91Ó°ÊÓ

The following data represent the ages of the winners of the Academy Award for Best Actor for the years \(2004-2009\) $$ \begin{array}{lc} \hline \text { 2004: Jamie Foxx } & 37 \\ \hline \text { 2005: Philip Seymour Hoffman } & 38 \\ \hline \text { 2006: Forest Whitaker } & 45 \\ \hline \text { 2007: Daniel Day-Lewis } & 50 \\ \hline \text { 2008: Sean Penn } & 48 \\ \hline \text { 2009: Jeff Bridges } & 60 \\ \hline \end{array} $$ (a) Compute the population mean, \(\mu\). (b) List all possible samples with size \(n=2 .\) There should be \({ }_{6} C_{2}=15\) samples. (c) Construct a sampling distribution for the mean by listing the sample means and their corresponding probabilities. (d) Compute the mean of the sampling distribution. (e) Compute the probability that the sample mean is within 3 years of the population mean age. (f) Repeat parts (b)-(e) using samples of size \(n=3\). Comment on the effect of increasing the sample size.

Step by step solution

01

Compute the population mean, \( \mu \)

First, add up all the ages. \( 37 + 38 + 45 + 50 + 48 + 60 = 278 \). Then, divide by the number of individuals, which is 6. The population mean is therefore \( \mu = \frac{278}{6} = 46.33 \).

02

List all possible samples with size \( n=2 \)

Using combinations calculation \( \binom{6}{2} = 15 \), list all possible sample pairs:1. (37, 38)2. (37, 45)3. (37, 50)4. (37, 48)5. (37, 60)6. (38, 45)7. (38, 50)8. (38, 48)9. (38, 60)10. (45, 50)11. (45, 48)12. (45, 60)13. (50, 48)14. (50, 60)15. (48, 60)

03

Construct a sampling distribution for the mean

Calculate sample means and their probabilities:1. (37, 38): \( \frac{37+38}{2} = 37.5 \)2. (37, 45): \( \frac{37+45}{2} = 41 \)3. (37, 50): \( \frac{37+50}{2} = 43.5 \)4. (37, 48): \( \frac{37+48}{2} = 42.5 \)5. (37, 60): \( \frac{37+60}{2} = 48.5 \)6. (38, 45): \( \frac{38+45}{2} = 41.5 \)7. (38, 50): \( \frac{38+50}{2} = 44 \)8. (38, 48): \( \frac{38+48}{2} = 43 \)9. (38, 60): \( \frac{38+60}{2} = 49 \)10. (45, 50): \( \frac{45+50}{2} = 47.5 \)11. (45, 48): \( \frac{45+48}{2} = 46.5 \)12. (45, 60): \( \frac{45+60}{2} = 52.5 \)13. (50, 48): \( \frac{50+48}{2} = 49 \)14. (50, 60): \( \frac{50+60}{2} = 55 \)15. (48, 60): \( \frac{48+60}{2} = 54 \)Each sample mean has equal probability of \( \frac{1}{15} \).

04

Compute the mean of the sampling distribution

Add all the sample means and divide by the number of samples to get the mean:\( \mu_x = \frac{37.5 + 41 + 43.5 + 42.5 + 48.5 + 41.5 + 44 + 43 + 49 + 47.5 + 46.5 + 52.5 + 49 + 55 + 54}{15} = 46.33 \).

05

Compute the probability that the sample mean is within 3 years of the population mean

The population mean is 46.33. We need to find means in the range [43.33, 49.33]. Sample means in this range: 43.5, 42.5, 48.5, 44, 43, 49, 47.5, 46.5, 49. Probability: \( \frac{9}{15} = 0.6 \)

06

List all possible samples with size \( n=3 \)

Using combinations calculation \( \binom{6}{3} = 20 \), list all possible sample triples:1. (37, 38, 45)2. (37, 38, 50)3. (37, 38, 48)4. (37, 38, 60)5. (37, 45, 50)6. (37, 45, 48)7. (37, 45, 60)8. (37, 50, 48)9. (37, 50, 60)10. (37, 48, 60)11. (38, 45, 50)12. (38, 45, 48)13. (38, 45, 60)14. (38, 50, 48)15. (38, 50, 60)16. (38, 48, 60)17. (45, 50, 48)18. (45, 50, 60)19. (45, 48, 60)20. (50, 48, 60)

