/*! 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 10 The following table shows the My... [FREE SOLUTION] | 91Ó°ÊÓ

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The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and F refers to feeling. $$\begin{array}{l|c|c|c} \hline \multirow{2}{*} {\text { Occupation }} & \multicolumn{2}{c} {\text { Personality Preference Type }} \\ \\)\cline { 2 - 5 } & \\( \mathrm{T} & \mathrm{F} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\ \hline \text { M.D. } & 77 & 82 & 159 \\ \hline \text { Lawyer } & 118 & 94 & 212 \\ \hline \text { Column Total } & 252 & 267 & 519 \\ \hline \end{array}$$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.

Short Answer

Expert verified
Occupation and personality preference type are not independent.

Step by step solution

01

Set up the Hypotheses

The null hypothesis \( H_0 \) is that occupation and personality preference type are independent. The alternative hypothesis \( H_a \) is that occupation and personality preference type are not independent.
02

Calculate Expected Frequencies

The expected frequency for each cell is calculated using the formula: \[ E_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}} \]. Compute the expected values as follows:- For Clergy (T): \[ E = \frac{148 \times 252}{519} \approx 71.73 \]- For Clergy (F): \[ E = \frac{148 \times 267}{519} \approx 76.27 \]- For M.D. (T): \[ E = \frac{159 \times 252}{519} \approx 77.25 \]- For M.D. (F): \[ E = \frac{159 \times 267}{519} \approx 81.75 \]- For Lawyer (T): \[ E = \frac{212 \times 252}{519} \approx 103.02 \]- For Lawyer (F): \[ E = \frac{212 \times 267}{519} \approx 108.98 \].
03

Compute Chi-Square Value

Use the formula \( \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \) to compute the chi-square statistic:- For Clergy (T): \[ \frac{(57 - 71.73)^2}{71.73} \approx 3.29 \]- For Clergy (F): \[ \frac{(91 - 76.27)^2}{76.27} \approx 2.83 \]- For M.D. (T): \[ \frac{(77 - 77.25)^2}{77.25} \approx 0.0008 \]- For M.D. (F): \[ \frac{(82 - 81.75)^2}{81.75} \approx 0.0008 \]- For Lawyer (T): \[ \frac{(118 - 103.02)^2}{103.02} \approx 2.18 \]- For Lawyer (F): \[ \frac{(94 - 108.98)^2}{108.98} \approx 2.17 \]. Sum these values to get \( \chi^2 \approx 11.47 \).
04

Determine Degrees of Freedom and Critical Value

The degrees of freedom for this table is calculated as \((r-1)(c-1)\), where \(r\) is the number of rows, and \(c\) is the number of columns. Here, DF = \((3-1)(2-1) = 2\). The critical value at \( \alpha = 0.01\) for DF = 2 is 9.21.
05

Compare Chi-Square Statistic with Critical Value

Since the computed \( \chi^2 = 11.47 \) is greater than the critical value 9.21, we reject the null hypothesis.
06

