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To calculate the value of the test statistic \( z \), we use the formula:

\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]

where:
- \(\bar{x} = 54\) (the sample mean),
- \(\mu = 58\) (the population mean under the null hypothesis),
- \(\sigma = 13\) (the population standard deviation),
- \(n = 58\) (the sample size).

Let's substitute these values into the formula:

\[
z = \frac{54 - 58}{13 / \sqrt{58}}
\]

Now, we calculate each step:
1. Calculate \( \bar{x} - \mu = 54 - 58 = -4 \).
2. Calculate the standard error \( \sigma / \sqrt{n} = 13 / \sqrt{58} \).
3. Finally, divide \(-4\) by the result from step 2.

I'll perform these calculations for precision.The calculated value of the test statistic \( z \) is approximately \(-2.34\), rounded to two decimal places.

Would you like further details or have any questions?

Here are five related questions for further understanding:
1. How do we interpret a left-tailed test in this context?
2. Why do we use a \( z \)-test instead of a \( t \)-test here?
3. How is the standard error calculated, and why is it important?
4. What is the next step after finding the \( z \)-value in hypothesis testing?
5. How does the level of significance (\( \alpha = 0.01 \)) affect the hypothesis test?

**Tip:** Always ensure you know if the test should be one-tailed or two-tailed, as it affects how you interpret the \( p \)-value and test direction.

A random sample of 58 second-graders in a school district is given a standardized math test with a mean score of 54 and standard deviation of 13. The nationwide average is 58. Test if the school district’s second-graders have weaker math skills at a 0.01 significance level using the P-value method.

Conduct a hypothesis test to compare the mean math score of second-graders in a school district to the nationwide average. This example uses a Z-test and the P-value method at a 0.01 significance level, suitable for college-level statistics students.

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