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When the sample evidence is sufficient to justify rejecting the null hypothesis in a goodness-of-fit test, can you tell exactly how the distribution of observed values over the specified categories differs from the expected distribution? Explain.

Short Answer

Expert verified
No, the test alone indicates a difference exists but not specifics. Examine residuals to see which categories differ.

Step by step solution

01

Understand the Hypothesis Test

For a goodness-of-fit test, we start with a null hypothesis (H_0) that assumes the observed frequencies are equal to the expected frequencies. An alternative hypothesis (H_1) states that there is at least one significant difference between observed and expected frequencies.
02

Analyze the Test Result

When sample evidence is sufficient to reject the null hypothesis in a goodness-of-fit test, it means the observed frequencies are significantly different from what was expected across the categories.
03

Evaluate the Goodness-of-Fit Test

The goodness-of-fit test, typically a Chi-Square test, tells us whether there is a significant difference overall, but it does not indicate which specific categories contribute to the difference.
04

Investigate Further for Detailed Analysis

To understand how each category contributes to the difference, examine the residuals or standardized residuals for each category. These show how much each observed frequency diverges from what was expected.

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Key Concepts

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

Understanding the Null Hypothesis
In any statistical test, including the goodness-of-fit test, the null hypothesis plays a central role. It is denoted as \(H_0\) and represents a statement which assumes that there is no effect or no difference; essentially, it is the default or starting assumption. In the context of a goodness-of-fit test, the null hypothesis asserts that the observed frequencies of outcomes in various categories match the expected frequencies if a given theoretical distribution is true.
The goal of conducting the test is to find out if there is enough evidence to reject this hypothesis. If the data significantly differ from what was expected, we may reject the null hypothesis. This suggests that at least one category differs significantly. However, just rejecting \(H_0\) doesn't give detailed information about specific discrepancies across categories, which is where further analysis is needed.
Chi-Square Test as a Tool for Goodness-of-Fit
The chi-square test is the most commonly used statistical test for evaluating goodness-of-fit, especially when we deal with categorical data. It assesses whether observed data align with expected data based on a particular hypothesis. This involves calculating the chi-square statistic, which summarizes the differences:
  • Expected Frequency: What you would expect if the null hypothesis were true.
  • Observed Frequency: What you actually measure or observe in your data.
Using the chi-square formula:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency. A higher chi-square statistic indicates a greater discrepancy between observed and expected frequencies. By comparing this result to a critical value from the chi-square distribution, we determine if observed frequencies significantly differ from expected frequencies. This overall test indicates a general difference but does not pinpoint where the differences lie.
Exploring Residuals to Identify Differences
Once a chi-square test leads to rejecting the null hypothesis, the next step is to understand where the differences occur. This is done by examining residuals. Residuals provide insights into the contribution of each category to the overall chi-square statistic, helping to identify specific areas of significant deviation.
There are different types of residuals, including raw residuals and standardized residuals. In particular:
  • Raw Residuals: Simple difference between observed and expected frequencies \(O_i - E_i\).
  • Standardized Residuals: Adjusted for the variance in expected frequency, calculated as \(\frac{O_i - E_i}{\sqrt{E_i}}\).
Standardized residuals are especially useful as they are on a standard scale, making them easier to interpret. Large standardized residuals indicate categories contributing significantly to the overall difference. By analyzing these values, researchers can derive detailed insights into which specific elements of the distribution differ from what was hypothesized.

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Most popular questions from this chapter

The following problem is based on information from an article by N. Keyfitz in The American Journal of Sociology (Vol. 53, pp. \(470-480)\). Let \(x=\) age in years of a rural Quebec woman at the time of her first marriage. In the year 1941 , the population variance of \(x\) was approximately \(\sigma^{2}\) \(=5.1 .\) Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance \(s^{2}=3.3 .\) Use a \(5 \%\) level of significance to test the claim that the current variance is less than \(5.1 .\) Find a \(90 \%\) confidence interval for the population variance.

