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A study reported in the Journal of Small Business Management concluded that selfemployed individuals do not experience higher job satisfaction than individuals who are not self-employed. In this study, job satisfaction is measured using 18 items, each of which is rated using a Likert-type scale with \(1-5\) response options ranging from strong agreement to strong disagreement. A higher score on this scale indicates a higher degree of job satisfaction. The sum of the ratings for the 18 items, ranging from \(18-90\), is used as the measure of job satisfaction. Suppose that this approach was used to measure the job satisfaction for lawyers, physical therapists, cabinetmakers, and systems analysts. The results obtained for a sample of 10 individuals from each profession follow. At the \(\alpha=.05\) level of significance, test for any difference in the job satisfaction among the four professions.

Short Answer

Expert verified
Use a one-way ANOVA at \( \alpha=0.05 \) to test for differences in job satisfaction; conclude based on the F-statistic and critical value comparison.

Step by step solution

01

Define the Hypotheses

We want to test if there is any significant difference in job satisfaction among the four professions. The null hypothesis \( H_0 \) states that there is no difference in job satisfaction among the groups, i.e., all group means are equal. The alternative hypothesis \( H_1 \) is that at least one group mean is different from the others. Mathematically, \( H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4 \) and \( H_1: \text{at least one } \mu \text{ is different}.\)
02

Determine the Significance Level and Test Statistic

The level of significance \( \alpha \) is given as 0.05. We will use a one-way ANOVA to test for differences in means among the groups. The test statistic for ANOVA is the F-ratio: \[ F = \frac{\text{Variance between the groups}}{\text{Variance within the groups}} \]
03

Calculate Group Means and Overall Mean

Calculate the mean job satisfaction score for each profession group. Then calculate the overall mean of all scores.
04

Calculate Variance Between Groups

The variance between groups is calculated by multiplying the sum of squares between the groups (SSB) by the mean square between the groups. SSB is calculated as follows: \[ \text{SSB} = \sum n_i (\bar{x}_i - \bar{x})^2 \] where \( n_i \) is the number of observations in group \( i \), \( \bar{x}_i \) is the mean of group \( i \), and \( \bar{x} \) is the overall mean.
05

Calculate Variance Within Groups

The variance within groups is calculated using the sum of squares within the groups (SSW). This is done by \[ \text{SSW} = \sum (x_{ij} - \bar{x}_i)^2 \] where \( x_{ij} \) is an observation in group \( i \).
06

Compute Total Sum of Squares

The total sum of squares (SST) is the sum of SSB and SSW: \[ \text{SST} = \text{SSB} + \text{SSW} \] This value summarizes the total variability in the data.
07

Calculate the F-statistic

The F-statistic is calculated by dividing the mean square between (MSB) by the mean square within (MSW). These are obtained by dividing SSB and SSW by their respective degrees of freedom: \[ F = \frac{MSB}{MSW} = \frac{\frac{SSB}{k-1}}{\frac{SSW}{N-k}} \] where \( k \) is the number of groups and \( N \) is the total number of observations.
08

Compare F-statistic to Critical Value

Use an F-distribution table or software to find the critical value of F at \( \alpha = 0.05 \), with \( k-1 \) and \( N-k \) degrees of freedom. Compare the calculated F-statistic to this critical value.
09

Make a Decision

If the calculated F-statistic is greater than the critical value, reject the null hypothesis \( H_0 \). Otherwise, do not reject \( H_0 \).
10

Conclusion

Based on the decision, conclude whether there is sufficient evidence to suggest a difference in job satisfaction among the professions at the 0.05 significance level.

