Venture-Capital Investments. Refer to Exercise 8.69.

a. Find a $99\%$ confidence interval for $\mu$

b. Why is the confidence interval you found in part (a) longer than the one in Exercise $8.69$?

c. Draw a graph similar to that shown in Fig. $8.6$on page $326$to display both confidence intervals.

d. Which confidence interval yields a more accurate estimate of$\mu$? Explain your answer.

$\sigma =\$2.04,n=18$

The formula used:$\overline{x}=\frac{1}{n}\times \sum _{i=1}^{n}xi$and$\overline{x}-z_{0.005}\times \frac{\sigma }{\sqrt{n}}$

Let $\mu$be the population's mean venture capital investment.

Given that population S.D. $\sigma =\$2.04$million and sample of size $n=18$

Let $x_i$be $i-th$sample observation on venture-capital investment $x_{i}i=1,2,\dots ,18$

From the data $\sum _{i=1}^{18}xi=113.97$

As a result, the average of all venture capital investment

$\therefore \overline{x}=\frac{1}{n}\times \sum _{i=1}^{n}xi=\frac{1}{18}\times 113.97=6.3317$

We have to find $99\%$ confidence interval of $\mu$

$\therefore$Confidence level $=99\%=100\times 0.99\%$

$\therefore 1-\alpha =0.99$

$\Rightarrow \alpha =0.01$

$\therefore z_{\frac{\alpha }{2}}=z_{\frac{0.01}{2}}=z_{0.005}=2.575$

$99\%$The confidence interval of $\mu$ is given by$\left(\overline{x}-z_{0.005}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{0.005}\times \frac{\sigma }{\sqrt{n}}\right)$

$$\begin{gathered} \therefore \overline{x}-z_{0.005}\times \frac{\sigma }{\sqrt{n}} \\ =6.3317-2.575\times \frac{2.04}{\sqrt{18}} \\ =6.3317-1.2381 \\ =5.0936 \end{gathered}$$

$$\begin{gathered} \overline{x}+z_{0.005}\times \frac{\sigma }{\sqrt{n}} \\ =6.3317+1.2381 \\ =7.5698 \end{gathered}$$

$\therefore 99\%$The confidence interval of $\mu$ is $(5.0936,7.5698)$

i.e., the average amount of venture capital investments in fiber optics is between $\$5.0936$million and$\$7.598$

The $100(1-\alpha )\%$confidence interval of the population mean $\mu$is given by

$\left(\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\right)$

Length of the confidence interval of $\mu$

=Upper confidence limit -lower confidence limit

$$\begin{gathered} =\left(\overline{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\right)-\left(\overline{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\right) \\ =2\times z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}} \end{gathered}$$

Now confidence level $1-\alpha$is increasing

$\iff$value of $\alpha$is decreasing

$\iff$value of $\frac{\alpha }{2}$is decreasing......(A)

$\iff$value of ${\mathrm{z}}_{\frac{\alpha }{2}}$is decreasing......(B)

$\iff$value of $2\times {\mathrm{z}}_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$is decreasing.....[Since particular sample size $n$is constant $\&$$\sigma$population S.D. (known) is also constant

Increasing the confidence level increases the length of the confidence interval.

As a result, the $99\%$confidence interval is longer than the $95\%$confidence interval reported in the $8.31$exercise.

The step's justification $\left(\mathrm{A}\right)Leftrightarrow\left(\mathrm{B}\right)$

i.e. "if $\alpha$decreases value of $z_{\alpha }$increases \& vice -versa"

$z_{\alpha }$is the upper $\alpha$-point of the standard normal distribution.

$z_{\alpha }$is the point for which $P\left(z\ge z_{\alpha }\right)=\alpha$

Where $z~N(0,1)$

i.e. $\alpha$is the area under the standard normal curve on the right side of $z_{\alpha }$

We've looked at $\alpha >{\alpha }^1$ and can see from figs. $1$ and $2$ that for $\alpha >{\alpha }^{\text{'}},z_{\alpha }<z_{{\alpha }^{\text{'}}}$ and vice versa.

Because shifting any point on the horizontal axis to the right reduces the area under the standard normal curve on the right side of that point.

A more exact estimate is obtained using the $95\%$confidence interval. The $95\%$ confidence interval is shorter than the $95\%$ confidence interval because the $95\%$ confidence interval is shorter. That is, we have pinned down (i.e. approximated) $\mu$ better in the case of a $95\%$ confidence interval.