/*! 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 55 Refer to the following setting. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ (a) A Type I error is possible. (b) A Type II error is possible. (c) Both a Type I and a Type II error are possible. (d) There is no chance of making a Type I or Type II error because the \(P\) -value is approximately \(0 .\) (e) There is no chance of making a Type I or Type II error because the calculations are correct. For these data, \(\chi^{2}=69.8\) with a \(P\) -value of approximately \(0 .\) Assuming that the researchers used a significance level of \(0.05,\) which of the following is true?

Short Answer

Expert verified
(a) A Type I error is possible.

Step by step solution

01

Understand the hypotheses

In the context of chi-square tests, the null hypothesis (\(H_0\)) often states there is no association between two categorical variables. Here, it means gender and the belief in marriage chances are independent.
02

Define Type I and Type II errors

A Type I error occurs when we reject a true null hypothesis. Conversely, a Type II error happens when we fail to reject a false null hypothesis.
03

Evaluate the P-value

A P-value is used to test the null hypothesis. If the P-value is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. Here, the P-value is approximately 0, which is less than 0.05.
04

Conclusion based on P-value

Since the P-value is less than 0.05, we reject the null hypothesis. This means there is evidence to suggest a relationship between gender and belief in marriage chances.
05

Determine possibility of errors

Based on the rejection of the null hypothesis, there is a chance of making a Type I error since we might have rejected a true null hypothesis. However, since we did not fail to reject the null hypothesis, a Type II error is not possible.

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.

Type I Error
In statistical hypothesis testing, a Type I error occurs when you reject a true null hypothesis. This is often regarded as a false positive. Imagine that the null hypothesis is true, indicating no association between the categorical variables under investigation—in this case, gender and belief in future marital status. - When performing a chi-square test, if the P-value is lower than the chosen significance level (like 0.05), you reject the null hypothesis.
- However, this rejection might be incorrect, thus resulting in a Type I error, where you conclude incorrectly that there is an association when in fact, there isn't. The consequence of a Type I error in social sciences could lead to believing there's a gender influence when none exists, potentially affecting policies or interventions that might have been based on this conclusion.
Type II Error
A Type II error, also known as a false negative, occurs when you fail to reject a false null hypothesis. In this scenario, you miss detecting an actual effect or association between the categorical variables assessed—in this case, gender and marriage belief. - This means accepting the null hypothesis when gender does indeed influence marital belief, but this influence remains undetected.
- A Type II error can occur for various reasons, such as insufficient sample size or variability in the data that isn't captured by the test. In this exercise, a Type II error is not considered possible because the null hypothesis was actually rejected based on a very low P-value. Thus, gender is suggested to influence beliefs about future marriage, and failing to detect this effect wasn't an issue in this case.
P-value
The P-value in statistical testing is a measure that helps you determine the strength of your evidence against the null hypothesis. Consequently, it helps you decide whether to reject or fail to reject the null hypothesis. - If the P-value is less than or equal to your significance level (commonly 0.05), it indicates strong evidence to reject the null hypothesis.
- A small P-value suggests that the observed data is unlikely under the assumption that the null hypothesis is true. In the given exercise, the P-value was approximately 0, much lower than 0.05. This indicates very strong evidence against the null hypothesis, leading to its rejection, and implying a significant relationship between gender and belief in marriage likelihood.
Categorical Variables
Categorical variables are those that represent distinct categories or groups. They are not numerical but rather qualitative in nature, like gender (male or female) in this context. - Categorical variables can be nominal with no order (like color) or ordinal where a clear order exists (like preference levels).
- The chi-square test, used in this scenario, specifically analyzes the relationship between two categorical variables. In the exercise, the two categorical variables were 'gender' and 'belief in marriage chances,' both non-numerical but critically important in examining if there's an association between them. Understanding these variables helps in analyzing survey or observational study data like that from the Longitudinal Study of Adolescent Health, where associations between different demographic factors are explored.

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

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.

Going Nuts The UR Nuts Company sells Deluxe and Premium nut mixes, both of which contain only cashews, brazil nuts, almonds, and peanuts. The Premium nuts are much more expensive than the Deluxe nuts. A consumer group suspects that the two nut mixes are really the same. To find out, the group took separate random samples of 20 pounds of each nut mix and recorded the weights of each type of nut in the sample. Here are the data: \({ }^{18}\) $$ \begin{array}{lcc} {\text { Type of mix }} \\ \text { Type of nut } & \text { Premium } & \text { Deluxe } \\ \text { Cashew } & 6 \mathrm{lb} & 5 \mathrm{lb} \\ \text { Brazil nut } & 3 \mathrm{lb} & 4 \mathrm{lb} \\ \text { Almond } & 5 \mathrm{lb} & 6 \mathrm{lb} \\ \text { Peanut } & 6 \mathrm{lb} & 5 \mathrm{lb} \end{array} $$ Explain why we can't use a chi-square test to determine whether these two distributions differ significantly.

Seagulls by the seashore Do seagulls show a preference for where they land? To answer this question, biologists conducted a study in an enclosed outdoor space with a piece of shore whose area was made up of \(56 \%\) sand, \(29 \%\) mud, and \(15 \%\) rocks. The biologists chose 200 seagulls at random. Each seagull was released into the outdoor space on its own and observed until it landed somewhere on the piece of shore. In all, 128 seagulls landed on the sand, 61 landed in the mud, and 11 landed on the rocks. Do these data provide convincing evidence that seagulls show a preference for where they land?

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.

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.