07

Compute the sample means for n=3

Calculate sample means:1. \( \frac{37+38+45}{3} = 40 \)2. \( \frac{37+38+50}{3} = 41.67 \)3. \( \frac{37+38+48}{3} = 41 \)4. \( \frac{37+38+60}{3} = 45 \)5. \( \frac{37+45+50}{3} = 44 \)6. \( \frac{37+45+48}{3} = 43.33 \)7. \( \frac{37+45+60}{3} = 47.33 \)8. \( \frac{37+50+48}{3} = 45 \)9. \( \frac{37+50+60}{3} = 49 \)10. \( \frac{37+48+60}{3} = 48.33 \)11. \( \frac{38+45+50}{3} = 44.33 \)12. \( \frac{38+45+48}{3} = 43.67 \)13. \( \frac{38+45+60}{3} = 47.67 \)14. \( \frac{38+50+48}{3} = 45.33 \)15. \( \frac{38+50+60}{3} = 49.33 \)16. \( \frac{38+48+60}{3} = 48.67 \)17. \( \frac{45+50+48}{3} = 47.67 \)18. \( \frac{45+50+60}{3} = 51.67 \)19. \( \frac{45+48+60}{3} = 51 \)20. \( \frac{50+48+60}{3} = 52.67 \)All sample means have equal probability: \( \frac{1}{20} \)

08

Compute the mean of the sampling distribution for n=3

Sum all sample means and divide by 20: \( \frac{40 + 41.67 + 41 + 45 + 44 + 43.33 + 47.33 + 45 + 49 + 48.33 + 44.33 + 43.67 + 47.67 + 45.33 + 49.33 + 48.67 + 47.67 + 51.67 + 51 + 52.67}{20} = 46.33 \)

09

Compute the probability that the sample mean for n=3 is within 3 years of the population mean

Find sample means in range [43.33, 49.33]: 45, 44, 43.33, 47.33, 45, 49, 48.33, 44.33, 43.67, 47.67, 45.33, 49.33, 48.67, 47.67. Probability: \( \frac{14}{20} = 0.7 \)

10

Comment on the effect of increasing the sample size

As the sample size increases from n=2 to n=3, the probability that the sample mean is within 3 years of the population mean increases from 0.6 to 0.7. This indicates that larger sample sizes yield more accurate estimates of the population mean.

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Key Concepts

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

Population Mean

The population mean, often represented by \( \mu \), is a crucial concept in statistics. It's the average of all the values in a population. For example, in our exercise, we were given the ages of Best Actor winners from 2004 to 2009: 37, 38, 45, 50, 48, and 60. To find the population mean, follow these steps:

• Add all the ages together: 37 + 38 + 45 + 50 + 48 + 60 = 278.
• Divide this sum by the number of observations (in this case, 6): \( \mu = \frac{278}{6} = 46.33 \).

This \( 46.33 \) years is our population mean. It's a measure that gives a central value for the age of these winners, providing a single figure that summarizes the entire group.

Sample Mean

When working with a sample, rather than the entire population, we calculate the sample mean, denoted by \( \bar{x} \). The sample mean helps us make inferences about the population mean. For instance, in our task, we looked at samples of size \( n=2 \) and \( n=3 \). Here's how you find the sample mean:

• Take a sample (e.g., ages 37 and 38).
• Add the values of the sample: 37 + 38 = 75.
• Divide by the number of observations in the sample: \( \bar{x} = \frac{75}{2} = 37.5 \).

Each sample mean has an equal chance of occurring since each sample is equally likely. Calculating multiple sample means and taking their average will give a good approximation of the population mean. This is illustrated in step by step calculations in the solution.

Probability

Probability is the measure of how likely an event is to occur. When working with sample means, probabilities help us understand the distribution of these means. For example, in the exercise, the probability of each sample mean was determined based on the number of possible samples. Here’s how it works:

• For size \( n=2 \), there are 15 possible samples, so the probability of each sample is \( \frac{1}{15} \) or approximately 0.067.
• For size \( n=3 \), with 20 possible samples, the probability is \( \frac{1}{20} \) or 0.05.

Additionally, we calculated the probability of sample means falling within 3 years of the population mean. As the sample size increases, the accuracy of our sample mean as an estimate of the population mean improves. For example, for \( n=2 \) samples, this probability was 0.6. For \( n=3 \) samples, it increased to 0.7, illustrating how larger samples provide better estimates of the population mean.