Chapter 14, Problem 14.9AYK

(a)

To determine

To find out what is the sampling distribution of $\overline{x}$ in many samples of sizes $50$ if in fact $\mu =12$ and make a sketch of the normal curve for this distribution.

Answer to Problem 14.9AYK

  $\overline{x}\sim N(\mu =12,\sigma =0.23)$ .

Explanation of Solution

In the question, it is given that people with less than $12$ grams of hemoglobin per deciliter of blood are anemic. Thus, a public health official in Jordan suspects that the mean $\mu$ for all children is less than $12$ g/dl because of inadequate health conditions. Therefore, we can say that the parameter defined in this scenario is $\mu$ , the mean for all children is less than $12$ g/dl in Jordan. And also the null and alternative hypotheses is defined as follows:

  $\begin{array}{l}{H}_0:\mu =12\text{ grams}\\ H_a:\mu <12\text{ grams}\end{array}$

Thus, we can say that $\overline{x}$ follows Normal distribution with,

  $\mu =12$

  $\begin{array}{l}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{1.6}{\sqrt{50}}\\ =0.23\end{array}$

Thus, we have $\overline{x}\sim N(\mu =12,\sigma =0.23)$ .

And the sketch of the normal curve for this distribution is as follows:

  

(b)

To determine

To mark the outcome on the sampling distribution and enter the hypotheses, n, $\overline{x}$ and $\sigma$ for the Jordan study into the P-value of a test of significance applet and find out what is the P-value for this study and what do you conclude.

Answer to Problem 14.9AYK

The P-value is $p=0.002$ .

Explanation of Solution

In the question, it is given that people with less than $12$ grams of hemoglobin per deciliter of blood are anemic. Thus, a public health official in Jordan suspects that the mean $\mu$ for all children is less than $12$ g/dl because of inadequate health conditions. Therefore, we can say that the parameter defined in this scenario is $\mu$ , the mean for all children is less than $12$ g/dl in Jordan. And also the null and alternative hypotheses is defined as follows:

  $\begin{array}{l}{H}_0:\mu =12\text{ grams}\\ H_a:\mu <12\text{ grams}\end{array}$

Also, we have,

  $\begin{array}{l}\overline{x}=11.3\\ \sigma =1.6g/dl\\ n=50\\ \mu =12\end{array}$

Now, let us calculate the z, test statistics as:

  $\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{11.3-12}{\frac{1.6}{\sqrt{50}}}\\ =-3.09\\ P-\text{value}=2\times P(z<-3.09)\\ \text{From the }z\text{ table we get,}\\ =0.002\end{array}$

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,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that null hypothesis is rejected.

(c)

To determine

To use the applet to obtain the P-value for this country and what do you conclude.

Answer to Problem 14.9AYK

The P-value is $p=0.3789$ .

Explanation of Solution

In the question, it is given that people with less than $12$ grams of hemoglobin per deciliter of blood are anemic. Thus, a public health official in Jordan suspects that the mean $\mu$ for all children is less than $12$ g/dl because of inadequate health conditions. Therefore, we can say that the parameter defined in this scenario is $\mu$ , the mean for all children is less than $12$ g/dl in Jordan. And also the null and alternative hypotheses is defined as follows:

  $\begin{array}{l}{H}_0:\mu =12\text{ grams}\\ H_a:\mu <12\text{ grams}\end{array}$

Also, we have,

  $\begin{array}{l}\overline{x}=11.8\\ \sigma =1.6g/dl\\ n=50\\ \mu =12\\ c=95\%\end{array}$

Now, using the applet to obtain the P-value for this country, we get,

  $\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{11.8-12}{\frac{1.6}{\sqrt{50}}}\\ =-0.88\\ P-\text{value}=2\times P(z<-0.88)\\ \text{From the }z\text{ table we get,}\\ =0.3789\end{array}$

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 null hypothesis.

(d)

To determine

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

Explanation of Solution

In the question, it is given that people with less than $12$ grams of hemoglobin per deciliter of blood are anemic. Thus, a public health official in Jordan suspects that the mean $\mu$ for all children is less than $12$ g/dl because of inadequate health conditions. Therefore, we can say that the parameter defined in this scenario is $\mu$ , the mean for all children is less than $12$ g/dl in Jordan. And also the null and alternative hypotheses is defined as follows:

  $\begin{array}{l}{H}_0:\mu =12\text{ grams}\\ H_a:\mu <12\text{ grams}\end{array}$

Now, in part (b) we have given P-value as $p=0.002$ and in part (c) the P-value is given as $p=0.3789$ . So, 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,

From the part (b) that it is low which says that the study outcome would be very unlikely that null hypothesis is true where as in part (c), the P-value is not low which is inconclusive so we fail to reject null hypothesis.

(e)

To determine

To find out is the outcome for either study statistically significant at the $\alpha =0.05$ level and explain briefly why reporting the actual P-value is more relevant to your conclusion than simply reporting whether the findings are significantly at the $\alpha =0.05$ level.

Answer to Problem 14.9AYK

Yes, the outcome for first study is statistically significant at the $\alpha =0.05$ level but second one is not.

Explanation of Solution

In the question, it is given that people with less than $12$ grams of hemoglobin per deciliter of blood are anemic. Thus, a public health official in Jordan suspects that the mean $\mu$ for all children is less than $12$ g/dl because of inadequate health conditions. Therefore, we can say that the parameter defined in this scenario is $\mu$ , the mean for all children is less than $12$ g/dl in Jordan. And also the null and alternative hypotheses is defined as follows:

  $\begin{array}{l}{H}_0:\mu =12\text{ grams}\\ H_a:\mu <12\text{ grams}\end{array}$

Now, in part (b) we have given P-value as $p=0.002$ and in part (c) the P-value is given as $p=0.3789$ . 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 statistically significant at the $\alpha =0.05$ level but 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 first study provided very strong evidence against null hypothesis.