Chapter 7.3, Problem 111E

To determine

To find: The standard deviations and describe why it is a good idea to choose larger standard deviation.

Answer to Problem 111E

Solution: The standard deviations are ${\sigma }_N=17.50015,{\sigma }_S=14.25831,$${\sigma }_E=16.07433$, and ${\sigma }_W=15.33142$. To overcome the unexpected variation (noise), it is advisable to choose a slightly larger value of $\sigma$. As the value of standard deviation $\left(\sigma \right)$ increase, the power decreases.

Explanation of Solution

Calculation: To calculate the standard deviation for four regions with reference to Exercise 7.85 and Exercise 7.86, on excel following steps were followed:

Step 1: Open the data set for exercise 7.85. The partial screenshot is shown below:

Step 2: Write the standard deviation formula $\text{=STDEV}\text{.S(B2:B31)}$ under the formula tab and then select the data for north as:

Step 3: Write the standard deviation formula $\text{=STDEV}\text{.S(B32:B61)}$ under the formula tab and then select the data for south as:

Step 4: Open the data set for exercise 7.86. The partial screenshot is shown below:

Step 5: Go to the Write the standard deviation formula $\text{=STDEV}\text{.S(B2:B31)}$ under the formula tab and then select the data for east as:

Step 6: Write the standard deviation formula $\text{=STDEV}\text{.S(B2:B31)}$ under the formula tab and then select the data for west as:

To see the impact of standard deviation on power statistic, consider arbitrary choose some different $\sigma$ and calculate the power for each $\sigma$ value. The power is represented as $P\left(z>t^{*}-\delta \right)$, where $\delta$ is the noncentrality factor which depends on $\sigma$.

$\text{For }{\sigma }_1=7.4$, the noncentrality parameter can be obtained as:

$\begin{array}{c}\text{The noncentrality parameter}(\delta )=\frac{{\mu }_1-{\mu }_2}{{\sigma }_1\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)}}\\ =\frac{5}{7.4\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$45$}\right.\right)}}\\ =3.21\end{array}$

The power can be calculated as:

$\begin{array}{c}\text{Power}=P\left(z>t^{*}-\delta \right)\\ =P\left(z>1.99-3.21\right)\\ =P\left(z>-1.22\right)\\ =0.8888\end{array}$

$\text{For }{\sigma }_2=7.7$, the noncentrality parameter can be obtained as:

$\begin{array}{c}\text{The noncentrality parameter}(\delta )=\frac{{\mu }_1-{\mu }_2}{{\sigma }_2\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)}}\\ =\frac{5}{7.7\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$45$}\right.\right)}}\\ =3.08\end{array}$

The power can be calculated as:

$\begin{array}{c}\text{Power}=P\left(z>t^{*}-\delta \right)\\ =P\left(z>1.99-3.08\right)\\ =P\left(z>-1.09\right)\\ =0.8621\end{array}$

$\text{For }{\sigma }_3=8$, the noncentrality parameter can be obtained as:

$\begin{array}{c}\text{The noncentrality parameter}(\delta )=\frac{{\mu }_1-{\mu }_2}{{\sigma }_2\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)}}\\ =\frac{5}{8\sqrt{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$45$}\right.\right)}}\\ =2.964\end{array}$

The power can be calculated as:

$\begin{array}{c}\text{Power}=P\left(z>t^{*}-\delta \right)\\ =P\left(z>1.99-2.964\right)\\ =P\left(z>-0.974\right)\\ =0.8350\end{array}$

Noise is the unexpected variation in experiments. To consider their effect, it is recommended to choose larger value of $\sigma$ than their experimental value.