Greenbelt High School seniors did exceptionally well on their SAT’s this year. The principal wants to know how well their students did compared to the average national scores. What statistical test should he use to get his answer?
The correct answer and explanation is :
The principal should use a one-sample t-test to compare Greenbelt High School seniors’ SAT scores to the national average.
Explanation:
A one-sample t-test is used to determine whether the mean of a single sample differs significantly from a known population mean. In this case, the principal wants to compare the mean SAT scores of Greenbelt High School seniors to the national average SAT score.
Steps to Conduct a One-Sample t-Test:
- State the Hypotheses:
- Null Hypothesis (H₀): There is no significant difference between Greenbelt High School’s average SAT score and the national average.
[
H₀: \mu_{\text{Greenbelt}} = \mu_{\text{National}}
] - Alternative Hypothesis (H₁): Greenbelt High School’s average SAT score is significantly different from the national average.
[
H₁: \mu_{\text{Greenbelt}} \neq \mu_{\text{National}}
]
- Collect Data:
- Obtain SAT scores for a sample of Greenbelt High School seniors.
- Determine the national average SAT score and its standard deviation.
- Calculate the Test Statistic:
The t-score is calculated using the formula:
[
t = \frac{\bar{x} – \mu}{s / \sqrt{n}}
]
where:
- (\bar{x}) = sample mean (Greenbelt High School’s average SAT score),
- (\mu) = national average SAT score,
- (s) = sample standard deviation,
- (n) = sample size.
- Compare to the t-Distribution:
- Find the critical t-value based on the sample size and significance level (typically α = 0.05).
- If the calculated t-value exceeds the critical value, reject the null hypothesis.
Conclusion:
If the test shows a significant difference, the principal can conclude that Greenbelt High School students performed significantly better (or worse) than the national average. Otherwise, their performance is statistically similar to the national norm.