91Ó°ÊÓ

Job satisfaction and income \(\quad\) A recent GSS was used to cross-tabulate income \((<\$ 15\) thousand, \(\$ 15-25\) thousand, \(\$ 25-40\) thousand, \(>\$ 40\) thousand \()\) in dollars with job satisfaction (very dissatisfied, little dissatisfied, moderately satisfied, very satisfied) for 96 subjects. a. For these data, \(X^{2}=6.0 .\) What is its \(d f\) value, and what is its approximate sampling distribution, if \(\mathrm{H}_{0}\) is true? b. For this test, the P-value is 0.74 . Interpret in the context of these variables. c. What decision would you make with a 0.05 significance level? Can you accept \(\mathrm{H}_{0}\) and conclude that job satisfaction is independent of income?

Step by step solution

01

Determine Degrees of Freedom

The degrees of freedom for a chi-squared test with a cross-tabulated table is calculated using the formula \( df = (r - 1)(c - 1) \), where \( r \) is the number of rows and \( c \) is the number of columns. Here, there are 4 income categories and 4 job satisfaction categories, so \( r = 4 \) and \( c = 4 \). Thus, \( df = (4 - 1)(4 - 1) = 3 \times 3 = 9 \).

02

State Approximate Sampling Distribution

If the null hypothesis \( H_0 \) is true, the sampling distribution of the chi-squared statistic \( X^{2} \) is approximately chi-squared with the previously calculated degrees of freedom. In this case, it is chi-squared with 9 degrees of freedom.

03

Interpret the P-value

The P-value of 0.74 suggests that there is a 74% probability of observing a chi-squared statistic as extreme as 6.0 or more extreme, assuming \( H_0 \) is true. Given that this P-value is much larger than common significance levels such as 0.05 or 0.01, it indicates weak evidence against \( H_0 \).

04

Make Decision at Significance Level

With a significance level of 0.05, if the P-value is greater than 0.05, we fail to reject the null hypothesis. Since 0.74 > 0.05, we do not reject \( H_0 \), meaning there isn't sufficient evidence to state that job satisfaction and income are dependent.

05

Answer Research Question

Based on our calculations and outcome of the chi-square test, we conclude that at a 0.05 significance level, we cannot reject the null hypothesis. Thus, we have no evidence to claim that job satisfaction and income are not independent.

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Key Concepts

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

Degrees of Freedom

In a chi-squared test, the concept of "Degrees of Freedom" ( 1df 3) is crucial for determining the appropriate chi-squared distribution. This concept essentially captures the idea of how much freedom is available for variation within the data due to constraints. The formula for calculating degrees of freedom in a chi-squared test with cross-tabulated data is given by: \[df = (r - 1)(c - 1)\]where:

• \( r \) is the number of rows (or categories of one variable), in this case, job satisfaction categories.
• \( c \) is the number of columns (or categories of the other variable), in this case, income categories.

For this exercise, we have 4 categories for both income and job satisfaction:

• Jobs categories: Very dissatisfied, Little dissatisfied, Moderately satisfied, Very satisfied.
• Income categories: Less than \(15k, \)15-\(25k, \)25-\(40k, More than \)40k.

Therefore, putting these into the formula, our calculation is:\[df = (4 - 1)(4 - 1) = 3 \times 3 = 9\]This means our chi-squared statistic follows a chi-squared distribution with 9 degrees of freedom.

P-value Interpretation

The "P-value" is a key metric used in statistical hypothesis testing to help determine the strength of the evidence against the null hypothesis ( H_0 ). In simple terms, it tells us how likely we are to observe our data, or something more extreme, if the null hypothesis is true. Here, a P-value of 0.74 suggests that if job satisfaction and income were truly independent (as H_0 posits), there is a 74% chance of observing the computed chi-squared statistic or a more extreme statistic by random variation alone.

• A P-value much larger than typical significance levels (e.g., 0.05 or 0.01) means there's weak evidence against the null hypothesis.
• In this context, our P-value = 0.74 is quite high, which suggests that the association between job satisfaction and income can easily be explained by random chance.

Thus, with such a high P-value, we lack strong evidence to refute the premise that income and job satisfaction are independent, according to our sample data.

Significance Level

The "Significance Level" in hypothesis testing is a threshold used to decide whether to reject the null hypothesis ( H_0 ). It's denoted by \( \alpha \) and often set at 0.05, giving it an approachable and commonly accepted criterion for making decisions.Here's how it fits in:

• A significance level (\( \alpha \)) of 0.05 means there's a 5% risk of rejecting the null hypothesis incorrectly (a Type I error).
• If our P-value is less than or equal to \( \alpha \), we reject ('strong evidence against H_0 '). Conversely, if it is greater, we don't reject ('not enough evidence against H_0 ').

In the example given, with a significance level of 0.05:

• The P-value (0.74) is greater than 0.05.
• This means we do not reject the null hypothesis, indicating no significant evidence that job satisfaction is related to income.

Thus, setting a significance level helps standardize the decision-making process in statistical analyses by providing a clear cutoff point for judging the strength of evidence.