Question

Suppose that college faculty with the rank of professor at public 2 -year institutions earn an average of \(\$ 71,802\) per year \(^{7}\) with a standard deviation of \(\$ 4000\). In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all 2-year institutions in the United States. a. Describe the sampling distribution of the sample \(\operatorname{mean} \bar{x}\) b. Within what limits would you expect the sample average to lie, with probability \(.95 ?\) c. Calculate the probability that the sample mean \(\bar{x}\) is greater than \(\$ 73,000 ?\) d. If your random sample actually produced a sample mean of \(\$ 73,000,\) would you consider this unusual? What conclusion might you draw?

Short answer

Since the probability of sample mean > $73,000 is 0.0104, which is low (below 0.05), we can consider the sample mean of $73,000 to be unusual. The conclusion we might draw is that the initial average salary stated might not be accurate or that the sample we've taken may not be representative of the entire population of professors at public 2-year institutions.

Step-by-step solution

a. Sampling Distribution of Sample Mean

Using the Central Limit Theorem, we know that the sampling distribution of the sample mean will be approximately normally distributed with: Mean: μ = \(71,802\) Standard deviation of the sampling distribution: σ/√(n) = \(4000 / √(60)\)

b. Limits for Sample Average with 0.95 Probability

To find the limits within which the sample average would lie with 0.95 probability, we need to calculate the z-score associated with the 0.95 probability for a standard normal distribution. The z-score for 0.95 probability is 1.96. Now, we will find the margin of error (ME): ME = Z * (σ/√(n)) = 1.96 * (4000 / √(60)) Next, calculate the lower and upper limits: Lower limit = Mean - ME = 71802 - ME Upper limit = Mean + ME = 71802 + ME

c. Probability that Sample Mean is Greater Than $73,000

To calculate the probability that the sample mean is greater than \(73,000, we need to find the z-score associated with \)73,000. Z = (\(73000 - Mean) / (σ/√(n)) = (\)73000 - 71802) / (4000 / √(60)) Now, we use the standard normal distribution table to find the probability for the given z-score. If the value isn't in the table, approximate the probability using nearby values. This will give us the probability of sample mean being less than \(73,000. To find the probability of it being greater than \)73,000, we will subtract it from 1. Probability of sample mean > $73,000 = 1 - P(Z)

d. Is a Sample Mean of $73,000 Unusual and What Conclusion Can We Draw?

To determine if a sample mean of \(73,000 is unusual, we look at its probability previously calculated. If the probability is low (usually below 0.05), then we can consider that the sample mean of \)73,000 is unusual. If the sample mean is found to be unusual, the conclusion we might draw is that the initial average salary stated might not be accurate or that the sample we've taken may not be representative of the entire population of professors at public 2-year institutions. However, if the sample mean is not unusual, the conclusion might be that the initial average salary stated could be accurate and the sample represents the population well.

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental concept in statistics that tells us about the behavior of sample means. It states that when you take a large enough sample size from a population with any distribution, the sampling distribution of the sample mean will approach a normal distribution.
This theorem is crucial especially when the original population distribution is unknown. In our example, since the sample size is 60, which is considered large, we can safely assume that the sample mean distribution is approximately normal.
This allows us to use normal probability tools to analyze the data, making it easier to find probabilities and confidence intervals.

Standard Normal Distribution

The standard normal distribution is a special normal curve with a mean of 0 and a standard deviation of 1. It serves as a reference to find probabilities and critical values for normal distributions.

• Z-scores: The standard normal distribution uses z-scores, which help convert any normal distribution to the standard form by adjusting for the mean and standard deviation.
• Confidence Intervals: For instance, a z-score of 1.96 corresponds to a 95% confidence interval. It indicates the range within which the sample mean would lie 95% of the time.

In the problem, the z-score helps determine the range for the sample average with 0.95 probability.

Sample Mean

The sample mean is the average of a set of observations from a sample and serves as an estimate of the population mean. Calculating it involves summing up all the sample values and dividing by the number of observations.
For our exercise, the sample mean (represented as \( \bar{x} \)) is used to determine if the observed data point, such as the sample mean of $73,000, is within expected limits. The calculated sample mean provides critical evidence when making inferences about the entire population.
Since every sample provides a slightly different mean, using many samples allows us to observe variability and gain insights into the population mean.

Probability Calculation

Probability calculations involving the sample mean typically require using the z-score formula:
\[ Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
In the exercise, the probability of the sample mean being greater than \( \(73,000 \) involves calculating the z-score for this value and finding its corresponding probability from the standard normal distribution table.
By subtracting this probability from 1, we determine the probability of the sample mean exceeding \)73,000. This method helps in decision-making regarding whether the result is statistically unusual or expected.