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 µ = 500 for the math SAT, test the null hypothesis. Did training significantly improve scores (α = .05, directional)?

Answer 1

Answer:

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

Step-by-step 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