The following 14 questions (Q18 to Q31) are based on the following example: A re

Question

The following 14 questions (Q18 to Q31) are based on the following example: A researcher wants to determine whether high school students who attend an SAT preparation course score significantly different on the SAT than students who do not attend the preparation course. For those who do not attend the course, the population mean is 1050 16 students who attend the preparation course average 1150/on the SAT, with a sample standard deviation of 100 On the basis of these data, can the researcher conclude that the preparation course has a significant difference on SAT scores? Set alpha equal to .01

Explanation / Answer

Solution:

Question 26

Here, we have to use one sample t test for population mean.

The null and alternative hypotheses for this test are summarised as below:

H0: µ = 1050 versus Ha: µ ? 1050

This is a two tailed test.

From given information, we have

Xbar = 1150, S = 100, n = 16, ? = 0.01, df = n – 1 = 16 – 1 = 15

Critical t value = -2.9467 and 2.9467

(by using t-table and df)

Test statistic formula is given as below:

Test statistic = t = (Xbar - µ)/[S/sqrt(n)]

Test statistic = t = (1150 – 1050)/[100/sqrt(16)]

Test statistic = t = 100/25 = 4

P-value = 0.0012

(by using t-table)

P-value < ? = 0.01

So, we reject the null hypothesis

There is a statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.

Correct answer: B.

Question 27

For the above scenario we reject the null hypothesis when it is not true. So, decision of rejection of null hypothesis is correct. There is a possibility of do not rejecting null hypothesis when there is significant difference. So, do not rejecting null when ‘it is false’ is a type II statistical error. So correct answer is B.

Question 28

We use the sample mean for the calculation of confidence interval. Sample mean for the above test is given as 1150. So, correct answer is B.

Question 29

We are given

Confidence level = 95%

df = n – 1 = 16 – 1 = 15

so, critical value = 2.1314

(By using t-table)

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