Respuesta :
Answer:
We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.
Step-by-step explanation:
We are given the following in the question: Â
Population mean, ÎĽ = 510
Sample mean, [tex]\bar{x}[/tex] = 502
Sample size, n = 60
Sample standard deviation, s = 30
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 510\\H_A: \mu \neq 510[/tex]
We use Two-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{502- 510}{\frac{30}{\sqrt{60}} } = -2.0655[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 59 degree of freedom } = \pm 1.671[/tex]
Since, Â Â Â Â Â Â Â
[tex]|t_{stat}| < |t_{critical}|[/tex]
We reject the null hypothesis and fail to accept it.
We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.
