The circumference of a circle is 20cm . what is the diameter of the circle a. 4cm b.5cm c. 10cm d20cm

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:

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!