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The student government of a large college polled a random sample of 325 male students and found that 221 were in favor of a new grading system. At the same time, 120 out of a random sample of 200 female students were in favor of the new system. Do the results indicate a significant difference in the proportion of male and female students who favor the new system? Test at the \(.05\) level of significance.

Short Answer

Expert verified
After conducting a two-proportion Z-test with a level of significance of \(.05\), we compared the test statistic to the critical value to determine whether to reject or fail to reject the null hypothesis. If we rejected the null hypothesis, this would indicate that there is a significant difference between the proportions of male and female students who favor the new grading system. If we failed to reject the null hypothesis, we cannot conclude that there is a significant difference between the proportions.

Step by step solution

01

State the hypotheses

The null hypothesis (H0) is that there is no significant difference in the proportion of male and female students who favor the new grading system, meaning that the difference between the proportions is zero. The alternative hypothesis (H1) is that there is a significant difference between the proportions, meaning that the difference between the proportions is not equal to zero. H0: \(p_m - p_f = 0\) (No significant difference) H1: \(p_m - p_f ≠ 0\) (Significant difference)
02

Determine the sample proportions

For this step, we will calculate the sample proportions of male and female students who favor the new grading system by dividing the number of students in favor by the total number of students in each sample. Sample proportion of male students: \(\hat{p}_m = \frac{221}{325}\) Sample proportion of female students: \(\hat{p}_f = \frac{120}{200}\) Now, let's determine the combined sample proportion (\(\hat{p}\)). \(\hat{p} = \frac{221+120}{325+200}\)
03

Calculate the test statistic

We will use the Z-test statistic for the difference between two proportions. The formula for the test statistic is: \( Z = \frac{(\hat{p}_m - \hat{p}_f) - (p_m - p_f)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_m}+\frac{1}{n_f})}} \), where \(n_m\) and \(n_f\) are the sample sizes for male and female students, respectively. Using the calculated sample proportions, sample sizes, and the assumption under H0 that \(p_m - p_f = 0\), calculate the Z-test statistic.
04

Determine the critical value and rejection region

Given the level of significance (α) is \(.05\), and since this is a two-tailed test, we will use the Z-distribution table to find the critical value and determine the rejection region. For a \(.05\) level of significance in a two-tailed test, the critical value is \(Z = ±1.96\). Therefore, the rejection region is when \(Z < -1.96\) or \(Z > 1.96\).
05

Compare the test statistic to the critical value and make a decision

Now that we have the test statistic and the rejection region, we can compare the two values and decide whether to reject or fail to reject the null hypothesis. If the calculated test statistic falls within the rejection region, we reject the null hypothesis and conclude that there is a significant difference between the proportions of male and female students who favor the new grading system. Otherwise, we fail to reject the null hypothesis and cannot conclude that there is a significant difference.
06

State the conclusion

After comparing the test statistic to the critical value, state the conclusion of the hypothesis test. If we rejected the null hypothesis, it means there is a significant difference between the proportions of male and female students who favor the new grading system. If we failed to reject the null hypothesis, it means we don't have enough evidence to claim a significant difference.

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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, symbolized as H0, is a fundamental concept in statistics that proposes no effect or no difference is present regarding the question under investigation. In the exercise, the null hypothesis is that there is no significant difference in the proportions of male (\( p_m \)) and female (\( p_f \)) students who favor the new grading system. Mathematically, it is expressed as H0: \( p_m - p_f = 0 \). The null hypothesis serves as a starting point for hypothesis testing and is assumed to be true until evidence suggests otherwise.

The null hypothesis is beneficial as it gives a specific statement that can be tested through statistical analysis. When facing such problems, one crucial improvement is to ensure clarity of what the null hypothesis implies so that it can be contrasted effectively with the alternative hypothesis.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis (H1 or Ha) is the hypothesis that there is an effect or a difference. It symbolizes what researchers aim to support. For our example, the alternative hypothesis posits that there is a significant difference between the proportion of male and female students who are in favor of the new grading system, expressed as H1: \( p_m - p_f eq 0 \).

