Sat Math Score Improvement Significance Analysis

Sat Math Score Improvement Significance Analysis

Tutor O Rama claims that their services will improve student SAT math scores by at least 50 points. The average score on the SAT math portion is μ = 350 with a standard deviation σ = 35. A sample of 100 students who completed the tutoring program has an average score of 385 points. The assignment asks whether this observed average is statistically significant at the 5% and 1% significance levels, and to explain why or why not.

Paper For Above instruction

In evaluating the effectiveness of Tutor O Rama's claim that their services will raise students' SAT math scores by at least 50 points, a statistical hypothesis test is appropriate. The key question is whether the observed sample mean significantly exceeds the population mean of 350 points, given the sample size and variability. This analysis employs a z-test for the sample mean, which is suitable due to the large sample size (n=100) and known population standard deviation.

The null hypothesis (H₀) posits that there is no significant increase in scores attributable to the tutoring service, meaning the mean score increase is less than or equal to 50 points above the population mean: H₀: μ ≤ 400. In contrast, the alternative hypothesis (H₁) suggests that the true mean score after tutoring exceeds this threshold: H₁: μ > 400.

Using the sample information, the test statistic (z) is calculated as follows:

z = (x̄ - μ₀) / (σ / √n)

where x̄ = 385, μ₀ = 350 + 50 = 400 (the targeted mean increase), σ = 35, and n = 100.

Substituting the values:

z = (385 - 400) / (35 / √100) = (-15) / (35 / 10) = (-15) / 3.5 ≈ -4.29

The computed z-value of approximately -4.29 indicates that the sample mean is 4.29 standard deviations below the hypothesized mean of 400. To determine significance, we compare this z-value to critical values at the 5% and 1% significance levels for a one-tailed test. The critical z-values are approximately 1.645 and 2.33, respectively.

Because the calculated z-value (-4.29) is less than both critical values (1.645 and 2.33), we fail to reject the null hypothesis in a one-tailed test that examines whether scores exceed 400. Instead, the negative z-score underscores that the sample mean does not support evidence that scores are significantly higher than the threshold of 400 points. Since the observed average is actually below the mean of 400, we conclude that the data does not provide significant evidence at either the 5% or 1% significance levels to confirm Tutor O Rama's claim that their program raises scores by at least 50 points.

Furthermore, if the question interprets the sample mean of 385 as a potential improvement over the population mean of 350, then the test would be framed to assess whether the sample mean of 385 is significantly greater than 350, which involves testing the null hypothesis that μ = 350 versus the alternative μ > 350. In this case, the test statistic would be:

z = (385 - 350) / (35 / √100) = 35 / 3.5 = 10

This z-value (10) is well beyond the critical z-values, indicating a significant increase from 350 to 385. Because the question specifically states that the average score after tutoring was 385 points, and the concern is whether this is statistically significant for an increase of at least 50 points (to 400), the first test is more aligned with the ask.

Overall, based on the provided data and hypothesis testing procedures, the conclusion is that the observed average of 385 is not statistically significant to assert an increase of at least 50 points at either the 5% or 1% levels. Instead, the data shows that the increase, if any, is not statistically conclusively above the claimed threshold of 400 points, and thus Tutor O Rama cannot confidently claim their services raise scores by at least 50 points based on this evidence.

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