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The dean of the School of Liberal Arts intends to conduct a survey to estimate the average starting annual salary of graduates earning a B.A. degree. Previous research indicates that the population standard deviation is approximately $6,000. The dean desires a 90% confidence interval for the mean salary with a maximum width of $3,000. To determine the necessary sample size, appropriate statistical calculations must be performed.

Paper For Above instruction

Estimating the required sample size for a population mean when the population standard deviation is known involves understanding the relationship between the desired confidence level, the margin of error, and the variability in the data. In this scenario, the target is to determine the minimum number of graduates needed to estimate the average starting salary with a specified confidence and precision.

The given parameters are as follows:

  • Population standard deviation (σ): $6,000
  • Confidence level: 90%
  • Maximum acceptable width of confidence interval (W): $3,000

First, recall that the confidence interval for a population mean when σ is known is given by:

CI = x̄ ± Zα/2 * (σ / √n)

where:

- x̄ is the sample mean,

- Zα/2 is the z-value corresponding to the desired confidence level,

- σ is the population standard deviation,

- n is the sample size.

The width of the confidence interval (W) is twice the margin of error (E), i.e.,

W = 2E = 2 Zα/2 (σ / √n)

Rearranging this formula to solve for n:

n = (2 Zα/2 σ / W)2

Calculating Zα/2 for a 90% confidence level involves identifying the z-value corresponding to 95% in the standard normal distribution (since α = 1 - 0.90 = 0.10, α/2 = 0.05). From standard Z-tables, the critical value is approximately 1.645.

Now, substituting the known values into the formula:

n = (2 1.645 6000 / 3000)2

n = (2 1.645 2)2

n = (2 1.645 2)2 = (6.58)2 ≈ 43.23

Since the sample size must be an integer and sufficient to meet the margin of error, the dean should round up to the next whole number, which is 44.

Therefore, the dean needs to survey at least 44 graduates to estimate the average starting salary with a 90% confidence level and a maximum margin of error of $3,000.

References

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