Chapter 14, Problem 14.10AYK

(a)

To determine

To find out what is the sampling distribution of the mean blood arsenic concentration $\overline{x}$ in many samples of $25$ adults if the claim is true and sketch the density curve of this distribution.

Answer to Problem 14.10AYK

The sampling distribution of the mean blood arsenic concentration $\overline{x}$ in many samples of $25$ adults is $\overline{x}\sim N(\mu =3.2,\sigma =0.3)$ .

Explanation of Solution

It is given in the question that Arsenic blood concentrations in healthy individuals are normally distributed with mean $\mu =3.2$ micro per deciliter and standard deviation $\sigma =1.5\mu g/dl$ . And they have taken a sample of adults and asked whether the data provide good evidence that the mean blood arsenic concentration in this population is elevated compared with the $3.2\mu g/dl$ mea of the population of healthy individuals. Thus, the parameter defined in this scenario is $\mu$ , the mean blood arsenic concentration in this population. And the null and alternative hypotheses is defined by:

  $\begin{array}{l}{H}_0:\mu =3.2\mu g/dl\\ H_a:\mu \ne 3.2\mu g/dl\end{array}$

Now, we know that the mean blood arsenic concentration $\overline{x}$ in many samples of $25$ adults follows the Normal distribution as the population is normal. So, the sampling distribution of the mean blood arsenic concentration $\overline{x}$ in many samples of $25$ adults is $\mu =3.2$ micro per deciliter. And we have,

  $\begin{array}{l}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{1.5}{\sqrt{25}}\\ =0.3\end{array}$

Thus, we have, $\overline{x}\sim N(\mu =3.2,\sigma =0.3)$ .

And the sketch of the density curve of this distribution is as follows:

  

(b)

To determine

To obtain P-value for this area and what would we conclude.

Answer to Problem 14.10AYK

The P-value is $p=0.6171$ .

Explanation of Solution

It is given in the question that Arsenic blood concentrations in healthy individuals are normally distributed with mean $\mu =3.2$ micro per deciliter and standard deviation $\sigma =1.5\mu g/dl$ . And they have taken a sample of adults and asked whether the data provide good evidence that the mean blood arsenic concentration in this population is elevated compared with the $3.2\mu g/dl$ mea of the population of healthy individuals. Thus, the parameter defined in this scenario is $\mu$ , the mean blood arsenic concentration in this population. And the null and alternative hypotheses is defined by:

  $\begin{array}{l}{H}_0:\mu =3.2\mu g/dl\\ H_a:\mu \ne 3.2\mu g/dl\end{array}$

And $\overline{x}\sim N(\mu =3.2,\sigma =0.3)$ .

Now, the first SRS gives,

  $\begin{array}{l}\overline{x}=3.35\\ \sigma =1.5\\ n=25\\ \mu =3.2\end{array}$

So, we have the test statistics value and P-value as:

  $\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{3.35-3.2}{\frac{1.5}{\sqrt{25}}}\\ =0.5\\ P-\text{value}=2\times P(z<0.5)\\ \text{From the }z\text{ table we get,}\\ p=0.6171\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that we fail to reject the null hypothesis and there is no evidence that there is statistical difference.

(c)

To determine

To find out what would be the P-value for this area and what would we conclude.

Answer to Problem 14.10AYK

The P-value is $p=0.0668$ .

Explanation of Solution

It is given in the question that Arsenic blood concentrations in healthy individuals are normally distributed with mean $\mu =3.2$ micro per deciliter and standard deviation $\sigma =1.5\mu g/dl$ . And they have taken a sample of adults and asked whether the data provide good evidence that the mean blood arsenic concentration in this population is elevated compared with the $3.2\mu g/dl$ mea of the population of healthy individuals. Thus, the parameter defined in this scenario is $\mu$ , the mean blood arsenic concentration in this population. And the null and alternative hypotheses is defined by:

  $\begin{array}{l}{H}_0:\mu =3.2\mu g/dl\\ H_a:\mu \ne 3.2\mu g/dl\end{array}$

And $\overline{x}\sim N(\mu =3.2,\sigma =0.3)$ .

Now, the second SRS gives,

  $\begin{array}{l}\overline{x}=3.75\\ \sigma =1.5\\ n=25\\ \mu =3.2\end{array}$

So, we have the test statistics value and P-value as:

  $\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{3.75-3.2}{\frac{1.5}{\sqrt{25}}}\\ =1.83\\ P-\text{value}=2\times P(z<1.83)\\ \text{From the }z\text{ table we get,}\\ p=0.0668\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that we fail to reject the null hypothesis and there is no evidence that there is statistical difference.

(d)

To determine

To explain briefly why these P-valuestell us that one outcome is strong evidence against the null hypothesis and that the other outcome is not.

Explanation of Solution

It is given in the question that Arsenic blood concentrations in healthy individuals are normally distributed with mean $\mu =3.2$ micro per deciliter and standard deviation $\sigma =1.5\mu g/dl$ . And they have taken a sample of adults and asked whether the data provide good evidence that the mean blood arsenic concentration in this population is elevated compared with the $3.2\mu g/dl$ mea of the population of healthy individuals. Thus, the parameter defined in this scenario is $\mu$ , the mean blood arsenic concentration in this population. And the null and alternative hypotheses is defined by:

  $\begin{array}{l}{H}_0:\mu =3.2\mu g/dl\\ H_a:\mu \ne 3.2\mu g/dl\end{array}$

And $\overline{x}\sim N(\mu =3.2,\sigma =0.3)$ .

Now, from the first SRS we get the P-value $p=0.6171$ and from the second SRS we get $p=0.0668$ . So, as we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected. Now, the first P-value is not statistically significant at the $\alpha =0.05$ level and the second P-value is not statistically significant at the $\alpha =0.05$ level. However, the actual P-values are more informative because we can conclude that the studies can provide very strong evidence against null hypothesis.