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a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are met. Statistics Class Times A professor wishes to see if students show a time preference for statistics classes. A sample of four statistics classes shows the enrollment. At \(\alpha=0.01,\) do the students show a time preference for the classes? $$ \begin{array}{l|cccc} \text { Time } & 8: 00 \mathrm{AM} & 10: 00 \mathrm{AM} & 12: 00 \mathrm{PM} & 2: 00 \mathrm{PM} \\ \hline \text { Students } & 24 & 35 & 31 & 26 \end{array} $$

Short Answer

Expert verified
We fail to reject the null hypothesis; no time preference is evident at \(\alpha=0.01\).

Step by step solution

01

State the Hypotheses and Identify the Claim

For this exercise, we are testing whether students have a preference for certain class times. Therefore, our null hypothesis (\(H_0\)) will be that there is no preference for any particular class time, meaning the class enrollments are equal. The alternative hypothesis (\(H_a\)) is that there is a preference for at least one class time. Specifically, \(H_0\): "The distributions of enrollments are equal" and \(H_a\): "There is a preference for different class times." The claim is represented by the alternative hypothesis \(H_a\).
02

Find the Critical Value

To find the critical value, we will use a chi-square distribution, as we are dealing with categorical data. The degrees of freedom (df) are calculated as \(df = k - 1\), where \(k\) is the number of classes (4 in this case). Thus, \(df = 4 - 1 = 3\). At a significance level \(\alpha = 0.01\), the critical value (\(\chi^2_{crit}\)) from the chi-square table with 3 degrees of freedom is approximately 11.345.
03

Compute the Test Value

First, calculate the expected frequency for each class time if there were no preference. Total students are \(24 + 35 + 31 + 26 = 116\), so the expected frequency for each class is \(116/4 = 29\). Then, calculate the chi-square test 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.\[ \chi^2 = \frac{(24-29)^2}{29} + \frac{(35-29)^2}{29} + \frac{(31-29)^2}{29} + \frac{(26-29)^2}{29} = \frac{25}{29} + \frac{36}{29} + \frac{4}{29} + \frac{9}{29} = 3.45\]
04

Make the Decision

Compare the computed chi-square test statistic (3.45) with the critical value (11.345). Since 3.45 is less than 11.345, we fail to reject the null hypothesis. This implies that the data does not provide sufficient evidence to conclude the students have a time preference for classes.
05

Summarize the Results

At the 0.01 level of significance, there is not enough evidence to support the claim that students show a time preference for statistics classes. The observed enrollment differences do not suggest significant time preferences among students.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a statement regarding a population parameter that is typically based on the idea of "no effect" or "no difference." In the context of hypothesis testing, it acts as a starting point for statistical investigation.

For the exercise where a professor wants to understand if students show a time preference for statistics classes, the null hypothesis states there's no preference in class times. This means each time slot should have equal enrollment. Mathematically, this is expressed as "The distributions of enrollments are equal."

When conducting a hypothesis test, we assume the null hypothesis is true until we have sufficient evidence to prove otherwise. Therefore, the null hypothesis is crucial for defining the framework of any hypothesis testing process.
Alternative Hypothesis
The alternative hypothesis, represented as \(H_a\), proposes an alternative to the null hypothesis. It suggests what we suspect might be true or would like to prove.

In our exercise, the alternative hypothesis expresses that students may prefer one time over others. This implies that at least one enrollment figure differs significantly from the expected, indicating a preference for certain class times. The phrasing might be: "There is a preference for different class times."

It's essential to clearly define this hypothesis, as it relates to the claim we're testing. Acceptance or rejection of the null hypothesis ultimately supports or refutes the alternative hypothesis.
Chi-Square Distribution
The chi-square distribution is used in hypothesis testing situations involving categorical data. It's particularly useful when you want to determine whether observed outcomes differ from expected outcomes under the null hypothesis.

For this exercise, the chi-square test helps compare the observed enrollments in each class time with what would be expected if there was no preference. The test statistic follows a chi-square distribution, which helps conclude whether the differences between observed and expected enrollments are significant.

