Chapter 26.7, Problem 12RE

(a)

To determine

Delineate whether it is usual that an investigator would get as many pairs as Person R with the treatment animal having the heavier cortex.

Answer to Problem 12RE

There is no evidence that there appears to be 50-50 chance for each animal in a pair to have heavier cortex.

Explanation of Solution

Let ‘T’ represent the treatment and ‘C’ represent the control.

The cortex weight of the animal in the treatment is 689 milligrams and the cortex weight of the animal in the control is 657 milligrams.

Based on the given information, the following values are known.

On observing the table, 52 pairs have higher value for the treatment and 7 have higher value for the control group.

$\begin{array}{l}\text{Sample}\text{size}\left(\text{Number of draws}\right)=59\\ \text{Observed count}=52\end{array}$

Let ‘0’ represent the tails.

Let ‘1’ represent the heads.

In this scenario, consider a box containing 2 tickets in which 1 ticket is labeled 1 and 1 ticket is labeled 0.

Therefore, the null hypothesis states that the number of higher values for the treatment is the sum of 59 random draws (with replacement) from a box containing a ticket labeled 1 and a ticket labeled 0.

On the other hand, the alternative hypothesis states that there are many higher values.

The expected value sum is as given below:

The expected value is obtained using the product of the percentage of 1’s in the box and the number of draws.

$\begin{array}{c}\text{Expected value sum}=\left\{\begin{array}{l}\text{Percentage of 1's}\times \\ \text{Number of draws}\end{array}\right\}\\ =50\%\times 59\\ =0.50\times 59\\ =29.5\end{array}$

Standard deviation of the box:

$\begin{array}{c}SD=\left(\text{Big number}-\text{small number}\right)\sqrt{\left(\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right)}\\ =\left(1-0\right)\sqrt{0.50\times \left(1-0.50\right)}\\ =\sqrt{0.25}\\ =0.5\end{array}$

Standard error of the sum:

$\begin{array}{c}\text{SE sum}=\sqrt{\text{Number of draws}}\times \text{SD box}\\ \text{=}\sqrt{59}\times 0.5\\ =3.8406\end{array}$

The known values are given below:

$\begin{array}{l}\mu =\text{Mean}=29.5\\ \sigma =3.8406\\ x=52\end{array}$

The formula for z-score is as follows:

$z=\frac{x-\mu }{\sigma }$

The z-score is obtained as given below:

$\begin{array}{c}z=\frac{52-29.5}{\text{3}\text{.8406}}\\ =5.86\end{array}$

The probability P is given below:

$\begin{array}{c}P\left(X>52\right)=P\left(Z>5.86\right)\\ =\frac{1}{2}\left(1-\left[P\left(Z<5.86\right)-P\left(Z<-5.86\right)\right]\right)\end{array}$

Using the standard normal table, the value corresponding to $P\left(Z<-5.86\right)$ is 0 and $P\left(Z<5.86\right)$ is 1.

Remaining calculation:

$\begin{array}{c}P\left(Z>5.86\right)=\frac{1}{2}\left(1-\left[1-0\right]\right)\\ =\frac{1}{2}\left(1-1\right)\\ =0\end{array}$

Since the P-value is less than any of the level of significance, it is unusual to obtain a difference in the sample percentage of 50% when there is difference in the population percentage and thus, the difference does not appear due to chance variation.

Therefore, there is evidence that there appears to be no 50-50 chance for each animal in a pair to have a heavier cortex.

(b)

To determine

Obtain the average and SD of the difference.

State whether there is evidence that the treatment has no effect.

Answer to Problem 12RE

There is no evidence that the treatment has no effect.

Explanation of Solution

Based on the given information, the following values are known:

$\text{Number of draws}=59$

The mean of the sample is as given below:

$\begin{array}{c}\text{Average sample}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{\left(689-657\right)+\left(656-623\right)+\mathrm{...}+\left(649-602\right)}{59}\\ =\frac{32+33+\mathrm{...}+47}{59}\\ =36.2373\end{array}$

The standard deviation is as given below:

$\begin{array}{c}SD\text{ sample}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(32-36.2373\right)}^2+{\left(33-36.2373\right)}^2+\mathrm{...}+{\left(47-36.2373\right)}^2}{59-1}}\\ =31.4831\end{array}$

The standard error of the sum is given below:

$\begin{array}{c}\text{SE}\text{Sum}=\sqrt{\text{Sample size }}\times \text{SD sample}\\ =\sqrt{59}\times 31.4831\\ =241.8263\end{array}$

The standard error of the average is the standard error of the sum divided by the number of draws, which is given below:

$\begin{array}{c}\text{SE avearge }=\frac{\text{SE sum}}{\text{Sample size}}\\ =\frac{241.8263}{59}\\ =4.0988\end{array}$

The test hypotheses are given below:

Let $\mu$ be the population mean.

Null hypothesis: $H_0:\mu =0$

Alternative hypothesis: $H_1:\mu \ne 0$

The formula for test statistic is as follows:

$z=\frac{\text{Observed}-\text{Expected}}{\text{SE}}$

Known values:

$\begin{array}{c}\mu =0\\ \sigma =4.0988\\ x=36.2373\end{array}$

The z-score is obtained as given below:

$\begin{array}{c}z=\frac{\text{36}\text{.2373}-\text{0}}{\text{4}\text{.0988}}\\ =8.84\end{array}$

The P-value is given below:

$P=P\left(Z>8.84\right)$

Using the standard normal table, the value corresponding to $P\left(Z>8.84\right)$ is 0.

Since the P-value is less than any of the level of significance, it is unusual to obtain a difference in the sample average of 36.2373 when there is difference in the population averages and thus, the difference does not appear due to chance variation.

Therefore, the treatment appears to have an effect.

(c)

To determine

Discuss about the point of precaution and also state whether it is a good idea.

Answer to Problem 12RE

The point of precaution is a good idea.

Explanation of Solution

In this context, a precaution was made that the dissection of a set of littermates were executed at random and the person performing the partition will not know which cage the rate came from.

The random order of the execution of the dissections is important because it is possible that the second cortex weight will always be more accurate than the first cortex weight due to the experience that is influencing the weighing.

 It is also important that the person performing the dissections does not know which littermates received which treatments. Hence, a person could influence the results when he/she knows which treatments were received by those littermates.

Therefore, the precautions were made to make the weighing more accurate and less influenced by bias.

Thus, the point of precaution is a good idea.