Let's break this problem down step by step. We are testing whether the proportion of graduates who get jobs within one year is different from the school's advertised rate of 86%, based on a random sample of 55 graduates, where 52 of them got a job within one year.

Step 1: Identify the test to use

Since we are dealing with proportions, we will use a one-proportion z-test to test the hypothesis.

Step 2: Define the null and alternative hypotheses

Let $p$ be the true proportion of graduates who get jobs within one year.

• Null Hypothesis $H_0$: $p = 0.86$
• Alternative Hypothesis $H_A$: $p \neq 0.86$

These are two-tailed hypotheses because the problem asks whether the sample proportion is different from 86%.

Step 3: Calculate the test statistic

The formula for the z-test statistic for a proportion is:

$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$

Where:

• $\hat{p}$ is the sample proportion
• $p_0$ is the hypothesized population proportion
• $n$ is the sample size

Given values:

• $p_0 = 0.86$
• $n = 55$
• $\hat{p} = \frac{52}{55} = 0.9455$ (sample proportion)

Now, we'll calculate the z-test statistic.

$z = \frac{0.9455 - 0.86}{\sqrt{\frac{0.86(1 - 0.86)}{55}}}$

Let me compute this.The test statistic $z$ is approximately $1.826$ (rounded to 3 decimal places).

Step 4: Calculate the p-value

Since this is a two-tailed test, we need to calculate the probability in both tails. The p-value for a two-tailed z-test is given by:

$\text{p-value} = 2 \times P(Z > |z|)$

Let me calculate the p-value.The p-value is approximately $0.0678$ (rounded to 4 decimal places).

Step 5: Decision Rule

We compare the p-value to the significance level $\alpha = 0.10$:

• If the p-value $\leq \alpha$, we reject the null hypothesis.
• If the p-value $> \alpha$, we fail to reject the null hypothesis.

Since the p-value $0.0678$ is less than $0.10$, we reject the null hypothesis.

Conclusion

The sample data suggest that the population proportion is significantly different than 86% at $\alpha = 0.10$, so there is sufficient evidence to conclude that the percent of graduates who get jobs within one year is different from 86%.

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Do you want any more details on this process, or have any questions? Here are five related questions to explore further:

1. How would the conclusion change if we used a significance level of $\alpha = 0.05$?
2. What is the power of this test for detecting a true proportion of 90%?
3. How would the test statistic and p-value change if 50 out of 55 graduates got jobs?
4. What assumptions must be satisfied for the z-test for proportions to be valid?
5. How can a confidence interval for the population proportion be constructed from the sample?

Tip: Always ensure that the sample size is large enough for the normal approximation to be valid when using a z-test for proportions.