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Chapter 14: Problem 15

Previous enrollment records at a large university indicate that of the total number of persons who apply for admission, \(60 \%\) are admitted unconditionally, \(5 \%\) are admitted on a trial basis, and the remainder are refused admission. Of 500 applications to date for the coming year, 329 applicants have been admitted unconditionally, 43 have been admitted on a trial basis, and the remainder have been refused admission. Do these data indicate a departure from previous admission rates? Test using \(\alpha=.05 .\)

Short Answer

Expert verified
Answer: Yes, there is a significant difference between the current admission data and the previous admission rates at a significance level of 0.05.

Step by step solution

01

Identify the Null and Alternative Hypotheses

Let's set up our null and alternative hypothesis: \(H_0:\) There is no significant difference between the current admission data and the previous admission rates. \(H_a:\) There is a significant difference between the current admission data and the previous admission rates.
02

Calculate the Expected Frequencies

Using the previous enrollment records, we will calculate the expected frequencies of students in each admission category. For unconditionally admitted, Expected frequency = Total number of applications * Previous percentage of unconditional admissions = \(500 * (60\%) = 500 * 0.6 = 300\) For trial basis admitted, Expected frequency = Total number of applications * Previous percentage of trial basis admissions = \(500 * (5\%) = 500 * 0.05 = 25\) For refused admission, Expected frequency is calculated from the remaining applications, = Total number of applications - unconditionally admitted - trial basis admitted = \(500 - 300 - 25 = 175\) The expected frequencies are (300, 25, 175).
03

Calculate the Chi-square Test Statistic

We will perform a Chi-square test to compare the observed and expected frequencies. The Chi-square test statistic is calculated as follows: \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\) where \(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies. Unconditionally admitted: \(\frac{(329 - 300)^2}{300} = 2.87\) Trial basis admitted: \(\frac{(43 - 25)^2}{25} = 12.88\) Refused admission: \(\frac{(128 - 175)^2}{175} = 12.62\) Summing these, we get the Chi-square test statistic: \(\chi^2 = 2.87 + 12.88 + 12.62 = 28.37\)
04

Determine the Critical Value and Degrees of Freedom

To determine the critical value, we first need to find the degrees of freedom (df). For this problem, the degrees of freedom is calculated as follows: df = Number of categories - 1 = 3 - 1 = 2 With df=2 and \(\alpha=0.05\), consulting a Chi-square distribution table, we get the critical value as \(5.991\).
05

Comparing the Test Statistic and Critical Value

Now, we compare the calculated Chi-square test statistic (\(\chi^2 = 28.37\)) with the critical value (\(5.991\)). Since \(28.37 > 5.991\), we reject the null hypothesis.
06

Conclusion

Rejecting the null hypothesis indicates that there is a significant difference between the current admission data and the previous admission rates. The current data departs from the previous admission rates at a significance level of \(\alpha = 0.05\).

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis ( H_0 ) in statistical testing signifies the assumption that there is no significant difference or effect in a situation.
In our context, it implies that the current admission data aligns with historical admission rates.
Essentially, the null hypothesis serves as the default claim we hold to be true until evidence suggests otherwise. It's like saying "nothing has changed" unless proven wrong by statistical evidence.
This forms the baseline for comparing observed data.
Exploring the Alternative Hypothesis
The alternative hypothesis ( H_a ) stands in opposition to the null hypothesis.
It suggests that there is a noticeable difference between the current admission records and past data.
This hypothesis is what researchers aim to prove through their analysis. It asserts that the existing patterns have shifted, and admits that new trends may be present.
Testing the alternative hypothesis involves seeking evidence to support these new observations.
The Role of Degrees of Freedom
Degrees of freedom (df) reflect the number of independent values or quantities that can vary in an analysis.
For a Chi-square test, it is calculated as the number of categories minus one.
In our problem, with three categories (unconditional admission, trial basis, and refusal), the degrees of freedom are 3 - 1 = 2. Calculating degrees of freedom is crucial for looking up critical values in statistical distribution tables.
It helps in assessing whether our test statistic is significant.
Significance Level in Hypothesis Testing
The significance level (\(\alpha\)) is the threshold at which we decide whether to accept or reject the null hypothesis.
It's usually set at 0.05, which implies a 5% risk of concluding that a difference exists when there is none.
This level acts as a boundary for making statistically supported decisions. If the probability of observing the data (given that the null hypothesis is true) is less than \(\alpha\), we reject the null hypothesis.
Thus, a significance level ensures objectivity in reporting findings.

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Most popular questions from this chapter

Use Table 5 in Appendix I to find the value of \(\chi^{2}\) with the following area \(\alpha\) to its right: a. \(\alpha=.05, d f=3\) b. \(\alpha=.01, d f=8\)

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Suppose that a consumer survey summarizes the responses of \(n=307\) people in a contingency table that contains three rows and five columns. How many degrees of freedom are associated with the chi-square test statistic?

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