Q 12RP. - Elementary Statistics

Question

Western Pygmy-Possum. The foraging behavior of the western pygmy-possum was investigated in the article "Strategies of a Small Nectarivorous Marsupial, the Western Pygmy-Possum, in Response to Seasonal Variation in Food Availability" (Journal of Mammalogy, Vol. 96, No. 6, pp. 1525-1535) by D. Morrant and S. Petit. The weights of adult male pygmy-possums in Australia are normally distributed with a mean of $8.5 \mathrm{g}$ and a standard deviation of $0.3 \mathrm{g}$

a. Sketch the normal curve for the pygmy-possum weights.

b. Find the sampling distribution of the sample mean for samples of size $4$ Draw a graph of the normal curve associated with $\overline{x}$

c. Repeat part (b) for samples of size $9$

Step-by-Step Solution

Answer

Part (b) For samples of size 4, the sampling distribution of the sample mean is mean $\left({\mu }_x\right) 8.5 \mathrm{g}$ and the standard deviation $\left({\sigma }_X\right)$ is equal to $0.15 \mathrm{g}$

Part (c) For samples of size $4$, the sampling distribution of the sample mean is mean $\left({\mu }_x\right) 8.5 \mathrm{g}$ and standard deviation $\left({\sigma }_X\right)$ is equal to $0.1 \mathrm{g}$

Part (a) The normal curve for the pygmy-possum weights. 

Part (a) Step 1: Given information

The weight of adult male pygmy-possums $\left(x\right)$ is normally distributed with mean $\left(\mu \right)8.5 \mathrm{g}$ and standard deviation $\left(\sigma \right) 0.3 \mathrm{g}$

Part (a) Step 2: Concept

The formula used: Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Explanation

To create the normal curve for the pygmy-possum weights, use MINITAB.

The weight of adult male pygmy-possums $\left(x\right)$ is assumed to be regularly distributed, with a mean of $\left(\mu \right) 8.5 \mathrm{g}$ and a standard deviation of $\left(\sigma \right) 0.3 \mathrm{g}$

Procedure for MINITAB: 

Step (1): Select Graph $>$ Probability Distribution Plot $>$ View Single $>$Ok from the drop-down menu.

Step (2): Select a Normal from the Distribution menu.

Step (3): Enter $\mathbf{8}\mathbf{.}\mathbf{5}$ in Mean and $\mathbf{0}\mathbf{.}\mathbf{3}$ in Standard deviation.

Step (4): Click OK.

MINITAB output: 

Part (b) Step 1: Calculation

For samples of size $4$, obtain the sampling distribution of the sample mean.

The sample distribution's mean $\left({\mu }_x\right)$ is $8.5 \mathrm{g}$, and the standard deviation is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.3}{\sqrt{4}} \\ =\frac{0.3}{2} \\ =0.15 \end{gathered}$$

For samples of size $4$, the sampling distribution of the sample mean is mean $\left({\mu }_x\right)8.5 \mathrm{g}$ and standard deviation $\left({\sigma }_{\overline{s}}\right)$ is equal to $0.15 \mathrm{g}$

Part (b) Step 2: Explanation

To create the normal curve for the sample distribution, use MINITAB.

Procedure for MINITAB: 

Step (1): Select Graph$>$ Probability Distribution Plot $>$ View Single $>$Ok from the drop-down menu.

Step 2: Select a Normal from the Distribution menu.

Step 3: Enter $\mathbf{8}\mathbf{.}\mathbf{5}$ in Mean and $\mathbf{0}\mathbf{.}\mathbf{15}$ in Standard deviation.

Step 4: Click OK.

MINITAB output:  

Procedure for MINITAB: 

Step (1): Select Graph

Part (c) Step 1: Calculation

For samples of size $9$, obtain the sampling distribution of the sample mean.

The sample distribution's mean $\left({\mu }_x\right)$ is $8.5 \mathrm{g}$, and the standard deviation is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.3}{\sqrt{9}} \\ =\frac{0.3}{3} \\ =0.1 \end{gathered}$$

For samples of size $4$, the sampling distribution of the sample mean is mean $\left({\mu }_x\right) 8.5 \mathrm{g}$ and standard deviation $\left({\sigma }_x\right)$ is $0.1 \mathrm{g}$

Part (c) Step 2: Calculation

Procedure for MINITAB: 

Step (1): Select Graph$>$Probability Distribution Plot $>$ View Single $>$Ok from the drop-down menu.

Step (2): Select a Normal from the Distribution menu.

Step (3): Enter $\mathbf{8}\mathbf{.}\mathbf{5}$ in Mean and $\mathbf{0}\mathbf{.}\mathbf{1}$ in Standard deviation.

Step (4): Click OK.

MINITAB output: