Final answer:
To construct the 95% confidence interval for the population mean salary, use the sample mean, standard deviation, and Z-score to calculate the margin of error and apply it to the sample mean. The 95% confidence interval is $56,549.71 to $67,450.29.
Step-by-step explanation:
To construct a 95% confidence interval for estimating the population mean μ (mu) when the sample mean (μ-bar) is $62,000 and the standard deviation (σ) is $17,582, we can use the formula:
μ-bar ± (Z* × (σ/√n)),
where Z* is the Z-score that corresponds to the desired confidence level (95% in this case), σ is the population standard deviation, and n is the sample size.
For a 95% confidence interval, the Z-score (Z*) is typically around 1.96, assuming we have a large enough sample size to use the Z-distribution as an approximation for the sampling distribution of the mean.
Plugging in the values we have:
- Sample mean (μ-bar): $62,000
- Standard deviation (σ): $17,582
- Sample size (n): 40
- Z-score (Z*): 1.96
Now, we calculate the margin of error (E):
E = 1.96 × ($17,582/√40) ≈ $5,450.29.
So the 95% confidence interval is:
62,000 ± $5,450.29 which is ($56,549.71 to $67,450.29).
Therefore, we are 95% confident that the true population mean salary for college graduates who took a statistics course lies between $56,549.71 and $67,450.29.