Question

Pulmonary Hypertension. In the paper "Persistent Pulmonary Hypertension of the Neonate and Asymmetric Growth Restriction" (Obstetrics \& Gynecology. Vol. $91$ No. $3$pp. $336-341)$. M. Williams et al. reported on a study of characteristics of neonates. Infants treated for pulmonary hypertension, called the PH group. were compared with those not so treated, called the control group. One of the characteristics measured was head circumference. The mean head circumference of the $10$ infants in the PH group was $34.2$ centimeters $\left(\mathrm{cm}\right)$

a. Assuming that head circumferences for infants treated for pulmonary hypertension are normally distributed with standard deviation $2.1\mathrm{cm}$, determine a $90\%$ confidence interval for the mean head circumference of all such infants.

b. Obtain the margin of error, $E$ for the confidence interval you found in part (a).

c. Explain the meaning of $E$in this context in terms of the accuracy of the estimate.

d. Determine the sample size required to have a margin of error of $0.5\mathrm{cm}$with a $95\%$confidence level.

Short answer

Part (a)$(20.91,24.01)$

Part (b)$1.55$

Part (c) Approximately $95\%$of the sample means (i.e. $95\%$of the drawn samples) are projected to differ from $\mu$by nearly $1.55$units.

Part (d) Sample size is$47$

Step-by-step solution

Part (a) Step 1: Given information

The mean head circumference of the $10$infants in the $PH$group was $34.2$ centimeters $\left(\mathrm{cm}\right)$

Part (a) Step 2: Concept

The formula used:$E=z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

Let $\mu$be the population s.d. and $\sigma =4.10$be the population mean % body fat.

We have to determine $95\%$confidence interval of $\mu$

$$\begin{gathered} \therefore \text{Confidence level}=0.95\times 100\% \\ \therefore 1-\alpha =0.95 \\ \alpha =0.05 \\ \therefore z_{\frac{\alpha }{2}}=z_{\frac{6.05}{2}}=z_{0.025}=1.96 \end{gathered}$$

$95\%$Confidence interval of $\mu$is given by

$\left(\overline{x}-z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}\right)=(\overline{x}-E,\overline{x}+E)$

Given that the sample mean $\overline{x}=22.46$

Sample size $n=27$

$$\begin{gathered} \therefore E=z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}} \\ =1.96\times \frac{4.10}{\sqrt{27}} \\ =1.55 \end{gathered}$$

$95\%$Confidence interval of $\mu$

$$\begin{gathered} =(\overline{x}-E,\overline{x}+E) \\ =(22.46-1.55,22.46+1.55) \\ =(20.91,24.01) \end{gathered}$$

i.e. we have a $95\%$confidence level that the average body fat percentage of all female graduate physical therapy students is between $20.91$and $24.01$

Part (b) Step 1: Calculation

Margin of error $E=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} =1.96\times \frac{4.10}{\sqrt{27}} \\ =1.55 \end{gathered}$$

Part (c) Step 1: Calculation

For a $95\%$confidence interval. We are $95\%$certain that our mistake in predicting the population mean $\mu$by sample mean $\overline{x}$is no more than $1.55\%$Or, to put it another way, if we took several simple random samples of size $n=27$from a population with a mean of $\mu$Approximately $95\%$of the sample means (i.e. $95\%$of the drawn samples) are predicted to depart from $\mu$by at most$1.55$units.

Part (d) Step 1: Calculation

We know that the formula for calculating the required sample size for a specific margin of error with a confidence level of $100(1-\alpha )\%$is given by

$n={\left\{\frac{z_{\alpha /2}\times \sigma }{E}\right\}}^2$

Here confidence level $=99\%=100\times 0.99\%$

$$\begin{gathered} \therefore 1-\alpha =0.99 \\ \Rightarrow \alpha =0.01 \\ \Rightarrow \frac{\alpha }{2}=0.05 \\ {z}_{\frac{\alpha }{2}}=z_{0.005}=2.57583 \end{gathered}$$

Required $E=1.55$

Population s.d., $\sigma =4.10$

Thus $n={\left\{\frac{z_{0.005}\times \sigma }{E}\right\}}^2$

$$\begin{gathered} ={\left\{\frac{2.57583\times 4.10}{1.55}\right\}}^2 \\ =46.42 \end{gathered}$$

$D47$

$\therefore$ required sample size is $47$