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Chapter 11: Problem 20

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. (Hint: This sample will be a random sample if it includes approximately \(19 \%\) seniors, \(23 \%\) juniors, \(27 \%\) sophomores, and \(31 \%\) freshmen.)

Short Answer

Expert verified
Based on the Chi-square goodness of fit test, we cannot reject the null hypothesis that this sample is a random sample. The observed distribution of the sample lies within the expected ratios for different classes.

Step by step solution

01

Calculate the Expected Frequencies

First, calculate the expected frequencies for each category. This can be obtained by multiplying the total number of students in the sample by the known distribution percent for each category. That is for seniors: \(200 \times 0.19 = 38\), juniors: \(200 \times 0.23 = 46\), sophomores: \(200 \times 0.27 = 54\) and freshmen: \(200 \times 0.31 = 62\).
02

Calculate the Observed Frequencies

Then, identify the observed frequencies for each category. These are the counts obtained from the sample and given in the problem. Here they are: seniors: 50, juniors: 46, sophomores: 55 and freshmen: 49.
03

Conduct a Chi-Square Goodness of Fit Test

Next step is to conduct a Chi-square goodness of fit test. The Chi-square statistic can be calculated using the formula: \(\chi^2 = \Sigma (O_i - E_i )^2 / E_i\), where \(O_i\) represents the observed frequency and \(E_i\) is the expected frequency. The calculated Chi-square value will then be compared to the Chi-square critical value to determine whether the difference between observations is statistically significant.
04

Calculate Chi-Square Value

Calculate the Chi-square value. For seniors: \((50 - 38)^2 / 38 = 3.79\), juniors: \((46 - 46)^2 / 46 = 0\), sophomores: \((55 - 54)^2 / 54 = 0.02\) and freshmen: \((49 - 62)^2 / 62 = 2.71\). Summing up all values yields a Chi-square value of \(6.52\)
05

Compare the Chi-Square Value with the Critical Value

The last step is to compare the calculated Chi-square value with the critical value for a \(2.5 \%\) significance level (which is 11.07 with 3 degrees of freedom). As our calculated value of 6.52 is less than the critical value, we do not have the evidence to reject the null hypothesis.

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Key Concepts

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

Statistical Significance
When we talk about statistical significance, we are wondering if a certain result or observation is due to actual effects rather than just chance. This concept is key to deciding if our observed data greatly deviates from what we expect. In the exercise, the significance level is set at 2.5%. This means there is a 2.5% risk of saying there's a significant effect, when actually there isn't.
In simpler words, it's a safety measure to prevent mistakes in deciding whether a pattern or change exists in our data.
  • If our Chi-square calculated value is larger than the critical value, it indicates that our observation is more than just chance and stands as statistically significant.
  • Conversely, if it's smaller (as in our exercise), it suggests that the observed frequencies might just be random and no significant change exists.
Understanding statistical significance guides us with confidence on whether to trust our data-driven decisions.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It's our starting assumption that states there is no effect or no difference present.
In our example, the null hypothesis claims that the observed student distribution is a true, random sample matching the university's population distribution. This 'null' position assumes that any deviation from the expected is purely by chance, not because of some specific reason.
We test this by comparing our Chi-square value to the critical value:
  • If the Chi-square value exceeds the critical value, we may have grounds to reject the null hypothesis, suggesting there's a significant difference in our sample compared to the population.
  • Otherwise, as in our scenario, the calculated Chi-square (6.52) is smaller than the critical (11.07), leading us to continue accepting the null hypothesis.
This means that based on our sample data, there's no strong evidence that suggests our students' sample isn't reflective of the population.
Expected Frequencies
Expected frequencies are the counts we anticipate based on a known distribution or theory. They form the benchmark against which we compare our observed results.
In the exercise, expected frequencies are calculated using percentages of seniors, juniors, sophomores, and freshmen in the known student population.
  • Seniors: Expected frequency is calculated as 19% of 200, resulting in 38.
  • Juniors: Expected frequency becomes 23% of 200, which is 46.
  • Sophomores: From 27% of 200, we find an expected frequency of 54.
  • Freshmen: 31% of 200 leads to an expected frequency of 62.
These frequencies embody the null hypothesis and give us a frame of reference to detect any unusual deviations.
Observed Frequencies
Observed frequencies are the actual data points we gather from a study or sample collection. In this context, these frequencies represent the number of seniors, juniors, sophomores, and freshmen found in our student sample. Here's a breakdown from the problem:
  • Seniors: 50 students.
  • Juniors: 46 students.
  • Sophomores: 55 students.
  • Freshmen: 49 students.
These values are directly gathered from the survey conducted by the student senate.
In a Chi-square Goodness of Fit Test, these observed frequencies are compared to expected frequencies. By measuring how these observed values deviate from what we expected, we determine if discrepancies exist between our sample and the underlying population.

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Most popular questions from this chapter

The makers of Flippin' Out Pancake Mix claim that one cup of their mix contains 11 grams of sugar. However, the mix is not uniform, so the amount of sugar varies from cup to cup. One cup of mix was taken from each of 24 randomly selected boxes. The sample variance of the sugar measurements from these 24 cups was \(1.47\) grams. Assume that the distribution of sugar content is approximately normal. a. Construct the \(98 \%\) confidence intervals for the population variance and standard deviation. b. Test at a \(1 \%\) significance level whether the variance of the sugar content per cup is greater than 1.0 gram.

Find the value of \(\chi^{2}\) for 4 degrees of freedom and A. \(.005\) area in the right tail of the chi-square distribution curve b. \(.05\) area in the left tail of the chi-square distribution curve

National Electronics Company buys parts from two subsidiaries. The quality control department at this company wanted to check if the distribution of good and defective parts is the same for the supplies of parts received from both subsidiaries. The quality control inspector selected a sample of 300 parts received from Subsidiary A and a sample of 400 parts received from Subsidiary \(\mathrm{B}\). These parts were checked for being good or defective. The following table records the results of this investigation. $$ \begin{array}{lcc} \hline & \text { Subsidiary A } & \text { Subsidiary B } \\ \hline \text { Good } & 284 & 381 \\ \text { Defective } & 16 & 19 \\ \hline \end{array} $$ Using a \(5 \%\) significance level, test the null hypothesis that the distributions of good and defective parts are the same for both subsidiaries.

Construct the \(98 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. $$ \text { a. } n=21, s^{2}=9.2 \quad \text { b. } n=17, s^{2}=1.7 $$

You are performing a goodness-of-fit test with four categories, all of which are supposed to be equally likely. You have a total of 100 observations. The observed frequencies are \(21,26,31\), and 22, respectively, for the four categories. a. Show that you would fail to reject the null hypothesis for these data for any reasonable significance level. b. The sum of the absolute differences (between the expected and the observed frequencies) for these data is 14 (i.e., \(4+1+6+3=14\) ). Is it possible to have different observed frequencies keeping the sum at 14 so that you get a \(p\) -value of \(.10\) or less?
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