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91Ó°ÊÓ

Psychologists' income In \(2003,\) the American Psychological Association conducted a survey (at research.apa.org) of a random sample of psychologists to estimate mean incomes for psychologists with various academic degrees and levels of experience. Of the 31 psychologists who received a masters degree in \(2003,\) the mean income was \(\$ 43,834\) with a standard deviation of \(\$ 16,870\) a. Construct a \(95 \%\) confidence interval for the population mean. Interpret. b. What assumption about the population distribution of psychologists' incomes does the confidence interval method make? c. If the assumption about the shape of the population distribution is not valid, does this invalidate the results? Explain.

Short Answer

Expert verified
a) The 95% confidence interval is \([37,682, 49,986]\). b) Assumes normal population distribution. c) If not normal, with small sample, results may be less valid.

Step by step solution

01

Identify the components for the confidence interval

To construct a 95% confidence interval for the population mean income, we need the sample mean \(\bar{x} = 43,834\), the sample standard deviation \(s = 16,870\), and the sample size \(n = 31\). The critical value for a 95% confidence level with 30 degrees of freedom (\(n-1\)) can be found using a t-distribution table: \(t_{\alpha/2} \approx 2.042\).
02

Calculate the standard error

The standard error (SE) of the mean is calculated using the formula \(SE = \frac{s}{\sqrt{n}}\). Substituting the values, \[SE = \frac{16,870}{\sqrt{31}} \approx 3,030.12\].
03

Compute the confidence interval

Use the formula for the confidence interval: \[\bar{x} \pm t_{\alpha/2} \times SE\]. Plugging in the values: \[43,834 \pm 2.042 \times 3,030.12\]. This results in a confidence interval of \([37,681.93, 49,986.07]\).
04

Interpret the confidence interval

The 95% confidence interval for the mean income of psychologists with a master's degree in 2003 is between \\(37,682 and \\)49,986. This means we are 95% confident that the true mean income lies in this interval.
05

Discuss the population distribution assumption

The method assumes that the population from which the sample is drawn is normally distributed. This is necessary for the t-distribution approximation to be valid, especially with small sample sizes.
06

Evaluate the impact of assumption invalidity

If the population distribution is not normal, and the sample size is small, the confidence interval may not accurately reflect the population parameter. However, for larger sample sizes, the Central Limit Theorem suggests that the sample mean will approximate normality, making the interval more reliable.

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Key Concepts

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

T-Distribution
When constructing confidence intervals, especially with small sample sizes, a t-distribution is often used rather than the normal distribution. This is because the t-distribution accounts for additional variability due to the smaller sample size.
T-distributions are a family of distributions that are similar in shape to the normal distribution but have thicker tails. This property allows them to be more forgiving with small sample sizes, capturing the increased uncertainty in estimating population parameters.
Key details about t-distributions are:
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • The degrees of freedom ( \(df\) ) for a t-distribution is usually calculated as ( \(n-1\) ) where ( \(n\) ) is the sample size.
  • For a larger \(df\), the t-distribution becomes closer to the standard normal distribution (Z-distribution).
In the given exercise, since the sample size ( \(n = 31\) ) is not very large, a t-distribution with 30 degrees of freedom was used to find the critical t-value, which was approximately 2.042 for a 95% confidence interval.
Standard Error
The standard error (SE) measures the variability of the sample mean from the actual population mean. It provides an estimate of the expected difference between the sample mean we calculate and the true population mean.
The formula for calculating the standard error of the mean is: \[SE = \frac{s}{\sqrt{n}}\] where ( \(s\) ) is the sample standard deviation and ( \(n\) ) is the sample size.
A smaller standard error indicates that our sample mean is a more accurate estimate of the population mean. In our example, the calculated standard error ( \(SE = 3,030.12\) ) reveals the average expected error in predicting the population mean using our sample data.
Key points about standard error:
  • Larger samples tend to have smaller standard errors, implying more precise estimates.
  • Standard error decreases as the sample size increases because it becomes a better representation of the population.
  • It plays a crucial role in constructing confidence intervals, helping determine how wide or narrow the interval will be.
Population Distribution Assumption
Constructing a confidence interval using a t-distribution assumes that the population from which the sample is drawn is normally distributed. When the sample size is small, this assumption becomes even more critical because the t-distribution relies on this normality to provide accurate confidence intervals.
However, when the population distribution is not known or is not normal, the reliability of the confidence interval may be questioned, especially with a smaller sample size like 31. Yet, thankfully, the Central Limit Theorem offers some reassurance:
  • With larger sample sizes (typically over 30), the sample means tend to form a normal distribution, even if the underlying population distribution is not normal.
  • This principle allows us to use normality-based methods like the t-distribution to estimate confidence intervals effectively.
If the normality assumption is violated in a smaller sample, then the confidence interval might not accurately capture the true population mean. However, with larger sample sizes, this concern becomes less significant, offering more reliability in the interval estimate.

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Most popular questions from this chapter

Do you like tofu? You randomly sample five students at your school to estimate the proportion of students who like tofu. All five students say they like it. a. Find the sample proportion who like it. b. Find the standard error. Does its usual interpretation make sense? c. Find a \(95 \%\) confidence interval, using the large-sample formula. Is it sensible to conclude that all students at your school like tofu? d. Why is it not appropriate to use the ordinary largesample confidence interval in part \(\mathrm{c}\) ? Use a more appropriate approach, and interpret the result.

Mean property tax A tax assessor wants to estimate the mean property tax bill for all homeowners in Madison, Wisconsin. A survey 10 years ago got a sample mean and standard deviation of \(\$ 1400\) and \(\$ 1000\). a. How many tax records should the tax assessor randomly sample for a \(95 \%\) confidence interval for the mean to have a margin of error equal to \(\$ 100 ?\) What assumption does your solution make? b. In reality, suppose that they'd now get a standard deviation equal to \(\$ 1500 .\) Using the sample size you derived in part a, without doing any calculation, explain whether the margin of error for a \(95 \%\) confidence interval would be less than \(\$ 100\), equal to \(\$ 100\), or more than \(\$ 100\). c. Refer to part b. Would the probability that the sample mean falls within \(\$ 100\) of the population mean be less than \(0.95,\) equal to \(0.95,\) or greater than \(0.95 ?\) Explain.

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Multiple choice: CI property 2 Other things being equal, increasing \(n\) causes the margin of error of a confidence interval to (a) increase, (b) decrease, (c) stay the same.

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