A company that sells an online course aimed at helping high-school students improve their SAT scores has claimed that SAT scores will improv

Question

A company that sells an online course aimed at helping high-school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the company’s course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, find the critical value in terms of improvement in SAT points, which would be needed prior to finding a beta.
A) Reject the null if SAT improvement is > 95 points.
B) Reject the null if SAT improvement is > 94.935 points.
C) Reject the null if SAT improvement is > 95.88 points.
D) Reject the null if SAT improvement is 85.065 points.

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Gabriella 2 months 2021-10-08T22:19:55+00:00 1 Answer 0 views 0

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    2021-10-08T22:21:11+00:00

    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:

    Z_{1-\alpha }= Z_{0.95}= 1.648

    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!

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