Question

Venture-Capital Investments. Data on investments in the high-tech industry by venture capitalists are compiled by Venture One Corporation and published in America's Network Telecom Investor Supplement. A random sample of $18$ venture-capital investments in the fiber optics business sector yielded the following data, in millions of dollars.

Determine and interpret a $95\%$ confidence interval for the mean amount, $\mu$, of all venture-capital investments in the fiber optics business sector. Assume that the population standard deviation is $\backslash (2.04$ million. (Note: The sum of the data is $\backslash )113.97$ million.)

Short answer

Part (a) $95\%$Confidence interval of the population mean $\mu$is $(5.3893,7.2741)$

Part (b) The average amount of all venture capital investments falls somewhere between $\$5.3893$ million and $\$7.27141$ million.

Step-by-step solution

Part (a) Step 1: Given information

5.60	6.27	5.96	10.51	2.04	5.48
5.74	5.58	4.13	8.63	5.95	6.67
4.21	7.71	9.21	4.98	8.64	6.66

Part (a) Step 2: Calculation

Let $\mu$be the population mean venture capital investment, and s.d. be the population $\sigma =\$2.04$million

A size $18$sample is picked at random.

Let $x_i$be the $i-th$sample observation on venture capita investment, $i=1,2,\dots ..,18$

Given that $\sum _{i=1}^{18}xi=113.97$

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

As a result, the average amount of venture capital investment in the sample is$\$6.3317$ million.

Confidence level $95\%=100*0.95\%$

$\therefore 1-\alpha =0.95$[comparing with the general form of [$100(1-\alpha )\%$the confidence interval ]

$$\begin{gathered} \Rightarrow \alpha =1-0.95 \\ \Rightarrow \alpha =0.05 \\ \therefore z_{\frac{\alpha }{2}}=z_{\frac{0.05}{2}}=z_{0.025}=1.96 \end{gathered}$$

Part (a) Step 3: Calculation

The $95\%$ confidence interval for the population mean $\mu$ is calculated as follows:

$$\begin{gathered} \left(\overline{x}-z_{0.025}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{0.025}\times \frac{\sigma }{\sqrt{n}}\right) \\ \therefore \overline{x}-z_{0.025}\times \frac{\sigma }{\sqrt{n}} \\ =6.3317-1.96\times \frac{2.04}{\sqrt{18}} \\ =6.3317-0.9424 \\ =5.3893 \end{gathered}$$

And

$$\begin{gathered} \overline{x}+z_{0.025}\times \frac{\sigma }{\sqrt{n}} \\ =6.3317+0.9424 \\ =7.2741 \end{gathered}$$

$\therefore 95\%$ Confidence interval of population mean $\mu$ is $(5.3893,7.2741)$

Part (b) Step 1: Calculation

We can state with $95\%$ confidence that the average amount of all venture capital investments falls somewhere between $\$5.3893$ million and $\$7.27141$ million.