Question

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.

Short answer

Part (a) $99\%$Confidence interval of $\mu$is $(5.0936,7.5698)$

Part (b) If $\alpha$decreases value of $z_{\alpha }$increases \& vice -versa

Part (c) 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

Part (d) In case of $95\%$ confidence interval we have pinned down (i.e. estimated) $\mu$ better.

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Concept

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

Part (a) Step 3: Calculation

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$\left(\overline{x}-z_{0.005}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{0.005}\times \frac{\sigma }{\sqrt{n}}\right)$

Part (a) Step 4: Calculation

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

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

$\therefore 99\%$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$\$7.598$

Part (b) Step 1: Calculation

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"

Part (c) Step 1: Calculation

$z_{\alpha }$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.

Part (c) Step 2: Explanation

Part (d) Step 1: Explanation

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.