/*! 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 81 You are performing a goodness-of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are performing a goodness-of-fit test with four categories, all of which are supposed to be equally likely. You have a total of 100 observations. The observed frequencies are \(21,26,31\), and 22, respectively, for the four categories. a. Show that you would fail to reject the null hypothesis for these data for any reasonable significance level. b. The sum of the absolute differences (between the expected and the observed frequencies) for these data is 14 (i.e., \(4+1+6+3=14\) ). Is it possible to have different observed frequencies keeping the sum at 14 so that you get a \(p\) -value of \(.10\) or less?

Short Answer

Expert verified
a. For the given data, the Chi-square test statistic is 0.64 and the associated p-value is well above .05 (or any reasonable significance level), indicating no evidence to reject the null hypothesis that the observed frequencies match the expected frequencies well enough to be attributed to random chance. b. While it may be possible to rearrange the observed frequencies to yield a lower p-value, without further information it is not possible to provide a definite set of observed frequencies which would result in a p-value of .10 or less.

Step by step solution

01

Set up the test and calculate expected frequencies

Firstly, set up the null hypothesis \(H_0:\) The observed frequencies fit the expected frequencies. In case of four equally likely categories, each should have 25 observations on average. So expected frequencies are 25 for each category.
02

Compute the Chi-square test statistic for the data

The Chi-square test statistic is defined as: \(\chi^2 = \Sigma \frac{(O_i - E_i)^2}{E_i}\) where \(O_i\) and \(E_i\) are the observed and expected frequencies. For the given data, the calculation is \(\chi^2 = \frac{(21-25)^2}{25} + \frac{(26-25)^2}{25} + \frac{(31-25)^2}{25} + \frac{(22-25)^2}{25} = 0.64\)
03

Find the p-value of the test

The p-value is the probability of observing a value of the test statistic as extreme as, or more extreme than, the value calculated from the data under the null hypothesis. For a Chi-square test with 3 degrees of freedom (df = number of categories - 1), the p-value associated with \(\chi^2 = 0.64\) is much greater than .05 (e.g., about .88, but it depends on the specific Chi-square distribution table). Thus, do not reject the null hypothesis at any reasonable significance level, i.e., the observed frequencies match the expected frequencies well enough that any differences could be attributed to random chance.
04

Consider modifying the frequencies

It is theoretically possible to rearrange the observed frequencies such that the sum of differences remains to be 14 to increase the \(\chi^2\) value and decrease the p-value. However, without further information or optimization algorithm it is not possible to provide a definitive set of observed frequencies. One possible way is to make one category to have very high frequency and all others very low.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chi-Square Test
The chi-square test is a statistical method used to determine if there's a significant difference between the expected and observed frequencies in one or more categories. It's especially useful when we want to see how well our observed data fit a particular distribution, like equal distribution across categories. The chi-square test calculates a test statistic, \( \chi^2 \), which assesses how much the observed frequencies differ from the expected ones. If the \( \chi^2 \) value is high, it suggests that there is a large difference between what was observed and what was expected, potentially indicating that some factor other than random chance affected the results.
To perform a chi-square test, we generally:
  • Determine the expected frequencies, based on the null hypothesis.
  • Calculate the chi-square statistic using the 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 for each category.
  • Compare the calculated \( \chi^2 \) value to a critical value from the chi-square distribution table, considering the chosen significance level and degrees of freedom.
Null Hypothesis
In statistical testing, the null hypothesis represents a general statement or default position that there is no relationship between two measured phenomena. For the goodness-of-fit test that we're discussing, the null hypothesis \( H_0 \) is that the observed frequencies fit the expected frequencies. This means we assume, initially, that the differences between what we observe in our categories and what we expect to see are due purely to random chance.
When performing hypothesis testing, the null hypothesis is what we test against. We determine if the data provide enough evidence to reject it. If not, we then conclude that there isn’t sufficient evidence to say the observed and expected frequencies differ significantly.
Expected Frequencies
Expected frequencies are the theoretical frequencies we anticipate in each category if the null hypothesis is true. In this exercise, with four equally likely categories, the expected frequency for each category is calculated by dividing the total number of observations by the number of categories.

