Final answer:
To calculate the 95% confidence interval with a sample mean of $62,000 and a standard deviation of $17,328 for 49 college graduates, use the Z-distribution. The calculated interval is $57,144.96 to $66,855.04, indicating where the actual population mean is likely to fall with 95% confidence.
Step-by-step explanation:
To construct a 95% confidence interval for the population mean μ given a sample mean of $62,000 and a standard deviation σ of $17,328, we would typically use the t-distribution since we're dealing with a sample of fewer than 30 graduates.
However, since the sample size here is 49, which is larger than 30, we can use the Z-distribution instead, which simplifies to the standard normal distribution when the sample size is sufficiently large.
The formula for a confidence interval is:
Confidence Interval =± Z* × (σ/√n
For a 95% confidence interval, we look up the Z-value that corresponds to the middle 95% of the normal distribution, which is approximately 1.96.
With a sample size (n) of 49, the formula becomes:
Confidence Interval = $62,000 ± 1.96 × ($17,328/√49
Calculating the margin of error ME:
ME = 1.96 × $17,328/7
= 1.96 × $2,476.85714
= $4,855.04
The 95% confidence interval is then:
$62,000 ± $4,855.04
= $57,144.96, $66,855.04
Therefore, we are 95% confident that the population mean salary μ lies between $57,144.96 and $66,855.04.