The alternative hypothesis is designed to be mutually exclusive to the null hypothesis; you either support one or the other based on the statistical evidence. After calculating the sample proportions and test statistics, the hypothesis test will aid in determining whether to reject H0 in favor of H1.
Z-test for Proportions
When comparing proportions from two independent samples, as in our textbook scenario, the Z-test is often used. This statistical test determines whether there is a significant difference between sample proportions by converting the difference into a Z-score. This score reflects how many standard deviations away from the expected value under the null hypothesis the observed difference lies.

To compute the test statistic, we use the formula: \( Z = \frac{(\hat{p}_m - \hat{p}_f) - (p_m - p_f)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_m}+\frac{1}{n_f})}} \), which takes into account the sample proportions, the sample sizes, and the combined sample proportion. In most cases, it is assumed under H0 that \( p_m - p_f = 0 \), simplifying the calculation.
Level of Significance
The level of significance, denoted as \( \alpha \), is a threshold that determines the likelihood of rejecting the null hypothesis when it is actually true (a type I error). In the provided exercise, the level of significance used is \( \alpha = 0.05 \), indicating a 5% risk of concluding a difference exists when there is none.

Choosing an appropriate \( \alpha \) is crucial as it influences the decision-making process in hypothesis testing. The smaller the \( \alpha \), the less likely you are to make a type I error. However, setting \( \alpha \) too low may increase the chance of a type II error, failing to reject a false null hypothesis. Therefore, a balance must be struck, with 0.05 being a commonly accepted standard.
Sample Proportions
Sample proportions represent the estimated probability of an event based on a sample. To compute the sample proportion, simply divide the number of 'successful' outcomes by the total sample size. For instance, in our exercise:
  • The sample proportion of male students who favor the new grading system (\( \hat{p}_m \)) is calculated as \( \frac{221}{325} \).
  • The sample proportion of female students who favor the new grading system (\( \hat{p}_f \)) is \( \frac{120}{200} \).
Knowing the sample proportions is vital as they are used to calculate the test statistic in the Z-test. Understanding how to determine these proportions accurately is essential for conducting correct hypothesis tests.
Critical Value
The critical value in hypothesis testing indicates the threshold at which the null hypothesis is rejected. To find the critical value, one relates the level of significance to a probability distribution, often the Z-distribution for Z-tests.

For a two-tailed test at a significance level of \( \alpha = 0.05 \), the critical values are commonly \( \pm1.96 \), representing the Z-scores that mark the rejection region boundaries. Any test statistic beyond these values suggests a statistically significant result. The Z-distribution table or standard normal distribution table helps to find these critical values based on the desired level of significance.

Identifying the Rejection Region

If the calculated Z-test statistic falls outside of the range between \( -1.96 \) and \( 1.96 \), the null hypothesis is rejected, indicating that the observed data are statistically unlikely under the assumption that the null hypothesis is correct.

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Most popular questions from this chapter

A manufacturer of transistors claims that its transistors will last an average of 1000 hours. To maintain this average, 25 transistors are tested each month. If the computed value of \(\mathrm{t}\) lies between \(-t_{.025}\) and \(t_{025}\), the manufacturer is satisfied with his claim. What conclusions should be drawn from a sample that has a mean \(\overline{\mathrm{x}}=1,010\) and a standard deviation \(\mathrm{s}=60\) ? Assume the distribution of the lifetime of the transistors is normal.

Past experience has shown that the scores of students who take a certain mathematics test are normally distributed with mean 75 and variance \(36 .\) The Mathematics Department members would like to know whether this year's group of 16 students is typical. They decide to test the hypothesis that this year's students are typical versus the alternative that they are not typical. When the students take the test the average score is 82 . What conclusion should be drawn?

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