to different sampling distributions in the decreasing order of their standard error.
A. $\sigma=1.6$, $\eta=16$
B. $\sigma=1.5$, $\eta=25$
C. $\sigma=0.6$, n=36
D. $\sigma=1.4$, $\eta=49$
Choose the correct answer from the options given below :
The standard error (SE) of a sampling distribution measures the variability of sample means. It is calculated using the formula:
$SE = \(\frac{\sigma}{\sqrt{n}}\)$
where $\(\sigma\)$ is the standard deviation and $\(n\)$ is the sample size.
To arrange the given options in decreasing order of their standard error, we calculate the SE for each pair:
| Option | Standard Deviation ($\(\sigma\)$) | Sample Size ($\(n\)$) | Standard Error (SE) Calculation | SE Value |
|---|---|---|---|---|
| A | 1.6 | 16 | $\(\frac{1.6}{\sqrt{16}} = \frac{1.6}{4}\)$ | 0.4 |
| B | 1.5 | 25 | $\(\frac{1.5}{\sqrt{25}} = \frac{1.5}{5}\)$ | 0.3 |
| C | 0.6 | 36 | $\(\frac{0.6}{\sqrt{36}} = \frac{0.6}{6}\)$ | 0.1 |
| D | 1.4 | 49 | $\(\frac{1.4}{\sqrt{49}} = \frac{1.4}{7}\)$ | 0.2 |
Now, we arrange the calculated SE values in decreasing order:
The decreasing order is A, B, D, C.
This order corresponds to Option 2 in the given choices. However, the provided correct answer is Option B, which lists the order as A, B, D, C. This matches our calculated order.
Therefore, the correct arrangement is A, B, D, C.
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