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$87,135 per year with a standard deviation of $49,063. What is the 90% confidence interval for the average salary of job openings for professors at KS universities?

a) $71,221 to $103,049
b) $64,072 to $110,198
c) $78,539 to $97,731
d) $92,406 to $83,864

1 Answer

6 votes

Answer:

The confidence interval calculation requires the sample size to determine the exact interval for the average salary, which is not provided. The given options suggest the interval, but without sample size, we cannot confirm the correct choice. Option d) is incorrect as a confidence interval cannot have a lower limit greater than the upper limit.

Step-by-step explanation:

To calculate the 90% confidence interval for the average salary of job openings for professors at KS universities, using the given mean salary of $87,135 and standard deviation of $49,063, we can use the formula for a confidence interval: CI = mean ± (z* × (SD/√n)), where z* is the z-score associated with the desired level of confidence and SD is the standard deviation. Assuming that the sample size is large enough to apply the Central Limit Theorem, we would use the z-score for 90% confidence, which is approximately 1.645. However, since the sample size isn't provided, we cannot perform this calculation accurately, and the question seems to be missing this crucial piece of information. Instead, we are given options to choose from.

Without the sample size, we cannot calculate the exact confidence interval but we can infer that option d) is incorrect because a confidence interval cannot have the lower limit greater than the upper limit. Options a), b), and c) could potentially be correct depending on the sample size which was not provided. The correct answer should reflect a range where the lower limit is less than the upper limit, consistent with the common properties of a confidence interval.

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