To compute the test statistic, use the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where \(O_i\) represents observed frequencies and \(E_i\) are expected frequencies. This calculation quantifies how far the observation diverges from expectation, which can then be compared to the critical value.
Critical Value
A critical value is a point on the test statistic's distribution that is compared with the computed test statistic to decide whether to reject the null hypothesis.

In the chi-square test for class time preference, the critical value is determined by the chi-square distribution with the appropriate degrees of freedom and significance level, \(\alpha\). For example, with 3 degrees of freedom and a significance level of 0.01, the chi-square critical value is approximately 11.345.

If your calculated chi-square statistic exceeds this critical value, you have enough evidence to reject the null hypothesis in favor of the alternative. If not, you do not reject the null hypothesis. Thus, the critical value serves as a decisive threshold in hypothesis testing.

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Most popular questions from this chapter

Perform the following steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are valid. This table lists the numbers of officers and enlisted personnel for women in the military. At \(\alpha=0.05\), is there sufficient evidence to conclude that a relationship exists between rank and branch of the Armed Forces? $$ \begin{array}{lrr} & \text { Officers } & \text { Enlisted } \\ \hline \text { Army } & 10,791 & 62,491 \\ \text { Navy } & 7,816 & 42,750 \\ \text { Marine Corps } & 932 & 9,525 \\ \text { Air Force } & 11,819 & 54,344 \end{array} $$

Perform the following steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are valid. The national Cesarean delivery rate for a recent year was \(32.2 \%\) (number of live births performed by Cesarean section). A random sample of 100 birth records from three large hospitals showed the following results for type of birth. Test for homogeneity of proportions using \(\alpha=0.10 .\) $$ \begin{array}{lccc} & \text { Hospital } & \text { Hospital } & \text { Hospital } \\ & \text { A } & \text { B } & \text { C } \\ \hline \text { Cesarean } & 44 & 28 & 39 \\ \text { Non-Cesarean } & 56 & 72 & 61 \end{array} $$

a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are met. A researcher wishes to see if the number of randomly selected adults who do not have health insurance is equally distributed among three categories (less than 12 years of education, 12 years of education, more than 12 years of education). A sample of 60 adults who do not have health insurance is selected, and the results are shown. At \(\alpha=0.05,\) can it be concluded that the frequencies are not equal? Use the \(P\) -value method. If the null hypothesis is rejected, give a possible reason for this. $$ \begin{array}{l|ccc} \text { Category } & \text { Less than } & &\text { More than } \\ & 12 \text { years } & 12 \text { years } & 12 \text { years } \\ \hline \text { Frequency } & 29 & 20 & 11 \end{array} $$

Perform the following steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are valid. The table shows the number of students (in thousands) participating in various programs at both two-year and four-year institutions. At \(\alpha=0.05,\) can it be concluded that there is a relationship between program of study and type of institution? $$ \begin{array}{lrr} & \text { Two-year } & \text { Four-year } \\ \hline \text { Agriculture and related sciences } & 36 & 52 \\ \text { Criminal justice } & 210 & 231 \\ \text { Foreign languages and literature } & 28 & 59 \\ \text { Mathematics and statistics } & 28 & 63 \end{array} $$

Perform the following steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are valid. According to a recent survey, \(59 \%\) of Americans aged 8 to 17 years would prefer that their mothers work outside the home, regardless of what they do now. A school district psychologist decided to select three random samples of 60 students each in elementary, middle, and high school to see how the students in her district felt about the issue. At \(\alpha=0.10\), test the claim that the proportions of the students who prefer that their mothers have jobs outside the home are equal. $$ \begin{array}{cccc} & \text { Elementary } & \text { Middle } & \text { High } \\ \hline \text { Prefer mothers work } & 29 & 38 & 51 \\ \text { Prefer mothers not work } & \frac{31}{60} & \frac{22}{60} & \frac{9}{60} \end{array} $$

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