A science teacher claims that the mean score of students in his class is
greater than 80 with a standard deviation of 18. If a sample of 87
students was selected with a mean score of 89 then check if there is
enough evidence to support this claim at a 0.05 level of significance.


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.

---

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!

A science teacher claims that the mean score of students in his class is greater than 80 with a standard deviation of 18. If a sample of 87 students was selected with a mean score of 89, check if there is enough evidence to support this claim at a 0.05 level of significance.

Explore how to perform a hypothesis test to verify a claim that the mean score of students is greater than 80. This solution involves a Z-Test using a sample size of 87 students with a mean score of 89 and a standard deviation of 18. Suitable for Grades 11-12 and college-level statistics students.

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