State the Conclusion

There is significant evidence at the \( \alpha = 0.01 \) significance level to suggest that occupation and personality preference type are not independent.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistics, used to determine if there exists enough evidence to reject a presumed null hypothesis (denoted as \( H_0 \)). In our example, the null hypothesis states that occupation and personality preference are independent, meaning one does not affect the other. The alternative hypothesis (\( H_a \)) claims the opposite, saying that these variables are not independent. This separation into null and alternative allows statisticians to systematically test and make decisions based on data. In hypothesis testing, we must also establish a significance level (commonly \( \, \alpha \)), which is the threshold for determining when to reject the null hypothesis. In this problem, the significance level is 0.01, indicating we are willing to accept a 1% chance of incorrectly rejecting \( H_0 \). This highlights the importance of careful interpretation in statistical analysis.
Understanding hypothesis testing provides a structured approach to deciding whether observed patterns in data are real or could have occurred by chance alone.
Degrees of Freedom
Degrees of Freedom (DF) are a crucial concept in statistical testing, including the chi-square test. They represent the number of values in a calculation that are free to vary. To compute the degrees of freedom in a chi-square test for independence, you use the formula \((r-1)(c-1)\), where \(r\) is the number of categories (or rows) and \(c\) is the number of levels or possibilities within each category (or columns).
In this problem, knowing we have three occupations (Clergy, M.D., Lawyer) and two personality preferences (T and F), the calculation is \((3-1)(2-1) = 2\). These two degrees of freedom are essential for determining the critical value from the chi-square distribution, helping us make an informed decision regarding our hypothesis.
Critical Value
The critical value is a point on the chi-square distribution that helps determine whether the observed statistic is significant. It acts like a benchmark; if our computed chi-square statistic exceeds the critical value, we have evidence to reject the null hypothesis. The critical value depends on the degrees of freedom and the significance level \(\alpha\). For our data, with 2 degrees of freedom and an alpha level of 0.01, the critical value is 9.21. This signifies that for our chi-square statistic to be considered significant, it must be greater than 9.21. In this instance, our computed chi-square of 11.47 surpasses the critical value, leading us to reject the null hypothesis and conclude that occupation and personality preferences are not independent. This analysis demonstrates how critical values serve as thresholds in hypothesis testing.
Personality Preferences
Personality preferences describe variations in how people think, feel, and behave, often categorized by models such as the Myers-Briggs Type Indicator (MBTI). This specific exercise deals with thinking (T) and feeling (F) as different personality preferences within the MBTI framework. Each occupation may have a tendency toward certain personality types, affecting patterns across different groups. By analyzing these preferences statistically, as in this chi-square test, we can explore deeper relationships and dependencies between human psychological attributes and professional disciplines. Understanding personality preferences provides insight into workplace dynamics and aids in developing environments that accommodate diverse character types.
Expected Frequencies
Expected frequencies are critical to performing a chi-square test of independence. They represent the frequency counts we would anticipate in each cell of our table, assuming the null hypothesis is true—meaning that the rows and columns are independent. Calculating expected frequencies involves using the formula: \[ E_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}} \] Applying this in our problem allows us to derive the expected counts for each occupation-personality preference pair. For example, for Clergy (T), the expected frequency is approximately 71.73, suggesting if there was no actual association, we'd expect about 71 clergy members to have a thinking preference. The discrepancies between these expected counts and the observed counts are critical for computing the chi-square statistic, ultimately informing us about the validity of our null hypothesis. Understanding expected frequencies helps us visualize what independence would look like and assess the extent of actual deviation in our data.

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Most popular questions from this chapter

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than \(5 ?\) What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the \(P\) -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Ecology: Fish The Fish and Game Department stocked Lake Lulu with fish in the following proportions: \(30 \%\) catfish, \(15 \%\) bass, \(40 \%\) bluegill, and \(15 \%\) pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the 500 fish in the sample were distributed as follows. \(\begin{array}{cccc}\text { Catfish } & \text { Bass } & \text { Bluegill } & \text { Pike } \\ 120 & 85 & 220 & 75\end{array}\) In the 5 -year interval, did the distribution of fish change at the 0.05 level?

For the study regarding mean cadence (see Problem 1), two-way ANOVA was used. Recall that the two factors were walking device (none, standard walker, rolling walker) and dual task (being required to respond vocally to a signal or no dual task required). Results of two-way ANOVA showed that there was no evidence of interaction between the factors. However, according to the article, "The ANOVA conducted on the cadence data revealed a main effect of walking device." When the hypothesis regarding no difference in mean cadence according to which, if any, walking device was used, the sample \(F\) was \(30.94,\) with \(d, f_{N}=2\) and \(d . f_{. D}=18\) Further, the \(P\) -value for the result was reported to be less than \(0.01 .\) From this information, what is the conclusion regarding any difference in mean cadence according to the factor "walking device used"?

A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was \(58.4 .\) Another random sample of 25 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was \(31.6 .\) Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?

Rothamsted Experimental Station (England) has studied wheat production since \(1852 .\) Each year, many small plots of equal size but different soil/fertilizer conditions are planted with wheat. At the end of the growing season, the yield (in pounds) of the wheat on the plot is measured. The following data are based on information taken from an article by G. A. Wiebe in the Journal of Agricultural Research (Vol. \(50,\) pp. \(331-357\) ). For a random sample of years, one plot gave the following annual wheat production (in pounds): $$\begin{array}{lllllll} 4.15 & 4.21 & 4.27 & 3.55 & 3.50 & 3.79 & 4.09 & 4.42 \\ 3.89 & 3.87 & 4.12 & 3.09 & 4.86 & 2.90 & 5.01 & 3.39 \end{array}$$ Use a calculator to verify that, for this plot, the sample variance is \(s^{2} \approx 0.332\) Another random sample of years for a second plot gave the following annual wheat production (in pounds): $$\begin{array}{llllllll} 4.03 & 3.77 & 3.49 & 3.76 & 3.61 & 3.72 & 4.13 & 4.01 \\ 3.59 & 4.29 & 3.78 & 3.19 & 3.84 & 3.91 & 3.66 & 4.35 \end{array}$$ Use a calculator to verify that the sample variance for this plot is \(s^{2} \approx 0.089\) Test the claim that the population variance of annual wheat production for the first plot is larger than that for the second plot. Use a \(1 \%\) level of significance.

For a chi-square goodness-of-fit test, how are the degrees of freedom computed?

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