A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a \(0.05\) level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. \(\begin{array}{rrcr}\text { Country I } & \text { Country II } & \text { Country III } & \text { Country IV } \\ 12.7 & 8.3 & 20.3 & 17.2 \\\ 9.2 & 17.2 & 16.6 & 8.8 \\ 10.9 & 19.1 & 22.7 & 14.7 \\ 8.9 & 10.3 & 25.2 & 21.3 \\ 16.4 & & 19.9 & 19.8\end{array}\)

The fan blades on commercial jet engines must be replaced when wear on these parts indicates too much variability to pass inspection. If a single fan blade broke during operation, it could severely endanger a flight. A large engine contains thousands of fan blades, and safety regulations require that variability measurements on the population of all blades not exceed \(\sigma^{2}=0.18 \mathrm{~mm}^{2}\). An engine inspector took a random sample of 61 fan blades from an engine. She measured each blade and found a sample variance of \(0.27\) \(\mathrm{mm}^{2}\). Using a \(0.01\) level of significance, is the inspector justified in claiming that all the engine fan blades must be replaced? Find a \(90 \%\) confidence interval for the population standard deviation.

ow reliable are mutual funds that invest in bonds? Again, this depends on the bond fund you buy (see reference in Problem 9). A random sample of annual percentage returns for mutual funds holding shortterm U.S. government bonds is shown below. \(\begin{array}{lllllll}4.6 & 4.7 & 1.9 & 9.3 & -0.8 & 4.1 & 10.5\end{array}\) $$ \begin{array}{llllll} 4.2 & 3.5 & 3.9 & 9.8 & -1.2 & 7.3 \end{array} $$ Use a calculator to verify that \(s^{2} \approx 13.59\) for the preceding data. A random sample of annual percentage returns for mutual funds holding intermediate-term corporate bonds is shown below. $$ \begin{array}{rrrrrrrr} -0.8 & 3.6 & 20.2 & 7.8 & -0.4 & 18.8 & -3.4 & 10.5 \\ 8.0 & -0.9 & 2.6 & -6.5 & 14.9 & 8.2 & 18.8 & 14.2 \end{array} $$ Use a calculator to verify that \(s^{2}=72.06\) for returns from mutual funds holding intermediate-term corporate bonds. Use \(\alpha=0.05\) to test the claim that the population variance for annual percentage returns of mutual funds holding short-term government bonds is different from the population variance for mutual funds holding intermediate- term corporate bonds. How could your test conclusion relate to the question of reliability of returns for each type of mutual fund?

An economist wonders if corporate productivity in some countries is more volatile than in other countries. One measure of a company's productivity is annual percentage yield based on total company assets. Data for this problem are based on information taken from Forbes Top Companies, edited by J. T. Davis. A random sample of leading companies in France gave the following percentage yields based on assets: \(\begin{array}{llllllllllll}4.4 & 5.2 & 3.7 & 3.1 & 2.5 & 3.5 & 2.8 & 4.4 & 5.7 & 3.4 & 4.1\end{array}\) \(\begin{array}{llllllllll}6.8 & 2.9 & 3.2 & 7.2 & 6.5 & 5.0 & 3.3 & 2.8 & 2.5 & 4.5\end{array}\) Use a calculator to verify that \(s^{2}=2.044\) for this sample of French companies. Another random sample of leading companies in Germany gave the following percentage yields based on assets: \(\begin{array}{rrrrrrrrr}3.0 & 3.6 & 3.7 & 4.5 & 5.1 & 5.5 & 5.0 & 5.4 & 3.2\end{array}\) \(\begin{array}{llllllllll}3.5 & 3.7 & 2.6 & 2.8 & 3.0 & 3.0 & 2.2 & 4.7 & 3.2\end{array}\) Use a calculator to verify that \(s^{2} \approx 1.038\) for this sample of German companies. Test the claim that there is a difference (either way) in the population variance of percentage yields for leading companies in France and Germany. Use a \(5 \%\) level of significance. How could your test conclusion relate to the economist's question regarding volatility (data spread) of corporate productivity of large companies in France compared with companies in Germany?

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