Answer:
Mean SAT score for Stevens High graduates are not the same as the national average.
Step-by-step explanation:
We are given the following information in question:
Population mean, μ = 510
Sample mean, [tex]\bar{x}[/tex] = 501
Sample size, n = 50
Alpha, α = 0.10
Sample standard deviation, s = 30
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{\sigma}{\sqrt{n-1}} }[/tex] Putting all the values, we have,
[tex]t_{stat} = \displaystyle\frac{501 - 510}{\frac{30}{\sqrt{49}} } = -2.1[/tex] Now,
[tex]t_{critical} \text{ at 0.10 level of significance, 49 degree of freedom } = \pm 1.6765[/tex] Since,
[tex]t_{stat} < t_{critical}[/tex]
We reject the null hypothesis and fail to accept it.
We accept the alternate hypothesis and mean SAT score for Stevens High graduates are not the same as the national average.