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be computed when population variance is known using $se = \\frac{\\sigm…

Question

be computed when population variance is known using
$se = \frac{\sigma}{\sqrt{n}}$ (for samples with replacement) and
$se = \frac{\sigma}{\sqrt{n}} \cdot \sqrt{\frac{n - n}{n - 1}}$ (for samples without replacement) where $\sigma$ is the population standard deviation.
example on the case above, three balls numbered 1, 2, and 3. two samples are drawn.
the standard deviation of the population is $\sigma = \sqrt{\frac{2}{3}}$.
let us solve for the se.
$se = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{\frac{2}{3}}}{\sqrt{2}} = \sqrt{\frac{2}{6}} = 0.577$

  1. the shape is approximately normal, provided the sample size is large enough, and regardless of the shape of parent distribution.

whats more
task 5
challen yourself!
a population consists of the numbers 3, 5, 7.

  1. enumerate all possible samples of size 2 with replacement.
  2. construct a sampling distribution table and compute for the mean of each sample.
  3. find the variance of the sample means.
  4. find the standard deviation of the sample means.
  5. find the population mean.
  6. find the variance of the population.
  7. find the standard deviation of the population.
  8. find the standard error of the mean.

Explanation:

Response
1. Enumerate all possible samples of size 2 with replacement.

Step1: Identify population elements

Population: {3, 5, 7}. Sample size \( n = 2 \), with replacement (so we can repeat elements).

Step2: List all combinations

  • (3, 3)
  • (3, 5)
  • (3, 7)
  • (5, 3)
  • (5, 5)
  • (5, 7)
  • (7, 3)
  • (7, 5)
  • (7, 7)

Step1: Calculate mean for each sample

Mean formula: \( \bar{x} = \frac{x_1 + x_2}{2} \)

  • (3, 3): \( \frac{3 + 3}{2} = 3 \)
  • (3, 5): \( \frac{3 + 5}{2} = 4 \)
  • (3, 7): \( \frac{3 + 7}{2} = 5 \)
  • (5, 3): \( \frac{5 + 3}{2} = 4 \)
  • (5, 5): \( \frac{5 + 5}{2} = 5 \)
  • (5, 7): \( \frac{5 + 7}{2} = 6 \)
  • (7, 3): \( \frac{7 + 3}{2} = 5 \)
  • (7, 5): \( \frac{7 + 5}{2} = 6 \)
  • (7, 7): \( \frac{7 + 7}{2} = 7 \)

Step2: Create sampling distribution table (frequency of each mean)

Sample Mean (\( \bar{x} \))Frequency (\( f \))
42
53
62
71

Step1: Recall variance formula for sampling distribution

\( \sigma_{\bar{x}}^2 = \frac{\sum f(\bar{x} - \mu_{\bar{x}})^2}{N} \) (where \( N \) is total number of samples, \( \mu_{\bar{x}} \) is mean of sample means)
First, find \( \mu_{\bar{x}} \):
\( \mu_{\bar{x}} = \frac{\sum f\bar{x}}{\sum f} = \frac{1(3) + 2(4) + 3(5) + 2(6) + 1(7)}{9} = \frac{3 + 8 + 15 + 12 + 7}{9} = \frac{45}{9} = 5 \)

Step2: Calculate \( (\bar{x} - \mu_{\bar{x}})^2 \) for each \( \bar{x} \)

  • For \( \bar{x} = 3 \): \( (3 - 5)^2 = 4 \), \( f = 1 \), \( f(\bar{x} - \mu_{\bar{x}})^2 = 1 \times 4 = 4 \)
  • For \( \bar{x} = 4 \): \( (4 - 5)^2 = 1 \), \( f = 2 \), \( f(\bar{x} - \mu_{\bar{x}})^2 = 2 \times 1 = 2 \)
  • For \( \bar{x} = 5 \): \( (5 - 5)^2 = 0 \), \( f = 3 \), \( f(\bar{x} - \mu_{\bar{x}})^2 = 3 \times 0 = 0 \)
  • For \( \bar{x} = 6 \): \( (6 - 5)^2 = 1 \), \( f = 2 \), \( f(\bar{x} - \mu_{\bar{x}})^2 = 2 \times 1 = 2 \)
  • For \( \bar{x} = 7 \): \( (7 - 5)^2 = 4 \), \( f = 1 \), \( f(\bar{x} - \mu_{\bar{x}})^2 = 1 \times 4 = 4 \)

Step3: Sum the squared terms

\( \sum f(\bar{x} - \mu_{\bar{x}})^2 = 4 + 2 + 0 + 2 + 4 = 12 \)

Step4: Compute variance

\( \sigma_{\bar{x}}^2 = \frac{12}{9} = \frac{4}{3} \approx 1.333 \)

Answer:

(3, 3), (3, 5), (3, 7), (5, 3), (5, 5), (5, 7), (7, 3), (7, 5), (7, 7)

2. Construct a sampling distribution table and compute for the mean of each sample.