Question

A company has developed a training procedure to improve scores on the

SAT. Following the training, 100 high school students take the SAT. The average math score is 517, with s = 90. Assuming u = 500 for the math SAT, test
the null hypothesis. Did the training significantly improve scores (a = .05,
directional)?

Answer 1

The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance

Null hypothesis is accepted at 0.05 level of significance

A company has developed a training procedure is significantly improve scores

Explanation

Step(i):-

Given sample mean (x⁻ ) = 517

sample standard deviation (s) =90

Mean of the Population (μ) =500

Null Hypothesis :- H₀ : (μ) =500

Alternative Hypothesis : H₁ : μ ≠ 500

Step(ii):-

Test statistic

t = \frac{x^{-} -mean}{\frac{s}{\sqrt{n} } }

t = \frac{517 -500}{\frac{90}{\sqrt{100} } }

t = 1.717

Level of significance

∝ = 0.05

t₀.₀₅ = 1.6604

The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance

Conclusion:-

Null hypothesis is accepted at 0.05 level of significance

A company has developed a training procedure is significantly improve scores