The monthly starting salaries of students who receive engineering degrees have a standard deviation of $600. What sample size should be selected so that there is 0.95 probability of estimating the mean monthly income within $150 or less?
To find the sample size required to estimate the mean monthly income within a certain range with a given probability, we can use the formula for the margin of error:
Margin of Error = (Z * Standard Deviation) / √(Sample Size)
Here, Z represents the z-score, which corresponds to the desired level of confidence. In this case, we want a 0.95 (or 95%) level of confidence, which corresponds to a z-score of 1.96.
For estimating the mean monthly income within $150 or less, the margin of error is $150. The standard deviation is given as $600.
Hence, we can rearrange the formula to solve for Sample Size:
Sample Size = (Z^2 * Standard Deviation^2) / Margin of Error^2
Substituting the known values:
Sample Size = (1.96^2 * 600^2) / 150^2
Calculating this expression gives us:
Sample Size ≈ 23.527
Since we cannot have a fraction of a student, we round up the sample size to the nearest whole number.
Therefore, the recommended sample size is 24.