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Chapter 11: Problem 9

No chi-square A school's principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night: $$ \begin{array}{lccccccc} \hline \text { Night: } & \text { Sunday } & \text { Monday } & \text { Tuesday } & \text { Wednesday } & \text { Thursday } & \text { Friday } & \text { Saturday } \\ \text { Average } & 130 & 108 & 115 & 104 & 99 & 37 & 62 \\ \text { time: } & & & & & & & \\ \hline \end{array} $$ Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

Short Answer

Expert verified
The chi-square test is inappropriate because it is designed for categorical data, not for comparing means of numerical data.

Step by step solution

01

Understanding Chi-square Test Requirements

The chi-square test for goodness of fit is used to compare observed frequencies with expected frequencies in categorical data. Each category should have a theoretical expected frequency. We consider these expected frequencies are distributed according to a known distribution, such as a uniform or binomial distribution.
02

Data Format Analysis

In the given exercise, each night has a different numerical value representing the average time spent on homework in minutes, which is quantitative data, not categorical. A chi-square test is not suited for continuous numerical data where means or averages are given.
03

Appropriate Test Identification

When dealing with means across different groups (such as day of the week), an ANOVA test or repeated measures analysis would be more appropriate for determining if there is a significant difference in the average time spent on homework each night, rather than a chi-square test.

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Key Concepts

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

ANOVA Test
The ANOVA (Analysis of Variance) test is a statistical procedure used to determine if there are any statistically significant differences between the means of three or more independent groups. In the context of our exercise, each night of the week can be considered a different group. The principal wants to know if the average homework time is about the same every night.

ANOVA helps us to test this by comparing the variance among the averages. Rather than handling one-to-one comparisons, which would be cumbersome with multiple groups, ANOVA provides a holistic analysis. If the ANOVA test finds a significant difference, it suggests at least one night differs in average homework time from the others.
  • One-way ANOVA: Tests differences among groups based on one independent variable.
  • Repeated measures ANOVA: Useful when the same subjects are used for each treatment, such as students tracked over different nights.
Understanding whether there's a significant difference can aid in identifying specific nights students might be spending more鈥攐r less鈥攖ime on homework. This can help in aligning school policies or homework loads more effectively.
Quantitative Data Analysis
Quantitative data analysis is a process focused on analyzing numerical data and uncovering patterns or insights. In our scenario, the principal needs to analyze average homework times, which is a classic example of quantitative data.

The data collected is numerical, representing the average minutes of homework time per night. To analyze this effectively, statistical tools such as ANOVA can be employed to identify variations across the nights. This kind of analysis gives meaning to the numbers, helping the principal make informed decisions.
  • Descriptive statistics: Summarize data using means, medians, and modes.
  • Inferential statistics: Make predictions or inferences about a population based on a sample, such as using the ANOVA test.
Quantitative analysis is essential as it turns raw data into actionable insights, helping educators to plan better and improve student workloads.
Categorical Data
Categorical data is qualitative, meaning it encompasses variables that describe categories or labels. This type of data divides information into groups that are distinct yet unordered. Examples include color, race, gender, or, in a classroom context, types of homework subjects.

In the given problem, each day of the week could initially seem like categorical data, but the focus is on numerical values of homework time per day, making it quantitative. Categorical data usually requires different statistical tests, like the chi-square test, which is not applicable when dealing with averages or numerical data.
  • Binary: Two categories, e.g., yes/no questions.
  • Nominal: Multiple categories without intrinsic ordering.
  • Ordinal: Categories with a logical order but not equidistant.
Understanding the difference between categorical and quantitative data helps in choosing the right method to analyze data effectively.
Goodness of Fit
Goodness of fit is a measure used in statistics to test how well a sample set matches the expected values according to a specified distribution. This concept is most often utilized with the chi-square test. The test observes how well the observed data adheres to the expected distribution across categories.

In the context of the exercise, if the principal had collected data that fit into categories (e.g., whether students did their homework or not per night), then a chi-square test might be relevant. However, the data is quantitative, focusing on time averages, making goodness of fit鈥攁nd therefore the chi-square test鈥攊napplicable.
  • The chi-square test for goodness of fit requires categorical data with expected frequencies.
  • It's unsuitable for averages or numerical data, hence not applicable for the principal's data collection method.
Understanding when to use goodness of fit is crucial in statistical testing, ensuring the right tests apply to the right type of data, thus providing accurate results.

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Most popular questions from this chapter

Is your random number generator working? Use your calculator's RandInt function to generate 200 digits from 0 to 9 and store them in a list. (a) State appropriate hypotheses for a chi-square test for goodness of fit to determine whether your calculator's random number generator gives each digit an equal chance to be generated. (b) Carry out a test at the \(\alpha=0.05\) significance level. For parts (c) and (d), assume that the students' random number generators are all working properly. (c) What is the probability that a student who does this exercise will make a Type I error? (d) Suppose that 25 students in an AP Statistics class independently do this exercise for homework. Find the probability that at least one of them makes a Type I error.

Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Exercises 23 through 25 refer to the following setting. Do students who read more books for pleasure tend to earn higher grades in English? The boxplots below show data from a simple random sample of 79 students at a large high school. Students were classified as light readers if they read fewer than 3 books for pleasure per year. Otherwise, they were classified as heavy readers. Each student's average English grade for the previous two marking periods was converted to a GPA scale where \(A+=4.3\), \(A=4.0, A-=3.7, B+=3.3,\) and so on. Reading and grades (1.3) Write a few sentences comparing the distributions of English grades for light and heavy readers.

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?
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