Question: Use the following information to answer the next 4 questions.
The national mean SAT score in math is $550$. Suppose a high
school principal claims that the mean SAT score in math at his
school is better than the national mean score. A random sample of
$72$ students finds a mean score of $574$. Assume that the
population standard deviation is $\sigma = 100$. Is the principal's claim
valid? Use a level of significance of $\alpha = 0.05$.
State the null and alternative hypotheses.
$\circ$ $H_0$: The students' mean SAT score is greater than $550$.
$\quad H_a$: The students' mean SAT score is $574$.
$\circ$ $H_0$: The students' mean SAT score is $550$.
$\quad H_a$: The students' mean SAT score is $574$.
$\circ$ $H_0$: The students' mean SAT score is greater than $550$.
$\quad H_a$: The students' mean SAT score is $550$.
$\circ$ $H_0$: The students' mean SAT score is $550$.
$\quad H_a$: The students' mean SAT score is greater than $550$.
position_hint: top-left
Compute the test statistic for this analysis. Round your answer to 3 decimal places. (Example: 0.398)
1. **State the problem:**
We want to test if the mean SAT math score at the principal's school is greater than the national mean of $550$ based on a sample of $72$ students with a sample mean of $574$ and population standard deviation $\sigma = 100$. The significance level is $\alpha = 0.05$.
2. **State the hypotheses:**
The null hypothesis $H_0$ assumes the mean score is equal to the national mean:
$$H_0: \mu = 550$$
The alternative hypothesis $H_a$ is that the mean score is greater than the national mean:
$$H_a: \mu > 550$$
3. **Formula for the test statistic:**
Since the population standard deviation is known, we use the $z$-test statistic:
$$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
where:
- $\bar{x} = 574$ is the sample mean
- $\mu_0 = 550$ is the population mean under $H_0$
- $\sigma = 100$ is the population standard deviation
- $n = 72$ is the sample size
4. **Calculate the test statistic:**
Calculate the denominator:
$$\frac{\sigma}{\sqrt{n}} = \frac{100}{\sqrt{72}}$$
Calculate $\sqrt{72}$:
$$\sqrt{72} \approx 8.485$$
So:
$$\frac{100}{8.485} \approx 11.785$$
Calculate the numerator:
$$\bar{x} - \mu_0 = 574 - 550 = 24$$
Calculate $z$:
$$z = \frac{24}{11.785} \approx 2.037$$
5. **Interpretation:**
The test statistic is approximately $2.037$. This value will be compared to the critical $z$ value for $\alpha = 0.05$ in a one-tailed test to decide if the principal's claim is valid.
**Final answer:**
$$\boxed{2.037}$$