Answer:
B) Reject the null if SAT improvement is > 94.935 points.
Step-by-step explanation:
Hello!
The claim is that the online course improves their students SAT scorer by more than 90 points on average students.
The variable of interest is X: Improvement of the SAT score of a student after taking the online course.
The population standard deviation is known as δ= 30
The statistic hypotheses are:
H₀: μ ≤ 90
H₁: μ > 90
α: 0.05
To find the critical value and determine the rejection region of this test you have to choose a statistic you are going to use for it.
In this example, there is no information about the distribution of the variable, but since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate de distribution of the sample mean to normal X[bar]≈N(μ;σ²/n)
With this, you can now use the standard normal distribution for the test.
This hypothesis test is one-tailed to the right, which means that you will reject the null hypothesis to the high values of Z.
There is only one critical value for this test:
[tex]Z_{1-\alpha }= Z_{0.95}= 1.648[/tex]
If Z < 1.648, you don't reject the null hypothesis.
If Z ≥ 1.648, you reject the null hypothesis.
To see what is the minimum value of the sample mean (X[bar]) that will lead to reject the null hypothesis I'll reverse the standardization using the known values of the population parameters and the critical value:
Z= (X[bar]- μ)/(δ/√n)
1.648= (X[bar]- 90)/(30/√10)
X[bar]= (1.648*3)+90
X[bar]= 94.944
This means that you will reject the null hypothesis to SAT improvements of average 94.94
I hope it helps!
