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Chapter 11: Problem 15

Of all students enrolled at a large undergraduate university, \(19 \%\) are seniors, \(23 \%\) are juniors, \(27 \%\) are sophomores, and \(31 \%\) are freshmen. A sample of 200 students taken from this university by the student senate to conduct a survey includes 50 seniors, 46 juniors, 55 sophomores, and 49 freshmen. Using a \(2.5 \%\) significance level, test the null hypothesis that this sample is a random sample.

Short Answer

Expert verified
To determine if the sample is random, calculate the observed frequencies, expected frequencies, chi-square value and compare it with the critical value with degrees of freedom. We then make a decision based on this comparison.

Step by step solution

01

Set Up Hypotheses

The null hypothesis (\(H_0\)) is that this sample is a random sample. The alternative hypothesis is that this sample is not a random sample.
02

Compute Expected Frequencies

The expected frequency for each category can be calculated by multiplying the total number of students in the sample (which is 200) by the proportion of students from the total population for each class: \n\nSeniors: \(200 * 0.19 = 38\) \nJuniors: \(200 * 0.23 = 46\) \nSophomores: \(200 * 0.27 = 54\) \nFreshmen: \(200 * 0.31 = 62\)
03

Compute Test Statistic

The test statistic, 'chi-square' (\(\chi^2\)), is computed by the formula: \n\n\(\chi^2 = \sum{{(O-E)^2/E}}\) \n\nwhere, O = observed frequency, E = expected frequency. Plugging in the values for seniors, juniors, sophomores, and freshmen, we can get the \(\chi^2\) value.
04

Compute Degrees of Freedom

The degrees of freedom for a chi-square test is given by \(df = k - 1\), where \(k\) is the number of categories/classes. Here, \(k = 4\) (Seniors, Juniors, Sophomores, Freshmen), so \(df = 4 - 1 = 3\).
05

Find the Critical Chi-Square

Look up the critical chi-square value in the chi-square distribution table for a given significance level(0.025) and degrees of freedom(3).
06

Decision

Compare the calculated chi-square value to the critical chi-square value. If the calculated chi-square value is less than the critical one, accept the null hypothesis, this indicates that the sample is indeed a random one. Otherwise, reject it.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial tool in statistics used to determine if there is enough evidence in a sample to infer that a certain condition holds true for the entire population. In this exercise, we're looking at whether a sample of students is truly representative or random as compared to the entire student body of the university.

We start by setting up two opposing hypotheses:
  • The **null hypothesis ( H_0 e)** asserts that the sample is a random sample.
  • The **alternative hypothesis ( H_1 e)** proposes that the sample is not a random sample.
Once we establish these, our goal is to assess statistically if we can support or reject the null hypothesis.
Degrees of Freedom
Degrees of freedom (df) is an essential concept to understand in hypothesis testing, particularly in the chi-square test. It refers to the number of values in a calculation that are free to vary.

For a chi-square test, the degrees of freedom are determined by the formula:
\[ df = k - 1 \]
where \( k \) is the number of categories or groups.
  • In this exercise, there are four categories: seniors, juniors, sophomores, and freshmen, so \( k = 4 \).
  • Thus, the degrees of freedom is \( df = 4 - 1 = 3 \).
Understanding degrees of freedom is vital as it influences the shape of the chi-square distribution and helps us find the critical value needed to make informed decisions.
Significance Level
The significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It sets the threshold for determining the strength of our evidence.

Common significance levels include 0.05, 0.01, and in this case, 0.025.
  • In this exercise, a 2.5% significance level is used, meaning we are willing to accept a 2.5% chance of incorrectly rejecting the null hypothesis.
Choosing a significance level is a critical decision and can affect the outcome of the test, balancing the risk of Type I and Type II errors.
Expected Frequency
Expected frequency refers to the calculated number of observations in each category if the null hypothesis is true. It is used to compare against the actual observed frequencies.

To find the expected frequency for each group, you multiply the total number of observations by the proportion expected for each category:
  • Seniors: \( 200 \times 0.19 = 38 \)
  • Juniors: \( 200 \times 0.23 = 46 \)
  • Sophomores: \( 200 \times 0.27 = 54 \)
  • Freshmen: \( 200 \times 0.31 = 62 \)
These expected frequencies form the basis of the chi-square test, allowing us to determine whether the observed sample significantly deviates from expectation.

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Most popular questions from this chapter

A chemical manufacturing company wants to locate a hazardous waste disposal site near a city of 50,000 residents and has offered substantial financial inducements to the city. Two hundred adults ( 110 women and 90 men) who are residents of this city are chosen at random. Sixty percent of these adults oppose the site, \(32 \%\) are in favor, and \(8 \%\) are undecided. Of those who oppose the site, \(65 \%\) are women; of those in favor, \(62.5 \%\) are men. Using a \(5 \%\) level of significance, can you conclude that opinions on the disposal site are dependent on gender?

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?

Construct the \(95 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=10, s^{2}=7.2\) b. \(n=18, s^{2}=14.8\)

A random sample of 100 persons was selected from each of four regions in the United States. These people were asked whether or not they support a certain farm subsidy program. The results of the survey are summarized in the following table. $$ \begin{array}{lccc} \hline & \text { Favor } & \text { Oppose } & \text { Uncertain } \\ \hline \text { Northeast } & 56 & 33 & 11 \\ \text { Midwest } & 73 & 23 & 4 \\ \text { South } & 67 & 28 & 5 \\ \text { West } & 59 & 35 & 6 \\ \hline \end{array} $$ Using a \(1 \%\) significance level, test the null hypothesis that the percentages of people with different opinions are similar for all four regions.

A sample of 30 observations selected from a normally distributed population produced a sample variance of \(5.8\). a. Write the null and alternative hypotheses to test whether the population variance is different from \(6.0 .\) b. Using \(\alpha=.05\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on a chi-square distribution curve. c. Find the value of the test statistic \(\chi^{2}\). d. Using a \(5 \%\) significance level, will you reject the null hypothesis stated in part a?
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