the observed value is 1.33 SEs above the expected value.

Explanation of Solution

Calculation:

From the given information, the box containing one 1’s, two 2’s, and one 5’s, the number of draws is 100.

For sum of the draws:

The average of the box is obtained as follows:

$\begin{array}{c}\text{Average of the box}=\frac{1+2+2+5}{4}\\ =\frac{10}{4}\\ =2.5\end{array}$

The standard deviation of the box is obtained as follows:

$\begin{array}{c}\text{SD of box}=\sqrt{\frac{\text{Sum}{\left(\text{Data value}-\text{Average}\right)}^2}{4}}\\ =\sqrt{\frac{{\left(1-2.5\right)}^2+{\left(2-2.5\right)}^2+{\left(2-2.5\right)}^2+{\left(5-2.5\right)}^2}{4}}\\ =\sqrt{\frac{9}{4}}\\ =1.5\end{array}$

The expected value for the sum is obtained as follows:

$\begin{array}{c}\text{Expected value for the sum}=\text{Number of draws}\times \text{Expected value of box}\\ =100\times 2.5\\ =250\end{array}$

The standard error of the sum is obtained as follows:

$\begin{array}{c}\text{Standard error of the sum}=\sqrt{\text{Number of draws}}\times \text{SD of box}\\ =\sqrt{100}\times 1.5\\ =15\end{array}$

In this case, the number of 1’s is 17, 2’s is 54, and 5’s is 29. Therefore, the observed sum is 270 $\left(=17\times 1+54\times 2+29\times 5\right)$.

The z-score for the observed sum is obtained as follows:

$\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{270-250}{15}\\ =1.33\end{array}$

Therefore, “for the sum of draws, the observed value is 1.33 SEs above the expected value”.

For number of 1’s:

In this case, the box containing single 1 and three 0’s, the number of draws is 100.

The average of the box is obtained as follows:

$\begin{array}{c}\text{Average of the box}=\frac{1+0+0+0}{4}\\ =\frac{1}{4}\\ =0.25\end{array}$

The standard deviation of the box is obtained as follows:

$\begin{array}{c}\text{SD of box}=\sqrt{\frac{\text{Sum}{\left(\text{Data value}-\text{Average}\right)}^2}{4}}\\ =\sqrt{\frac{{\left(1-0.25\right)}^2+{\left(0-0.25\right)}^2+{\left(0-0.25\right)}^2+{\left(0-0.25\right)}^2}{4}}\\ =\sqrt{\frac{0.75}{4}}\\ =0.4330\end{array}$

The expected value for the sum is obtained as follows:

$\begin{array}{c}\text{Expected value for the sum}=\text{Number of draws}\times \text{Expected value of box}\\ =100\times 0.25\\ =25\end{array}$

The standard error of the sum is obtained as follows:

$\begin{array}{c}\text{Standard error of the sum}=\sqrt{\text{Number of draws}}\times \text{SD of box}\\ =\sqrt{100}\times 0.4330\\ =4.33\end{array}$

In this case, the number of 1’s is 17.

The z-score for the number of 1’s is obtained as follows:

$\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{17-25}{4.33}\\ =-1.8\end{array}$

In this case, it can be observed that “for the number of 1’s, the observed value is –1.8 SEs above the expected value”.

For number of 2’s:

Here, change the value of 2’s to 1’s.

In this case, the box containing two 1’s and two 0’s, the number of draws is 100.

The average of the box is obtained as follows:

$\begin{array}{c}\text{Average of the box}=\frac{1+1+0+0}{4}\\ =\frac{2}{4}\\ =0.5\end{array}$

The standard deviation of the box is obtained as follows:

$\begin{array}{c}\text{SD of box}=\sqrt{\frac{\text{Sum}{\left(\text{Data value}-\text{Average}\right)}^2}{4}}\\ =\sqrt{\frac{{\left(1-0.5\right)}^2+{\left(1-0.5\right)}^2+{\left(0-0.5\right)}^2+{\left(0-0.5\right)}^2}{4}}\\ =\sqrt{\frac{1}{4}}\\ =0.5\end{array}$

The expected value for the sum is obtained as follows:

$\begin{array}{c}\text{Expected value for the sum}=\text{Number of draws}\times \text{Expected value of box}\\ =100\times 0.5\\ =50\end{array}$

The standard error of the sum is obtained as follows:

$\begin{array}{c}\text{Standard error of the sum}=\sqrt{\text{Number of draws}}\times \text{SD of box}\\ =\sqrt{100}\times 0.5\\ =5\end{array}$

In this case, the number of 2’s is 54.

The z-score for the number of 1’s is obtained as follows:

$\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{54-50}{5}\\ =0.8\end{array}$

In this case, it can be observed that “for the number of 2’s, the observed value is 0.8 SEs above the expected value”.