## A high school principle currently encourages students to enroll in a specific SAT prep program that has a reputation of improving score by 5

Question

A high school principle currently encourages students to enroll in a specific SAT prep program that has a reputation of improving score by 50 points on average. A new SAT prep program has been released and claims to be better than their current program. The principle is thinking of advertising this new program to students if there is enough evidence at the 5% level that their claim is true. The principle tests the following hypotheses: H0:μ=50 points HA:μ>50 points where μ is the true mean in difference of scores after the new SAT prep course is taken and before the new SAT prep course is taken. He randomly assigns 71 students to take this new SAT program. The difference in scores resulted in an average of 52.0011 points with a standard deviation of 10.9874 points. Step 1 of 2: What is the value of the test statistic for this test? Round your answer to four decimal places.

in progress 0
20 mins 2021-10-14T02:53:02+00:00 1 Answer 0

Z= 1.5346

Step-by-step 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!