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After deducting grants based on need, the average cost to attend the University of Southern California (USC) is \(\$ 27,175\) (U.S. News \& World Report, America 's Best Colleges, \(2009 \text { ed. }) .\) Assume the population standard deviation is \(\$ 7400 .\) Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? b. What is the probability that the sample mean will be more than \(\$ 27,175 ?\) c. What is the probability that the sample mean will be within \(\$ 1000\) of the population mean? d. How would the probability in part (c) change if the sample size were increased to \(100 ?\)

Short Answer

Expert verified
a. 954.77, b. 0.5, c. 0.70, d. Probability increases to 0.82 with a sample size of 100.

Step by step solution

01

Calculate the Standard Error (SE) of the Mean

To calculate the standard error of the mean, we use the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( \sigma = 7400 \) is the population standard deviation, and \( n = 60 \) is the sample size. Substituting these values into the formula gives: \[ SE = \frac{7400}{\sqrt{60}} \approx 954.77 \] Therefore, the standard error of the mean is approximately \( 954.77 \).
02

Calculate Probability of Sample Mean Being More Than $27,175

The sample has a mean of \( 27,175 \). We want to know the probability that the sample mean will be greater than this. Since the sample mean is equal to the population mean, the probability will be equal to 0.5 due to symmetry in the normal distribution.
03

Calculate Probability of Sample Mean within $1000 of Population Mean

We want the sample mean to be within \( 1000 \) of \( 27,175 \), so the bounds are \( 27,175 \pm 1000 \) or \( 26,175 \) and \( 28,175 \). We calculate the corresponding z-scores: \[ z = \frac{(X - \mu)}{SE} \] For \( X = 26,175 \), \[ z = \frac{(26175 - 27175)}{954.77} \approx -1.05 \] For \( X = 28,175 \), \[ z = \frac{(28175 - 27175)}{954.77} \approx 1.05 \] Using a standard normal distribution table, find the probability between \( z = -1.05 \) and \( z = 1.05 \). This probability is approximately 0.70.
04

Recalculate for Sample Size of 100

With \( n = 100 \), find the new SE: \[ SE = \frac{7400}{\sqrt{100}} = 740 \] Using the same boundaries as before, calculate the new z-scores: \[ z = \frac{(26175 - 27175)}{740} \approx -1.35 \] and \[ z = \frac{(28175 - 27175)}{740} \approx 1.35 \] We find the probability between \( z = -1.35 \) and \( z = 1.35 \), which is approximately 0.82. Thus, increasing the sample size increases the probability to 0.82.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Standard Error
In statistics, the standard error (SE) helps us understand how much our sample mean is expected to fluctuate. It provides a measure of the accuracy of our sample mean as an estimate of the population mean. The formula for standard error is given by:
\[ SE = \frac{\sigma}{\sqrt{n}} \]
Here, \( \sigma \) represents the population standard deviation, and \( n \) is the sample size. The formula shows that the standard error decreases as the sample size increases. This is because larger samples tend to provide more reliable estimates of the population mean. In our example, with a population standard deviation of \( 7400 \) and a sample size of \( 60 \), the standard error is calculated as approximately \( 954.77 \). This means that sample means will typically differ from the true population mean by about \( 954.77 \) on average.
Calculating Probabilities Using Standard Normal Distribution
Probability calculations are often made easier using the standard normal distribution, especially when dealing with sample means. The normal distribution is symmetrical, which is why a probability can often be easily determined. In statistics, this probability often relates to a particular event or range of outcomes. For example, knowing that the sample mean will be more than the population mean simply implies that there's a 50% chance, thanks to the symmetry of the normal distribution.
When calculating the probability that a sample mean falls within a certain range, such as within \\(1000 of the population mean, we use z-scores. These scores tell us how many standard errors away a value is from the mean. Once we know the z-scores for the bounds,
- For example, when calculating within \\)1000 of \$27,175, the z-scores were calculated as approximately \(-1.05\) and \(1.05\). - We can then determine the probability that the sample mean falls between these z-scores using a standard normal distribution table.
The calculated probability was approximately 0.70, indicating a 70% chance that the sample mean is within this range.
Introduction to Z-scores
Z-scores are a key component in probability calculations involving the normal distribution. A z-score tells us the number of standard deviations a data point is from the mean. It's a way to standardize scores across different scales. This can be extremely helpful in comparing different data points to each other or assessing how unusual a data point is.
The z-score formula is:
\[ z = \frac{(X - \mu)}{SE} \]
Where:
  • \( X \) is the data point, or sample mean.
  • \( \mu \) is the population mean.
  • \( SE \) is the standard error.
By knowing your z-score, you can easily determine the probability of a particular score using the standard normal distribution. As shown in the exercise, the z-scores helped determine the probability of a sample mean falling in a certain range, demonstrating their practical utility in statistical inference.

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Most popular questions from this chapter

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Suppose a simple random sample of size 50 is selected from a population with \(\sigma=10\) Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate) a. The population size is infinite. b. The population size is \(N=50,000\) c. The population size is \(N=5000\) d. The population size is \(N=500\).

The average price of a gallon of unleaded regular gasoline was reported to be \(\$ 2.34\) in northern Kentucky (The Cincinnati Enquirer, January 21,2006 ). Use this price as the population mean, and assume the population standard deviation is \(\$ .20\). a. What is the probability that the mean price for a sample of 30 service stations is within \(\$ .03\) of the population mean? b. What is the probability that the mean price for a sample of 50 service stations is within \(\$ .03\) of the population mean? c. What is the probability that the mean price for a sample of 100 service stations is within \(\$ .03\) of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within \(\$ .03\) of the population mean?

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