# Solved (Free): For the proportion of infants subjected to fetal surgery for congenital anomalies, the distribution of gestational ages at birth is approximately normal with unknown

#### ByDr. Raju Chaudhari

Mar 14, 2021

For the proportion of infants subjected to fetal surgery for congenital anomalies, the distribution of gestational ages at birth is approximately normal with unknown mean $\mu$ and standard deviation $\sigma$. A random sample of 15 such infants has mean gestational age =29.6 weeks and standard deviation s= 3.6 weeks

a) Construct a 95% confidence interval for the true population mean $\mu$.
b) What is the length of this interval?
c) How large a sample would be required for the 95% confidence interval to have length 3 weeks? Assume that the population standard deviation $\sigma$ is known and that $\sigma = 3.6$ weeks.
d) How large a sample would be needed for the 95% confidence interval to have length 2 weeks?

#### Solution

a) 95% confidence interval for the true population mean $\mu$

Given that sample size $n = 15$, sample mean $\overline{X}= 29.6$, sample standard deviation $s = 3.6$.

The confidence level is $1-\alpha = 0.95$.

##### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

##### Step 2 Given information

Given that sample size $n =15$, sample mean $\overline{X}=29.6$, sample standard deviation $s=3.6$.

##### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}
where $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, and $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

##### Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t_{\alpha/2,n-1} = t_{0.025,15-1}= 2.145$.

##### Step 5 Compute the margin of error

The margin of error for mean is

 \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.145 \frac{3.6}{\sqrt{15}} \\ & = 1.994. \end{aligned}

##### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is
 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 29.6 - 1.994 & \leq \mu \leq 29.6 + 1.994\\ 27.606 &\leq \mu \leq 31.594. \end{aligned}
Thus, $95$% confidence interval estimate for population mean is $(27.606,31.594)$.

b) The length of this interval is the difference between 31.594 and 27.606. That is $31.594 -27.606 = 3.988$ weeks.

c) The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$. The population standard deviation is $\sigma = 3.6$. The length of the confidence interval $L = 3$ weeks.

Since $\sigma$ is known we use the confidence interval $(\overline{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\overline{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}})$.

The length of the confidence interval is
 \begin{aligned} L &= \overline{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}}-\bigg(\overline{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\bigg)\\ & = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \end{aligned}

The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.

Required sample size can be obtained by solving following equation

 \begin{aligned} & 3 = 2\times z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\ \Rightarrow & 3= 2\times 1.96\frac{3.6}{\sqrt{n}}\\ \Rightarrow & n=\big(\frac{2\times 1.96\times 3.6}{3}\big)^2\\ \Rightarrow & n = 22.128 \end{aligned}
Thus, we require the sample size be $n=23$, to have a length of the confidence interval less than or equal to 3 weeks.

d) The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$. The population standard deviation is $\sigma = 3.6$. The length of the confidence interval $L = 2$ weeks.

Since $\sigma$ is known we use the confidence interval $(\overline{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\overline{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}})$.

The length of the confidence interval is
 \begin{aligned} L &= \overline{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}}-\bigg(\overline{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\bigg)\\ & = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \end{aligned}

The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.

Required sample size can be obtained by solving following equation

 \begin{aligned} & 2 = 2\times z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\ \Rightarrow & 2= 2\times 1.96\frac{3.6}{\sqrt{n}}\\ \Rightarrow & n=(1.96\times 3.6)^2\\ \Rightarrow & n = 49.787 \end{aligned}
Thus, we require the sample size be $n=50$, to have a length of the confidence interval less than or equal to 2 weeks.