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Isabel Myers was a pioneer in the study of personality types. The following information is taken from \(A\) Guide to the Development and Use of the Myers- Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts. (a) Let \(p\) represent the proportion of all actors who are extroverts. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief interpretation of the meaning of the confidence interval you have found. (c) Do you think the conditions \(n p > 5\) and \(n q >5 \) are satisfied in this problem? Explain why this would be an important consideration.

Short Answer

Expert verified
(a) Point estimate \( \hat{p} = 0.629 \). (b) 95% CI: (0.509, 0.749). (c) Conditions satisfied as both \( np \) and \( nq \) are greater than 5.

Step by step solution

01

Define the Point Estimate

In statistics, a point estimate is a single value given as an estimate of a population parameter. In this case, the point estimate of the proportion \( p \) of all actors who are extroverts is given by the sample proportion \( \hat{p} \). Here, \( \hat{p} \) is calculated by dividing the number of extroverts by the total number of actors in the sample.\[\hat{p} = \frac{39}{62}\approx 0.629\] Thus, \( \hat{p} \) is approximately 0.629.
02

Calculate the Confidence Interval

To construct a 95% confidence interval for \( p \), we need \( \hat{p} \), the standard error (SE), and the critical value for the normal distribution (\( z \)). First, calculate the standard error using \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \).\[SE = \sqrt{\frac{0.629 \times (1-0.629)}{62}} \approx 0.061\]The critical value for a 95% confidence interval is approximately 1.96. Thus, the confidence interval is:\[\hat{p} \pm z \times SE = 0.629 \pm 1.96 \times 0.061 \approx (0.509, 0.749)\]So, the 95% confidence interval for \( p \) is approximately (0.509, 0.749).
03

Interpret the Confidence Interval

The 95% confidence interval for the proportion \( p \) suggests that we are 95% confident that the true proportion of extroverted actors in the population is between 50.9% and 74.9%. This does not mean that 95% of sample proportions will fall within this interval, but rather that 95% of intervals calculated this way will contain the true \( p \).
04

Check Conditions

The conditions \( np > 5 \) and \( nq > 5 \) must be satisfied to use the normal approximation for the confidence interval. Here,\[ np = 62 \times 0.629 \approx 39 \] \[ nq = 62 \times (1-0.629) \approx 23 \] Both these values are greater than 5, indicating that the sample size is sufficiently large for the normal approximation to be valid. This is crucial because it ensures the reliability of the confidence interval.

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Key Concepts

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

Point Estimate
A point estimate is like finding the center of your guess in statistics. It's a single number that gives us a clue about a population parameter. When dealing with proportions, this is known as the sample proportion \(\hat{p}\).

In our exercise, we are interested in the proportion of extroverted actors in a population. The point estimate for this is simply calculated by taking the number of extroverted actors in the sample and dividing it by the total number of actors in the sample.
  • Example: \(\hat{p} = \frac{39}{62} \approx 0.629\).
This tells us that in our sample, about 62.9% of actors are extroverts. Having this point estimate is the first step towards understanding the larger population, as it provides a direct and intuitive snapshot of your collected data.
Statistical Conditions
Before calculating a confidence interval, it's crucial to check certain statistical conditions. These ensure that the methods we use yield sensible and reliable results. In the context of proportions, the conditions often involve ensuring that the sample size is large enough.

Specifically, the two conditions are:
  • The condition \(np > 5\): It checks that the number of successes (extroverted actors) is sufficiently large.
  • The condition \(nq > 5\): It checks that the number of failures (non-extroverted actors) is large enough.
For our exercise, we have:
  • \(np = 62 \times 0.629 \approx 39\)
  • \(nq = 62 \times (1-0.629) \approx 23\)
Both are greater than 5, confirming that the sample size is adequately large. Meeting these conditions allows us to use a normal approximation when constructing a confidence interval, which is important for making the intervals accurate and dependable.
Standard Error
Think of the standard error as a measure of how much our sample proportion is likely to vary if we took different samples from the same population. It's like estimating the wiggle room around our point estimate.

The smaller the standard error, the closer we expect our sample point estimate to be to the true population proportion. The standard error for a proportion is calculated using the formula:
  • \[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
where \(\hat{p}\) is the sample proportion and \(n\) is the sample size. In our scenario:
  • \[SE = \sqrt{\frac{0.629 \times (1-0.629)}{62}} \approx 0.061 \]
This tells us that our estimate of 0.629 has a standard error of 0.061. With the standard error known, we can build a confidence interval to see what range the true proportion might fall into. It's an essential part of ensuring our inference is grounded in statistical reliability.

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Most popular questions from this chapter

The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to \(1987 .\) Some of these data are published in the book The Story of Old Faithful, by G. D. Marler (Yellowstone Association Press). Let \(x_{1}\) be a random variable that represents the time interval (in minutes) between Old Faithful's eruptions for the years 1948 to \(1952 .\) Based on 9340 observations, the sample mean interval was \(\bar{x}_{1}=63.3\) minutes. Let \(x_{2}\) be a random variable that represents the time interval in minutes between Old Faithful's eruptions for the years 1983 to \(1987 .\) Based on 25,111 observations, the sample mean time interval was \(\bar{x}_{2}=72.1\) minutes. Historical data suggest that \(\sigma_{1}=9.17\) minutes and \(\sigma_{2}=12.67\) minutes. Let \(\mu_{1}\) be the population mean of \(x_{1}\) and let \(\mu_{2}\) be the population mean of \(x_{2}\) (a) Which distribution, normal or Student's \(t,\) do we use to approximate the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? Explain. (b) Compute a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the \(99 \%\) confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959

Diagnostic Tests: Total Calcium Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in mg/dl). $$ \begin{array}{ccccccc} 9.3 & 8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0 \\ 9.9 & 11.2 & 12.1 & & & & \end{array} $$ (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\) (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Interpretation Based on your results in part (b), does it seem that this patient still has a calcium deficiency? Explain.

In order to use a normal distribution to compute confidence intervals for \(p,\) what conditions on \(n p\) and \(n q\) need to be satisfied?

Answer true or false. Explain your answer. The point estimate for the population mean \(\mu\) of an \(x\) distribution is \(\bar{x},\) computed from a random sample of the \(x\) distribution.

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

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