A school reports that 86% of its graduates get jobs within one year. You take a random sample of 55 graduates, of whom 52 got a job within one year. Is this enough evidence to show that a significantly different percent of graduates get jobs than advertised by the school? (Use =0.10)
For this study, we should use

The null and alternative hypotheses would be:
:      (please enter a decimal)
:      (Please enter a decimal)

The test statistic =  (please show your answer to 3 decimal places.)

The p-value =  (Please show your answer to 4 decimal places.)

The p-value is

Based on this, we should  the null hypothesis.

As such, the final conclusion is that ...
The sample data suggest that the populaton proportion is significantly different than 86% at  = 0.10, so there is sufficient evidence to conclude that the percent of graduates get jobs within one year is different than 86%
The sample data suggest that the population proportion is not significantly different than 86% at  = 0.10, so there is not sufficient evidence to conclude that the percent of graduates get jobs within one year is different than 86%.

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%.

---

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.

A school reports that 86% of its graduates get jobs within one year. You take a random sample of 55 graduates, of whom 52 got a job within one year. Is this enough evidence to show that a significantly different percent of graduates get jobs than advertised by the school? (Use =0.10)

In this problem, we perform a one-proportion z-test to determine if the percentage of graduates who get jobs within one year differs from the school's reported 86%. With a random sample of 55 graduates, where 52 got jobs, and using a significance level of 0.10, we calculate the test statistic and p-value. This is a key problem for students learning hypothesis testing in statistics.

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