Problem 24 - Basic Statistics for Business and Economics with Student

Question

A recent article in USA Today reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At the .02 significance level, can we conclude that a larger proportion of students at your school have jobs?

Step-by-Step Solution

Answer

Yes, there's enough evidence to conclude a larger proportion of students have jobs.

Step 1: Define Hypotheses

First, we set up our hypotheses. The null hypothesis \( H_0 \) is that the proportion of students with jobs at your school is equal to 1/3, which is the same as the general population (\( P = 0.3333 \)). The alternative hypothesis \( H_a \) is that the proportion of students with jobs at your school is greater than 1/3 (\( P > 0.3333 \)).

Step 2: Collect Sample Data

From the survey, we know that 80 students out of 200 have jobs. This gives a sample proportion \( \hat{p} = \frac{80}{200} = 0.4 \).

Step 3: Calculate the Test Statistic

We use the formula for the test statistic for a proportion: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]where \( \hat{p} = 0.4 \), \( p_0 = 0.3333 \), and \( n = 200 \). Substituting these values, the calculation becomes: \[ Z = \frac{0.4 - 0.3333}{\sqrt{\frac{0.3333(1-0.3333)}{200}}} = \frac{0.0667}{0.0325} \approx 2.05 \]

Step 4: Determine Critical Value and Decision Rule

Since the level of significance is \( \alpha = 0.02 \) and it is a right-tailed test, we find the critical value from the standard normal distribution table corresponding to \( \alpha = 0.02 \), which is approximately 2.05.

Step 5: Compare Test Statistic with Critical Value

Compare the calculated \( Z \)-value of 2.05 with the critical value of 2.05. Since they are equal, the test statistic is on the boundary of the acceptance region of null hypothesis, conventionally this would mean rejecting or technical difference might be checked by using more precision.

Step 6: Conclusion Based on Test

Since we usually reject the null hypothesis if the test statistic exceeds the critical value, we have just enough statistical evidence to suggest the proportion of students with jobs at your school is greater than one in three at a 0.02 significance level.

Key Concepts

Proportion Test

A proportion test is a statistical method used to test an assertion or claim about the proportion of a population. In this particular problem, we are testing if the proportion of graduates who have found jobs is different from a stated parameter (one-third in this case).
The test utilizes sample data to determine if there is enough evidence to reject the null hypothesis in favor of an alternative hypothesis. This requires calculating the sample proportion, which is the number of successes (in this case, graduates with jobs) divided by the total sample size.

• Null Hypothesis ( $H_0$): This is a statement of no effect or no difference. It proposes that the population proportion equals a specific value, such as $P = 0.3333$.
• Alternative Hypothesis ( $H_a$): This statement suggests a deviation from the null. Here, the claim is that the proportion at your school is greater than one-third, $P > 0.3333$.

Performing a proportion test allows researchers to objectively evaluate sample data against established benchmarks.

Significance Level

The significance level, usually denoted by \(\alpha\), is a critical aspect of hypothesis testing. It defines the threshold for rejecting the null hypothesis. When you see a significance level like \(\alpha = 0.02\), it indicates a 2% risk of concluding that a difference exists when there is none.

This level of significance determines the 'critical value,' which helps in making the decision to reject or accept the null hypothesis.

• A smaller \(\alpha\) (e.g., 0.01) means a stricter criterion for rejection, leading to fewer type I errors (false positives).
• A larger \(\alpha\) (e.g., 0.05) gives more leeway and increases the confidence in rejecting \(H_0\).

In context, using \(\alpha = 0.02\) affects how we interpret the calculated test statistic against standard distribution tables, guiding us on whether or not the observed sample proportion significantly deviates from the hypothesized population proportion.

Test Statistic

The test statistic is a standardized value that makes it possible to decide whether to support the null hypothesis or not. Here, it's denoted as \(Z\) and is calculated from the sample data using a specific formula for proportion tests: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \].

Breaking down the components:

• \(\hat{p}\) is the sample proportion, calculated as the number of successes (graduates with jobs) divided by the sample size.
• \(p_0\) is the hypothesized population proportion, in this case, 0.3333.
• \(n\) is the total sample size.

The result, a Z-score, reflects how many standard deviations the sample proportion is from the hypothesized population proportion.

Comparing the calculated \(Z\) against critical values from the standard normal distribution table helps in determining the conclusion of the hypothesis test. If the test statistic falls into the critical region, we reject the null hypothesis, assuming statistical significance.

Normal Distribution

Normal distribution is a crucial concept in hypothesis testing, particularly for large samples. It's often referred to as the 'bell curve' because of its symmetrical shape. In this test, it's used as a model for the distribution of proportions when the sample size is large enough.

The assumption is that with sufficient sample size, the distribution of the sample proportion can be approximated by the normal distribution due to the Central Limit Theorem (CLT).

• For our test, the assumption holds as the sample size is 200, which is typically sufficient.
• The distribution is characterized by a mean (expected value) and a standard deviation, which helps determine how sample data relates to population parameters.

Utilizing the standard normal distribution, we derive critical values for different significance levels. This aids in comparing the test statistic to determine if the observed result can be attributed to chance or signifies a real effect.