/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Myers-Briggs: Marriage Counselin... [FREE SOLUTION] | 91Ó°ÊÓ

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Myers-Briggs: Marriage Counseling Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences, which 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) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation 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
Yes, the interval (0.671, 0.789) suggests more couples share common preferences.

Step by step solution

01

Verify Sample Size Requirement

To use a normal distribution to approximate the difference between proportions, the sample sizes should satisfy the rule of thumb: each group should have at least 5 expected successes and 5 expected failures. For \(n_1 = 375\) and \(p_{1,\text{hat}} = \frac{289}{375} = 0.77\), we have successes \(375 \times 0.77 = 288.75\) and failures \(375 \times 0.23 = 86.25\).For \(n_2 = 571\) and \(p_{2,\text{hat}} = \frac{23}{571} = 0.04\), we have successes \(571 \times 0.04 = 22.84\) and failures \(571 \times 0.96 = 548.16\). Both groups meet the requirement.
02

Calculate Proportions and Standard Error

The estimated proportions are \(\hat{p}_1 = 0.77\) and \(\hat{p}_2 = 0.04\).The standard error for the difference in proportions is given by:\[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} = \sqrt{\frac{0.77 \times 0.23}{375} + \frac{0.04 \times 0.96}{571}} \= \sqrt{0.00047267 + 0.00006723} = \sqrt{0.0005399} \approx 0.023\]
03

Find the 99% Z-Score

For a 99% confidence interval, the critical value \(Z\) from the standard normal distribution is approximately 2.576.
04

Calculate Confidence Interval

Calculate the confidence interval for the difference in population proportions:\[ (\hat{p}_1 - \hat{p}_2) \pm Z \times SE \= (0.77 - 0.04) \pm 2.576 \times 0.023 \= 0.73 \pm 0.059 \= (0.671, 0.789) \\]
05

Interpret the Confidence Interval

The 99% confidence interval for \(p_{1} - p_{2} \) is (0.671, 0.789), which means we are 99% confident that the difference in the proportion of married couples with two or more common personality preferences is between 0.671 and 0.789 compared to those with none in common.Since the interval includes only positive numbers, this suggests that significantly more couples share two or more personality preferences than none.

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Key Concepts

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

Confidence Intervals
Confidence intervals are a fundamental concept in statistical inference. They provide a range of values within which we can expect a population parameter, like a proportion, to lie with a certain level of confidence. In the context of marriage counseling and the Myers-Briggs personality types, the confidence interval for the difference in proportions indicates how much more frequently one characteristic occurs compared to another in the population.
To calculate a confidence interval, we first need the point estimate (like the difference in sample proportions) and the standard error of that estimate. Then, we determine the critical value corresponding to our desired confidence level (e.g., 99%). Using these, we construct the interval by adding and subtracting the product of the critical value and the standard error from the point estimate.
In the original exercise, the confidence interval was calculated to be between 0.671 and 0.789, meaning that we are 99% confident that the difference in the proportion of married couples with common preferences is between these limits.
Normal Distribution
The normal distribution, often referred to as the "bell curve," is a continuous probability distribution that is symmetrical around its mean. Being able to assume that a distribution is normal allows statisticians to use various formulas and tables to calculate probabilities and confidence intervals.
In practical applications like the Myers-Briggs exercise, the normal distribution is used to approximate the sampling distribution of the sample proportion differences, provided that certain conditions are met. For this, each sample in the exercise actually met the requirement of having at least 5 expected successes and failures in order for a normal approximation to be considered valid.
  • The central limit theorem underlies this by positing that, given a sufficiently large sample size, the distribution of the sample mean will be approximately normal, regardless of the distribution of the population.
  • This approximation is crucial as it offers a straightforward means to derive further statistical insights, such as constructing confidence intervals.
Population Proportion
Population proportion refers to the fraction of the total population that exhibits a particular characteristic. For instance, in the Myers-Briggs exercise, the population proportion speaks about the number of married couples who share two or more preferences versus none. These proportions are denoted by symbols like \( p_1 \) and \( p_2 \).
To determine these proportions, sample data is gathered and analyzed. Here, * The sample proportion \( \hat{p} \) is calculated by dividing the number of successes by the total number of observations.* The difference in these sample proportions \( \hat{p}_1 - \hat{p}_2 \) provides insight into whether one characteristic is more common than another in the broader population.
Population proportions help in predicting and influencing real-world outcomes, making them vital in fields, including marriage counseling.
Myers-Briggs Personality Types
The Myers-Briggs Type Indicator (MBTI) is a psychological tool that categorizes individuals into one of 16 distinct personality types, based on preferences across four dichotomies: Introversion/Extroversion, Sensing/Intuition, Thinking/Feeling, and Judging/Perceiving.
In the context of marital relationships, the MBTI can be instrumental in identifying compatibility and potential conflicts based on shared or divergent personality preferences among couples. This insight can allow counseling to be tailored to address specific interpersonal dynamics.
  • This personality typing system can foster better communication and insight among partners.
  • By identifying shared preferences, couples can build on mutual understandings and shared strengths.
  • On the other hand, differences highlighted by the MBTI can also serve as opportunities for growth and adaptation in relationships.
Understanding these personality dynamics is especially useful in counseling to provide couples with effective strategies to improve their communication and relationship dynamics.

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Most popular questions from this chapter

Archaeology: Cultural Affiliation "Unknown cultural affiliations and loss of identity at high elevations." These words are used to propose the hypothesis that archaeological sites tend to lose their identity as altitude extremes are reached. This idea is based on the notion that prehistoric people tended \(n o t\) to take trade wares to temporary settings and/or isolated areas (Source: Prehistoric New Mexico: Background for Survey, by D. E. Stuart and R. P. Gauthier, University of New Mexico Press). As elevation zones of prehistoric people (in what is now the state of New Mexico) increased, there seemed to be a loss of artifact identification. Consider the following information. $$ \begin{array}{lcc} \hline \text { Elevation Zone } & \text { Number of Artifacts } & \text { Number Unidentified } \\ \hline 7000-7500 \mathrm{ft} & 112 & 69 \\ 5000-5500 \mathrm{ft} & 140 & 26 \\ \hline \end{array} $$ Let \(p_{1}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(7000-7500\) feet in the given archaeological area. Let \(p_{2}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(5000-5500\) feet in the given archaeological area. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval in the context of this problem. Does the confidence interval contain all positive numbers? all negative numbers? both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the comparison of the population proportion of unidentified artifacts at high elevations \((7000-7500\) feet \()\) with the population proportion of unidentified artifacts at lower elevations (5000-5500 feet)? How does this relate to the stated hypothesis?

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Basic Computation: Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent distributions that are mound-shaped. A random sample of size \(n_{1}=36\) from the first distribution showed \(\bar{x}_{1}=15\), and a random sample of size \(n_{2}=40\) from the second distribution showed \(\bar{x}_{2}=14\) (a) Check Requirements If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Check Requirements Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4\). What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (e) If you have an appropriate calculator or computer software, find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Interpretation Based on the confidence intervals you computed, can you be \(95 \%\) confident that \(\mu_{1}\) is larger than \(\mu_{2} ?\) Explain.

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