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Chapter 8: Problem 13

Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences that are described at length in the book \(A\) Guide to the Development and Use of the Myers-Briggs Type Indicator, by Myers and McCaulley (Consulting Psychologists Press). Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. Myers took a random sample of 375 married couples and found that 289 had two or more personality preferences in common. In another random sample of 571 married couples, it was found that only 23 had no preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have two or more personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have no personality perferences in common. (a) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (b) Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?

Short Answer

Expert verified
The 99% confidence interval for \( p_1 - p_2 \) is (0.6752, 0.7856), indicating more couples have two or more preferences in common than none.

Step by step solution

01

Calculate Sample Proportions

First, we need to find the sample proportions for both groups. For the group with two or more preferences in common, the sample proportion \( \hat{p}_1 \) is \( \frac{289}{375} \approx 0.7707 \). For the group with no preferences in common, the sample proportion \( \hat{p}_2 \) is \( \frac{23}{571} \approx 0.0403 \).
02

Calculate the Difference of Sample Proportions

The difference between the two sample proportions is \( \hat{p}_1 - \hat{p}_2 = 0.7707 - 0.0403 = 0.7304 \).
03

Calculate Standard Error

The standard error \( SE \) for the difference in proportions is given by the formula: \[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \] Plugging in our values: \( SE = \sqrt{\frac{0.7707(1-0.7707)}{375} + \frac{0.0403(1-0.0403)}{571}} \approx 0.0214 \).
04

Find the Critical Value

For a 99% confidence interval, the critical value \( z \) associated with a standard normal distribution is approximately 2.576.
05

Calculate Confidence Interval

The 99% confidence interval is given by: \[ (\hat{p}_1 - \hat{p}_2) \pm z \times SE \]Substituting the values we have: \[ 0.7304 \pm 2.576 \times 0.0214 \]This gives the interval \( (0.6752, 0.7856) \).
06

Interpret the Confidence Interval

The interval \( (0.6752, 0.7856) \) indicates that we are 99% confident the true difference in population proportions \( p_1 - p_2 \) falls within this range. Since the interval only contains positive numbers, it suggests that significantly more married couples share two or more personality preferences compared to none.

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Key Concepts

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

Personality Types
Personality types refer to the different patterns of behavior, thoughts, and feelings that individuals exhibit. Isabel Myers, along with Katharine Cook Briggs, developed the Myers-Briggs Type Indicator (MBTI), a popular tool that identifies these patterns. The MBTI categorizes personalities based on four dichotomies, including:
  • Extraversion (E) vs. Introversion (I)
  • Sensing (S) vs. Intuition (N)
  • Thinking (T) vs. Feeling (F)
  • Judging (J) vs. Perceiving (P)
Understanding personality types helps to explain how individuals perceive the world and make decisions. In the context of the exercise, knowing whether married couples share personality preferences can have implications on relationship compatibility and dynamics. Couples sharing preferences may find it easier to understand each other, whereas those with fewer preferences in common might face more conflicts.
Population Proportion
Population proportion is a concept used in statistics to describe the fraction of a population that exhibits a certain characteristic. It is denoted by the symbol \( p \) and is key to understanding how widespread or common a particular trait is within an entire population. One important thing to remember is that population proportions are estimates made based on sample data from larger populations. When we talk about the population proportion of married couples with two or more shared personality preferences, we're estimating what this proportion would be if we could assess every married couple in existence. Similarly, the population proportion of couples sharing no preferences gives insight into how common such a scenario is among all couples.
Sample Proportion
Sample proportion is the estimate of the population proportion, calculated from a sample of the larger group. It is represented by \( \hat{p} \). In our exercise, we calculate two sample proportions:
  • The proportion of married couples with two or more shared personality preferences is \( \hat{p}_1 = 0.7707 \).
  • The proportion of married couples with no shared preferences is \( \hat{p}_2 = 0.0403 \).
These sample proportions are derived from sample data鈥�289 out of 375 couples having shared preferences for \( \hat{p}_1 \), and 23 out of 571 couples for \( \hat{p}_2 \). The sample proportions act as best estimates for their respective population proportions.
Standard Error
Standard error is a statistical metric that reflects the extent of variability or "spread" in a statistic from sample to sample. It is critical for constructing confidence intervals and assessing the reliability of statistical estimates.In the exercise, we focus on the standard error of the difference between sample proportions. The formula is:\[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]This formula takes into account the sample sizes as well, which comprise 375 couples for \( \hat{p}_1 \) and 571 for \( \hat{p}_2 \). Our calculated standard error gives us an idea of how precise our sample-based estimate of the difference in proportions truly is. Here, the standard error was found to be approximately 0.0214, indicating that estimates based on these proportions are likely to be quite reliable.

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Most popular questions from this chapter

The following data represent crime rates per 1000 population for a random sample of 46 Denver neighborhoods (Reference: The Piton Foundation, Denver, Colorado). $$ \begin{array}{lrrrrrr} 63.2 & 36.3 & 26.2 & 53.2 & 65.3 & 32.0 & 65.0 \\ 66.3 & 68.9 & 35.2 & 25.1 & 32.5 & 54.0 & 42.4 \\ 77.5 & 123.2 & 66.3 & 92.7 & 56.9 & 77.1 & 27.5 \\ 69.2 & 73.8 & 71.5 & 58.5 & 67.2 & 78.6 & 33.2 \\ 74.9 & 45.1 & 132.1 & 104.7 & 63.2 & 59.6 & 75.7 \\ 39.2 & 69.9 & 87.5 & 56.0 & 154.2 & 85.5 & 77.5 \\ 84.7 & 24.2 & 37.5 & 41.1 & & & \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 64.2\) and \(s \approx 27.9\) crimes per 1000 population. (b) Let us say the preceding data are representative of the population crime rates in Denver neighborhoods. Compute an \(80 \%\) confidence interval for \(\mu\), the population mean crime rate for all Denver neighborhoods. (c) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 57 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. (d) Another neighborhood has a crime rate of 75 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. (e) Repeat parts (b), (c), and (d) for a \(95 \%\) confidence interval. (f) In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(7.2\).

Consider a \(90 \%\) confidence interval for \(\mu .\) Assume \(\sigma\) is not known. For which sample size, \(n=10\) or \(n=20\), is the critical value \(t_{c}\) larger?

Why do we use \(1 / 4\) in place of \(p(1-p)\) in formula (22) for sample size when the probability of success \(p\) is unknown? (a) Show that \(p(1-p)=1 / 4-(p-1 / 2)^{2}\). (b) Why is \(p(1-p)\) never greater than \(1 / 4\) ?

How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from \(20^{\circ} \mathrm{F}\) to \(45^{\circ} \mathrm{F}\). A random sample of prices (\$) for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol. 25 , Issue 157, No. 2). Brand names include American Camper, Cabela's, Camp 7, Caribou, Cascade, and Coleman. \(\begin{array}{rrrrrrrrrr}80 & 90 & 100 & 120 & 75 & 37 & 30 & 23 & 100 & 110 \\ 105 & 95 & 105 & 60 & 110 & 120 & 95 & 90 & 60 & 70\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx \$ 83.75\) and \(s \approx \$ 28.97\). (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a \(90 \%\) confidence interval for the mean price \(\mu\) of all summer sleeping bags.

In a random sample of 519 judges, it was found that 285 were introverts (see reference of Problem 5). (a) Let \(p\) represent the proportion of all judges who are introverts. Find a point estimate for \(p\). (b) Find a \(99 \%\) 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.
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