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Key Concepts

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

Job Satisfaction
Job satisfaction is a crucial aspect of workplace psychology and management practices. It refers to the level of contentment employees feel regarding their job roles and environment. In the study mentioned, job satisfaction is measured using a Likert-type scale from 1 to 5, where higher scores indicate higher satisfaction.
Each individual rates their satisfaction based on 18 different items related to their job, resulting in a total score ranging from 18 to 90. By examining these scores, we can assess how satisfied individuals are in their respective professions, such as lawyers, physical therapists, cabinetmakers, and systems analysts.
Understanding job satisfaction helps employers improve work conditions and could lead to enhanced productivity, which benefits both employees and organizations.
  • Measures personal and professional contentment
  • Uses a scale to quantify satisfaction levels
  • A key predictor of retention and productivity
Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences or draw conclusions about a population based on sample data. It starts with a clear statement, known as the null hypothesis ( H_0 ), which assumes no effect or difference exists.
In the provided exercise, the null hypothesis states that there is no difference in job satisfaction between the four professional groups. The alternative hypothesis ( H_1 ) suggests that there is indeed a difference.
Through testing, researchers aim to determine if observational data provide enough evidence to reject the null hypothesis in favor of the alternative. This process involves comparing calculated test statistics to predetermined thresholds, or critical values, at a specific level of significance, such as 0.05.
  • Enables decision-making based on data
  • Compares observed results against expectations
  • Aims to establish statistical validity
F-distribution
The F-distribution is a fundamental part of variance analysis, specifically in ANOVA tests. It is used to compare variances between samples to determine if there are significant differences among group means. A crucial property of the F-distribution is that it varies based on two different degrees of freedom settings – between groups and within groups.
The calculation of the F-statistic involves dividing the variance estimate between groups by the variance estimate within groups. The resulting value is then compared to critical values from the F-distribution table.
The F-distribution helps determine whether observed variations are likely due to actual differences or just random chance. In our job satisfaction study, it helps identify if differences in scores among the four professions are statistically significant.
  • Compares group variances to assess differences
  • Dependent on sample size and group count
  • Outputs are used to support hypothesis testing
Statistical Significance
Statistical significance addresses whether the results of a study provide enough evidence to reject the null hypothesis. It is a measure of the strength of evidence that allows researchers to distinguish between true effects and random noise.
In statistical tests like ANOVA, results are considered statistically significant if the calculated F-statistic exceeds the critical value from the F-distribution at the predetermined significance level, often set at 0.05.
This concept ensures that findings are not merely due to random sampling error but reflect a genuine effect observed in the data. In the given study, statistical significance would mean there's truly a difference in job satisfaction across professions.
  • Distinguishing true differences from random chance
  • Determined by p-values and critical thresholds
  • Vital for making informed scientific conclusions

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Most popular questions from this chapter

The following data are from a completely randomized design. \(\begin{array}{ccc} & \text { Treatment } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 162 & 142 & 126 \\ 142 & 156 & 122 \\ 165 & 124 & 138 \\ 145 & 142 & 140 \\ 148 & 136 & 150 \\ 174 & 152 & 128\end{array}\) \(\begin{array}{lccc}\text { Sample mean } & 156 & 142 & 134 \\ \text { Sample variance } & 164.4 & 131.2 & 110.4\end{array}\) a. Compute the sum of squares between treatments. b. Compute the mean square between treatments. c. Compute the sum of squares due to error d. Compute the mean square due to error. e. Set up the ANOVA table for this problem. f. \(\quad\) At the \(\alpha=.05\) level of significance, test whether the means for the three treatments are equal.

Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment (US Airways Attaché, December 2003 ). Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1 ) and the annual credit card charges for dining out (population 2 ). Using the difference data, the sample mean difference was \(d=\$ 850,\) and the sample standard deviation was \(s_{d}=\$ 1123\) a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. b. Use a .05 level of significance. Can you conclude that the population means differ? What is the \(p\) -value? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the \(95 \%\) confidence interval estimate of the difference between the population means?

In early 2009 , the economy was experiencing a recession. But how was the recession affecting the stock market? Shown are data from a sample of 15 companies. Shown for each company is the price per share of stock on January 1 and April 30 (The Wall Street Journal, May 1,2009 ). a. What is the change in the mean price per share of stock over the four-month period? b. Provide a \(90 \%\) confident interval estimate of the change in the mean price per share of stock. Interpret the results. c. What was the percentage change in the mean price per share of stock over the fourmonth period? d. If this same percentage change were to occur for the next four months and again for the four months after that, what would be the mean price per share of stock at the end of the year \(2009 ?\)

Consider the following hypothesis test. \(H_{0}: \mu_{1}-\mu_{2}=0\) \(H_{\mathrm{a}}: \mu_{1}-\mu_{2} \neq 0\) The following results are for two independent samples taken from the two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\ n_{1}=80 & n_{2}=70 \\ \bar{x}_{1}=104 & \bar{x}_{2}=106 \\ \sigma_{1}=8.4 & \sigma_{2}=7.6\end{array}\) a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

The Consumer Reports Restaurant Customer Satisfaction Survey is based upon 148,599 visits to full-service restaurant chains (Consumer Reports website). One of the variables in the study is meal price, the average amount paid per person for dinner and drinks, minus the tip. Suppose a reporter for the Sun Coast Times thought that it would be of interest to her readers to conduct a similar study for restaurants located on the Grand Strand section in Myrtle Beach, South Carolina. The reporter selected a sample of eight seafood restaurants, eight Italian restaurants, and eight steakhouses. The following data show the meal prices (\$) obtained for the 24 restaurants sampled. Use \(\alpha=.05\) to test whether there is a significant difference among the mean meal price for the three types of restaurants.

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