Answer:
Z= 1.5346
Explanation:
Hello!
There is a new SAT prep course and the principal of a high school will encourage his students to take it if there is significant evidence that with this course scores are improved by 50 points on average.
Study variable X: "score on SAT of a student that took the new SAT prep course."
n= 71 students
X[bar]= 52.0011 points
S= 10.9874 points
Since you have a big enough sample, and no distribution of the variable is informed, I'll apply the Central Limit Theorem to approximate the sample mean (X[bar]) distribution to normal. This way I can use the Z-statistic for the test.
Hypothesis:
H₀: μ = 50
H₁: μ > 50
α: 0.05
Z= (X[bar] - μ)/(σ/√n) ≈ N(0;1)
Z= (52.001 - 50)/(10.9874/√71)
Z= 1.5346
This is a one-tailed test.
The critical value is

If the value of Z ≥ 1.64 then you reject the null hypothesis.
If the value of Z < 1.64 then you do not reject the null hypothesis.
Since the value is less than the critical value, the decision is to no reject the null hypothesis. You can assume that there is not enough statistical evidence to say that the new SAT course improves the SAT scores by 50 points.
I hope this helps!