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To estimate a population mean, the sample size needed to provide a margin of error of 3 or less with a 95% confidence when the population standard deviation equals 11 is
(Round your answer to the nearest integer.)

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Answer:

The minimum sample needed to provide a margin of error of 3 or less is 52.

Step-by-step explanation:

The confidence interval for population mean (μ) is:

[tex]\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]

The margin of error is:

[tex]MOE= z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]

Given:

MOE = 3

σ = 11

The critical value for 95% confidence interval is: [tex]z_{\alpha /2}=1.96[/tex]

**Use the z-table for critical values.

Compute the sample size (n) as follows:

[tex]MOE= z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\3=1.96\times \frac{11}{\sqrt{n}}\\n=(\frac{1.96\times11}{3} )^{2}\\=51.65\\\approx52[/tex]

Thus, the minimum sample needed to provide a margin of error of 3 or less is 52.

Ver imagen warylucknow

The sample size needed to provide a margin of error of 3 or less with a 95% confidence to estimate the population mean is 52

What is the margin of error for large samples?

Suppose that we have:

  • Sample size n > 30
  • Sample standard deviation = s
  • Population standard deviation = [tex]\sigma[/tex]
  • Level of significance = [tex]\alpha[/tex]

Then the margin of error(MOE) is obtained as

  • Case 1: Population standard deviation is known

Margin of Error = [tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

  • Case 2: Population standard deviation is unknown

[tex]MOE = Z_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]

where [tex]Z_{\alpha/2}[/tex] is critical value of the test statistic at level of significance

For the considered case, we have:

  • [tex]\sigma = 11[/tex]
  • MOE ≤ 3
  • Confidence level = 95%
  • Level of significance = α = 100% - 95% = 5% = 0.05
  • At this level of significance, for two tailed distribution(assuming values can be here both above and below mean, so not restricted on one side), the critical value is [tex]Z_{0.05/2} = \pm 1.96[/tex]

Let the sample size needed be n, then:

[tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}} \leq 3\\\\\pm 1.96 \times \dfrac{11}{\sqrt{n}} \leq 3\\\\n \geq (\dfrac{\pm 1.96 \times 11}{3})^2 \approx 51.65 \\\\n \geq 52[/tex]

Thus, the sample size needed to provide a margin of error of 3 or less with a 95% confidence to estimate the population mean is 52

Learn more about margin of error here:

https://brainly.com/question/13220147

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