Chapter 8, Problem 13P

A random sample is selected from a normal population with a mean of $\mu =100$ and a standard deviation of $\sigma =20$ . After a treatment is administered to the individuals in the sample, the sample mean is found to be $M=96$ .

a. How large a sample is necessary for this sample mean to be statistically significant? Assume a two-tailed test with $\alpha =.05$.

b. If the sample mean were $M=98$ , what sample size is needed to be significant for a two-tailed test with $\alpha =.05$ ?

To determine

1. How large a sample is necessary for the sample mean to be statistically significant if the sample mean $=96$?
2. How large a sample is needed for the sample mean to be statistically significant if the sample mean $=98$ ?

Answer to Problem 13P

Solution:

1. The sample size required for the sample mean $=96$ to be statistically significant is $97$.
2. The sample size required for the sample mean $=98$ to be statistically significant is $385$.

Explanation:

1. To determine a sample size that is necessary for the sample mean $=96$ to be statistically significant, we will make use of $z$ statistic.

   We are given:

   $\mu =100$, $\sigma =20$, $M=96$ and $n=?$

   We know that the two-tailed critical value at $5$% significance level $=\pm 1.96$. Now in order to determine the sample size for the sample mean to be statistically significant we will have

   $z$ - score $=\pm 1.96$. Therefore, we have:

   $z=\frac{M-\mu }{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   Let's take $z=-1.96$

   $-1.96=\frac{96-100}{\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   $-1.96=\frac{-4\times \sqrt{n}}{20}$

   $-1.96\times 20=-4\times \sqrt{n}$

   $-39.2=-4\times \sqrt{n}$

   $\sqrt{n}=\frac{-39.2}{-4}$

   $\sqrt{n}=9.8$

   $n={9.8}^2=96.04\approx 97$

   Therefore, the sample size required for the sample mean $=96$ to be statistically significant is $97$.

2. To determine a sample size that is necessary for the sample mean $=98$ to be statistically significant, we will make use of $z$ statistic.

   We are given:

   $\mu =100$, $\sigma =20$, $M=98$ and $n=?$

   We know that the two-tailed critical value at $5$% significance level $=\pm 1.96$. Now in order to determine the sample size for the sample mean to be statistically significant we will have

   $z$ - score $=\pm 1.96$. Therefore, we have:

   $z=\frac{M-\mu }{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   Let's take $z=-1.96$

   $-1.96=\frac{98-100}{\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   $-1.96=\frac{-2\times \sqrt{n}}{20}$

   $-1.96\times 20=-2\times \sqrt{n}$

   $-39.2=-2\times \sqrt{n}$

   $\sqrt{n}=\frac{-39.2}{-2}$

   $\sqrt{n}=19.6$

   $n={19.6}^2=384.16\approx 385$

   Therefore, the sample size required for the sample mean $=98$ to be statistically significant is $385$.

Conclusion:

1. The sample size required for the sample mean $=96$to be statistically significant at $\alpha =0.05$ is $97$.
2. The sample size required for the sample mean $=98$to be statistically significant at $\alpha =0.05$ is $385$.

Explanation of Solution

1. To determine a sample size that is necessary for the sample mean $=96$ to be statistically significant, we will make use of $z$ statistic.

   We are given:

   $\mu =100$, $\sigma =20$, $M=96$ and $n=?$

   We know that the two-tailed critical value at $5$% significance level $=\pm 1.96$. Now in order to determine the sample size for the sample mean to be statistically significant we will have

   $z$ - score $=\pm 1.96$. Therefore, we have:

   $z=\frac{M-\mu }{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   Let's take $z=-1.96$

   $-1.96=\frac{96-100}{\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   $-1.96=\frac{-4\times \sqrt{n}}{20}$

   $-1.96\times 20=-4\times \sqrt{n}$

   $-39.2=-4\times \sqrt{n}$

   $\sqrt{n}=\frac{-39.2}{-4}$

   $\sqrt{n}=9.8$

   $n={9.8}^2=96.04\approx 97$

   Therefore, the sample size required for the sample mean $=96$ to be statistically significant is $97$.

2. To determine a sample size that is necessary for the sample mean $=98$ to be statistically significant, we will make use of $z$ statistic.

   We are given:

   $\mu =100$, $\sigma =20$, $M=98$ and $n=?$

   We know that the two-tailed critical value at $5$% significance level $=\pm 1.96$. Now in order to determine the sample size for the sample mean to be statistically significant we will have

   $z$ - score $=\pm 1.96$. Therefore, we have:

   $z=\frac{M-\mu }{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   Let's take $z=-1.96$

   $-1.96=\frac{98-100}{\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}$

   $-1.96=\frac{-2\times \sqrt{n}}{20}$

   $-1.96\times 20=-2\times \sqrt{n}$

   $-39.2=-2\times \sqrt{n}$

   $\sqrt{n}=\frac{-39.2}{-2}$

   $\sqrt{n}=19.6$

   $n={19.6}^2=384.16\approx 385$

   Therefore, the sample size required for the sample mean $=98$ to be statistically significant is $385$.

Conclusion:

1. The sample size required for the sample mean $=96$to be statistically significant at $\alpha =0.05$ is $97$.
2. The sample size required for the sample mean $=98$to be statistically significant at $\alpha =0.05$ is $385$.