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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.

Short Answer

Expert verified
The correct answer is (d).

Step by step solution

01

Understand the Problem Context

We have four types of food: Asian, Mexican, pizza, and hamburgers. Each type of food is chosen by a different number of students in the sample of 100: Asian (18), Mexican (22), Pizza (39), Hamburgers (21). The goal is to test if these food choices are equally popular among students.
02

Define the Null Hypothesis

The null hypothesis in a test of proportions asserts that there is no difference in the preferences. Hence, the null hypothesis should state that each type of food is equally preferred, or each type has the same proportion of preference in the entire population. This implies that if each type is equally popular, they would each have a proportion of 0.25.
03

Identify the Correct Null Hypothesis Expression

Option (d) states the null hypothesis correctly as it expresses that the proportions of the entire student population who prefer each type of food are equal: \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25\). This indicates that all types have the same proportion (0.25) assuming equal preference.

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Key Concepts

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

Null Hypothesis
In statistics, the null hypothesis is a baseline statement that doesn’t assume any effect or difference in a particular scenario. It is a critical aspect of hypothesis testing. In the context of the school cafeteria, the null hypothesis suggests that the students' preferences for different food types are the same.
This means that for each category—Asian, Mexican, pizza, and hamburgers—the proportion of students favoring each type of food is equal.
In our case, the null hypothesis would state that each type of food is equally popular among students, indicated by a proportion of 0.25 for each category in this sample. Determining whether this hypothesis holds true is what the chi-square test seeks to find out.
This forms the foundation for testing whether the observed differences in student preferences are statistically significant or just due to random variation.
Test of Proportions
A test of proportions is used to determine if there are statistically significant differences in the distribution of categorical data. It allows us to compare the observed outcomes in a sample with what we would expect if the null hypothesis were true.
For instance, in the example of the cafeteria manager, we have a sample of 100 students. Each food preference (Asian, Mexican, pizza, hamburgers) acts as a category with an expected proportion of 0.25 under the null hypothesis.
The Chi-Square Goodness of Fit Test is an appropriate method for challenging the null hypothesis here. It measures how well the sample data fits with the expected outcome.
The idea is simple: if the actual preferences deviate significantly from what was expected (0.25 for each food type), then we might reject the null hypothesis. This means the differences in preferences were significant, not just random.
Student Preferences
Learning about student preferences helps the cafeteria manager make informed decisions about food ordering and menu planning. The goal is to understand which types of food are popular among students and to cater to these preferences.
In our scenario, 18 students preferred Asian food, 22 preferred Mexican food, 39 liked pizza, and 21 chose hamburgers. These numbers reflect how students ranked their food choices, presenting a snapshot of their preferences.
Identifying these preferences accurately allows the cafeteria manager to allocate resources efficiently and reduce waste, ensuring that the correct types of food are available according to student demand. By analyzing these preferences, the school can better meet the nutritional and taste needs of its students, contributing to a satisfying cafeteria experience.
Random Sampling
Random sampling is a fundamental concept in statistics, used to ensure that the results from a study can be generalized to a larger population. By randomly selecting 100 students, the cafeteria manager aims to capture a representative slice of the entire student body’s preferences.
This method eliminates bias and increases the validity of the test, as each student in the school has an equal chance of being selected. As such, the sample becomes a miniature version of the whole school, allowing us to infer how all the students might respond.
It's crucial for drawing reliable conclusions from statistical analyses. In this case, random sampling supports the manager in determining which food types generally appeal to the entire student population, based on an unbiased sample.

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Most popular questions from this chapter

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: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ The \(P\) -value for a chi-square test for goodness of fit is \(0.0129 .\) Which of the following is the most appropriate conclusion? (a) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is convincing evidence that the food choices are equally popular. (b) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is not convincing evidence that the food choices are equally popular. (c) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is convincing evidence that the food choices are not equally popular. (d) Because 0.0129 is less than \(\alpha=0.05,\) fail to reject \(H_{0}\). There is not convincing evidence that the food choices are equally popular. (e) Because 0.0129 is less than \(\alpha=0.05\), fail to reject \(H_{0}\). There is convincing evidence that the food choices are equally popular.

Where do young adults live? A survey by the National Institutes of Health asked a random sample of young adults (aged 19 to 25 years), "Where do you live now? That is, where do you stay most often?" Here is the full two-way table (omitting a few who refused to answer and one who claimed to be homeless):" $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Parents' home } & 923 & 986 \\ \text { Another person's home } & 144 & 132 \\ \text { Own place } & 1294 & 1129 \\ \text { Group quarters } & 127 & 119 \\ \hline \end{array} $$ (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). Minitab output from a chi-square test is shown below. (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

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.

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.

Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ The appropriate null hypothesis for performing a chi-square test is that (a) equal proportions of female and male teenagers are almost certain they will be married in 10 years. (b) there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers' opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population.

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