Question

A random sample of 90 observations produced a mean x = 25.9 and a standard deviation s = 2.7.

a. Find an approximate 95% confidence interval for the population mean$\mathit{\mu }$

b. Find an approximate 90% confidence interval for$\mathit{\mu }$

c. Find an approximate 99% confidence interval for$\mathit{\mu }$

Short answer

The standard deviation is commonly utilized as a measurement of an asset's comparative volatility. The standard deviation is determined as the square root of the variation by calculating the departure of every observation point from the mean.

Step-by-step solution

(a) The data is given below

$$\begin{gathered} \text{Sample size}\left(\text{n}\right)=90 \\ \overline{X}=25.9 \\ \text{s}=2.7 \end{gathered}$$

Here, the population standard deviation is known in this case, we use the t interval.

$$\begin{gathered} \text{Degrees of freedom = n}-1 \\ =90-1 \\ =89 \end{gathered}$$

The formula is given below:

$\overline{X}-E,\overline{X}+E$

Given the error in the margin

$$\begin{gathered} E\text{=}\frac{tc\times s}{\sqrt{n}} \\ \text{tc}=1.986 \end{gathered}$$

$\begin{array}{rcl}\text{E}& =& \frac{1.986\times 2.7}{\sqrt{90}}\\ & =& \frac{5.3622}{9.486833}\\ & =& 0.565226\end{array}$

The confidence interval is: $\left(\text{25.9}-\text{0.565226, 25.9 + 0.565226}\right)=\text{>}\left(25.3348,26.4652\right)$

(b) The data is given below

The formula is given below:

$\overline{X}-E,\overline{X}+E$

Given the error in the margin

$$\begin{gathered} E\text{=}\frac{tc\times s}{\sqrt{n}} \\ \text{tc}=1.662 \end{gathered}$$

$\begin{array}{rcl}\text{E}& =& \frac{1.662\times 2.7}{\sqrt{90}}\\ & =& \frac{4.4874}{9.486833}\\ & =& 0.473013\end{array}$

The confidence interval is:$\left(\text{25.9}-\text{0.473013, 25.9 + 0.473013}\right)\text{= >}\left(25.4270,26.3730\right)$

(c) The data is given below

The formula is given below:

$\overline{\chi }-E,\overline{\chi }+E$

Given the error in the margin

$\begin{array}{rcl}\mathit{E}& \mathbf{=}& \frac{\mathbf{t}\mathbf{c}\mathbf{\times }\mathbf{s}}{\sqrt{\mathbf{n}}}\\ \mathit{t}\mathit{c}& \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{662}\\ & & \\ \mathbf{\text{E}}& \mathbf{=}& \frac{\mathbf{2}\mathbf{.}\mathbf{632}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{7}}{\sqrt{\mathbf{90}}}\\ & \mathbf{=}& \frac{\mathbf{7}\mathbf{.}\mathbf{1064}}{\mathbf{9}\mathbf{.}\mathbf{486833}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{74908}\\ & & \end{array}$

The confidence interval is: (25.1509, 26.6491)