Question

Use the computer to generate 500 samples, each containing n = 25 measurements, from a population that contains values of x equal to 1, 2, . . 48, 49, 50 Assume that these values of x are equally likely. Calculate the sample mean $\left(\overline{\chi }\right)$ and median m for each sample. Construct relative frequency histograms for the 500 values of $\left(\overline{\chi }\right)$and the 500 values of m. Use these approximations to the sampling distributions of $\left(\overline{\chi }\right)$and m to answer the following questions:

a. Does it appear that and m are unbiased estimators of the population mean? [Note:$\mathbf{\mu }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{5}$]

b. Which sampling distribution displays greater variation?

Short answer

A histogram is a graphic display that divides a set of statistical points into user-defined categories. The histogram is a well-known charting instrument. It is used to summarize continuous or discrete facts on interval data.

Step-by-step solution

Step-by-Step Solution Step 1: (a) Unbiased population mean estimators

Using the command “sample," we may create 500 + 25 = 12500 integers ranging from 1 to 50.

The sample means will then be calculated using the 25 integers created for each of the 500 samples. Following that, we compute the mean (anticipated number) of all sample means stated in b.

The calculation is given below:

$\text{a}=\text{sample}(\text{1}:\text{5}0,\text{125}00\text{replace}=\text{T})\text{for}(\text{iin1}:\text{5}00)$

role="math" localid="1651466137316" $$\begin{gathered} \left\{\text{b}\right[\text{i}]=\text{mean}(\text{a}\left[\text{i}\right],\text{a}[\text{i}+\text{5}00),\text{a}[\text{i}+\text{1}000),\text{a}[\text{i}+\text{15}00),\text{a}[\text{i}+\text{2}000),\text{a}[\text{i}+\text{25}00),\text{a}[\text{i}+\text{3}000), \\ \text{a}[\text{i}+\text{35}00),\text{a}[\text{i}+\text{4}000),\text{a}[\text{i}+\text{45}00),\text{a}[\text{i}+\text{5}000),\text{a}[\text{i}+\text{55}00),\text{a}[\text{i}+\text{6}000),\text{a}[\text{i}+\text{65}00), \\ \text{a}[\text{i}+\text{7}000),\text{a}[\text{i}+\text{75}00),\text{a}[\text{i}+\text{8}000),\text{a}[\text{i}+\text{85}00),\text{a}[\text{i}+\text{9}000),\text{a}[\text{i}+\text{1}0000),\text{a}[\text{i}+\text{1}0\text{5}00], \\ \text{a}[\text{i}+\text{11}000),\text{a}[\text{i}+\text{115}00),\text{a}[\text{i}+\text{12}000) \end{gathered}$$

Mean (b)

In my instance, the mean of the sample mean was 25.438, which was closer to the population mean $\text{ji}=\text{25}.\text{5}$, and therefore the sample mean seems to be an unbiased approximation of the population mean.

Median

The sample medians will next be calculated using the 25 integers created for each of the 500 samples. Following that, we compute the average (estimated return) of all test medians indicated in c. (i in 1:500)

role="math" localid="1651466172593" $$\begin{gathered} \text{{c}\left[\text{i}\right]=\text{median}\left(\text{ai}\right],\text{a}[\text{i}+\text{5}00),\text{a}[\text{i}+\text{1}000),\text{a}[\text{i}+\text{15}00),\text{a}[\text{i}+2000)\text{a}[\text{i}+2500), \\ \text{a}[\text{i}+\text{3}000),\text{a}[\text{i}+\text{35}00),\text{a}[\text{i}+\text{4}000),\text{a}[\text{i}+\text{45}00),\text{a}[\text{i}+\text{5}000),\text{a}[\text{i}+\text{55}00), \\ \text{a}[\text{i}+\text{6}000),\text{a}[\text{i}+\text{65}00),\text{a}[\text{i}+\text{7}000),\text{a}[\text{i}+\text{75}00),\text{a}[\text{i}+\text{8}000),\text{a}[\text{i}+\text{85}00), \\ \text{a}[\text{i}+\text{9}000),\text{a}[\text{i}+\text{1}0000),\text{a}[\text{i}+\text{1}0\text{5}00),\text{a}[\text{i}+\text{11}000),\text{a}[\text{i}+\text{115}00),\text{a}[\text{i}+\text{1}2000) \end{gathered}$$

Mean (c)

In my example, the mean of the overall sample medians M was 25.598, which is near the population mean $\text{y}=\text{25}.\text{5},$ therefore the sample median M seems to be an unbiased approximation of the population mean.

(b) The variance in the sample distribution is higher

In part, they calculate the variance of the sample means as well as the sample medians (a) Variance (b) variance (c) I then calculated that the variation of the sample means was 197.3208, while the variance of the test medians was 21.62465, indicating that the volatility of the sample meansis higher in the sampling distribution.