To check if there is enough evidence to support the science teacher's claim that the mean score is greater than 80, we can perform a hypothesis test. Specifically, we'll use a one-sample t-test (though for large sample sizes like 87, we can approximate this using the z-test).

Step 1: Define the hypotheses

• Null Hypothesis (H₀): The mean score is 80 or less. $\mu \leq 80$
• Alternative Hypothesis (H₁): The mean score is greater than 80. $\mu > 80$

This is a right-tailed test since we are testing if the mean is greater than 80.

Step 2: Set the level of significance

The level of significance $\alpha$ is 0.05.

Step 3: Compute the test statistic

We use the formula for the z-test statistic:

$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

Where:

• $\bar{x} = 89$ is the sample mean.
• $\mu = 80$ is the population mean (under the null hypothesis).
• $\sigma = 18$ is the population standard deviation.
• $n = 87$ is the sample size.

Let's calculate the z-value.

$z = \frac{89 - 80}{\frac{18}{\sqrt{87}}}$The calculated z-value is approximately 4.66.

Step 4: Determine the critical value and p-value

For a significance level of $\alpha = 0.05$, we can find the critical z-value for a right-tailed test. The critical z-value at 0.05 significance level is approximately 1.645.

Since the calculated z-value (4.66) is much greater than the critical value (1.645), we reject the null hypothesis.

Step 5: Conclusion

There is enough evidence at the 0.05 level of significance to support the science teacher's claim that the mean score is greater than 80.

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Would you like more details on any of the steps? Here are 5 related questions for further exploration:

1. How do we determine the critical z-value for different levels of significance?
2. What would happen if the sample size were smaller (e.g., n < 30)?
3. How do you calculate the p-value associated with a z-test?
4. What is the difference between a t-test and a z-test?
5. How would this result change if the standard deviation was unknown?

Tip: Always check if the sample size is large enough to approximate a normal distribution when using the z-test!