For example, with 100 observations and four categories, the expected frequency in each category would be \( \frac{100}{4} = 25 \).
When conducting a chi-square test, these expected frequencies provide a benchmark to measure how the actual, observed frequencies deviate. This comparison is crucial to determining whether any differences are statistically significant or could be chalked up to random variation.
P-Value
The p-value in hypothesis testing measures the probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is true. It's a key metric to deciding whether we reject or fail to reject the null hypothesis in favor of an alternative.
In this exercise's goodness-of-fit test, the p-value is derived from the chi-square statistic and the degrees of freedom (which are the number of categories minus one). A larger p-value suggests that the differences between the observed and expected frequencies are probably due to random chance. In contrast, a small p-value indicates that such differences are unlikely to have occurred by random chance, and we might consider rejecting the null hypothesis in favor of some influence other than randomness.
  • A p-value below a predetermined significance level (e.g., 0.05) typically suggests rejecting the null hypothesis.
  • If the p-value is much greater than the significance level, it means the data doesn't show significant deviation from the null hypothesis.
In this context, a high p-value (close to 0.88) implies the null hypothesis is a plausible model for the data, suggesting no significant deviation from expected frequencies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To make a goodness-of-fit test, what should be the minimum expected frequency for each category? What are the alternatives if this condition is not satisfied?

Clasp your hands together. Which thumb is on top? Believe it or not, the thumb that you place on top is determined by genetics. If either of your parents has the gene that tells you to place your left thumb on top and passes it on to you, you will place your left thumb on top. The left-thumb gene is called the dominant gene, which means that if either parent passes it on to you, you will have that trait. If you place your right thumb on top, you received the recessive gene from both parents. If both parents have both the left and right thumb genes (the case denoted Lr), Mendelian genetics gives the probabilities listed in the following table about the children's genes. $$ \begin{array}{l|ccc} \hline & \multicolumn{3}{|c} {\text { Lr (Left-Thumbed, but Also Received }} \\\ \text { Child's genes } & \text { LL (Left-Thumbed) } & \text { a Right- Thumbed Gene) } & \text { rr (Right-Thumbed) } \\ \hline \text { Probability } & .25 & .50 & .25 \\ \hline \end{array} $$ Suppose that a random sample of 65 children whose both parents had Lr genes were tested for the genes. The following table lists the results of this experiment.$$ \begin{array}{l|rrr} \hline \text { Child's genes } & \text { LL } & \text { Lr } & \text { rr } \\\ \hline \text { Frequency } & 14 & 31 & 20 \\ \hline \end{array} $$ Test at a \(5 \%\) significance level whether the genes received by the sample of children are significantly different from what Mendelian genetics predicts.

A drug company is interested in investigating whether the color of their packaging has any impact on sales. To test this, they used five different colors (blue, green, orange, red, and yellow) for the boxes of an over-the- counter pain reliever, instead of their traditional white box. The following table shows the number of boxes of each color sold during the first month. $$ \begin{array}{l|ccccc} \hline \text { Box color } & \text { Blue } & \text { Green } & \text { Orange } & \text { Red } & \text { Yellow } \\ \hline \text { Number of boxes sold } & 310 & 292 & 280 & 216 & 296 \\ \hline \end{array} $$ Using a \(1 \%\) significance level, test the null hypothesis that the number of boxes sold of each of these five colors. is the same.

What is a goodness-of-fit test and when is it applied? Explain.

A sample of seven passengers boarding a domestic flight produced the following data on weights (in pounds) of their carry-on bags. \(\begin{array}{lllllll}46.3 & 41.5 & 39.7 & 31.0 & 40.6 & 35.8 & 43.2\end{array}\) a. Using the formula from Chapter 3, find the sample variance, \(s^{2}\), for these data. b. Make the \(98 \%\) confidence intervals for the population variance and standard deviation. Assume that the population from which this sample is selected is normally distributed. c. Test at a \(5 \%\) significance level whether the population variance is larger than 20 square pounds.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.