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Chapter 11: Problem 9

How is the expected frequency of a category calculated for a goodness-of-fit test? What are the degrees of freedom for such a test?

Short Answer

Expert verified
The expected frequency in a goodness-of-fit test is calculated by dividing the total number of observations by the number of categories. The degrees of freedom are defined as the number of categories minus one.

Step by step solution

01

Understanding The Goodness-of-Fit Test And Expected Frequency

The Goodness-of-Fit test is a statistical hypothesis test to determine if the observed frequencies differ from the theoretical or expected frequencies. For each category, the expected frequency is calculated by taking the total number of observations and dividing it by the number of categories.
02

Calculation of Expected Frequency

So, if total observations are denoted by N and the number of categories by k, the expected frequency (E) for each category is calculated as: \[ E_i = \frac{N}{k} \] where \(i\) is a category from \(1, 2, ..., k\).
03

Understanding Degrees of Freedom in a Goodness-of-Fit Test

Degrees of freedom (df) refer to the number of values involved in the calculation that have the freedom to vary. In a goodness-of-fit test, the degrees of freedom are calculated as the number of categories minus 1.
04

Calculation of Degrees of Freedom

So, if the number of categories is denoted by k, the degrees of freedom (df) are calculated as: \[ df = k - 1 \]

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Most popular questions from this chapter

Explain how the expected frequencies for cells of a contingency table are calculated in a test of independence or homogeneity.

Over the last 3 years, Art's Supermarket has observed the following distribution of payment methods in the checkout lines: cash (C) \(41 \%\), check (CK) \(24 \%\), credit or debit card (D) \(26 \%\), and other (N) \(9 \%\). In an effort to minimize costly credit and debit card fees, Art's has just begun offering a \(1 \%\) discount for cash payment in the checkout line. The following table lists the frequency distribution of payment methods for a random sample of 500 customers after the discount went into effect. $$ \begin{array}{l|cccc} \hline \text { Payment method } & \mathrm{C} & \mathrm{CK} & \mathrm{D} & \mathrm{N} \\ \hline \text { Number of customers } & 240 & 104 & 111 & 45 \\ \hline \end{array} $$ Test at a \(1 \%\) significance level whether the distribution of payment methods in the checkout line changed after the discount went into effect.

In a Pew Research Center poll conducted December \(3-8,2013\), American adults age 18 and older were asked if Christmas is more a religious or a cultural holiday for them. Of the respondents, \(51 \%\) said Christmas is a religious holiday for them, \(32 \%\) said it is a cultural holiday, and \(17 \%\) gave other answers (www.pewforum.org). Assume that these results are true for the 2013 population of adults. Recently, a random sample of 1200 American adults age 18 and older was taken, and these adults were asked the same question. Their responses are presented in the following table. $$ \begin{array}{l|ccc} \hline \text { Response } & \text { Religious Holiday } & \text { Cultural Holiday } & \text { Other } \\ \hline \text { Frequency } & 660 & 408 & 132 \\ \hline \end{array} $$ Test at a \(2.5 \%\) significance level whether the distribution of recent opinions is significantly different from that of the 2013 opinions.

Construct the \(98 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=21, s^{2}=9.2\) b. \(n=17, s^{2}=1.7\)

A sample of 21 observations selected from a normally distributed population produced a sample variance of \(1.97 .\) a. Write the null and alternative hypotheses to test whether the population variance is greater than \(1.75\). b. Using \(\alpha=.025\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on a chi-square distribution curve. c. Find the value of the test statistic \(\chi^{2}\). d. Using a \(2.5 \%\) significance level, will you reject the null hypothesis stated in part a?
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