Q 8.9RP. - Elementary Statistics

Question

A variable of a population has a mean of $266$ and a standard deviation of 16. Ten observations of this variable have a mean of $262$ I and a sample standard deviation of $20.4$ Obtain the observed value of the

a. standardized version of $\hat{x}$

b. Studentized version of $\overline{x}$

Step-by-Step Solution

Answer

Part (a) $-0.77$

Part (b) $-0.60$

Part (a) Step 1: Given information

The population mean $\mu =266$ and population S.D $\sigma =16$

Sample size $m=10$

Sample mean $\overline{x}=262.1$

$s=20.4$

Part (a) Step 2: Concept

The formula used:  Standardized version of $z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$, 

Part (a) Step 3: Calculation

Standardized version of $\overline{x}, z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

$$\begin{gathered} =\frac{262.1-266}{\frac{16}{\sqrt{10}}} \\ =\frac{-3.9\times \sqrt{10}}{16} \\ \approx -0.77 \end{gathered}$$

As a result, the standardized version of $\overline{x}$ is $-0.77$

Part (b) Step 1: Calculation

Studentized version of $\overline{x}, t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

$$\begin{gathered} =\frac{262.1-266}{\frac{20.4}{\sqrt{10}}} \\ =\frac{-3.9\times \sqrt{10}}{20.4} \\ \approx -0.60 \end{gathered}$$

Therefore, the studentized version of $\overline{x}$ is $-0.60$