Question

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 (μ = 1050). The 16 students who attend the preparation course average 1150 on the SAT, with a sample standard deviation of 300. 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 .05.

Q1: The appropriate statistical procedure for this example would be a

A. z-test

B. t-test

Q2: The most appropriate null hypothesis (in words) would be

A. There is no 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.

B. 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.

C. The students who took the SAT prep course did not score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.

D. The students who took the SAT prep course did score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.

Q3: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

B. μSATprep = 1150

C. μSATprep  1050

D. μSATprep  1050

Q4: Based on your evaluation of the null in and your conclusion, as a researcher you would be more concerned with a

A. Type I statistical error

B. Type II statistical error

Calculate the 99% confidence interval. Steps:

Q5: The mean you will use for this calculation is

A. 1050

B. 1150

Q6: What is the new critical value you will use for this calculation?

Q7: As you know, two values will be required to complete the following equation:

__________    __________

Q8: Which of the following is a more accurate interpretation of the confidence interval you just computed?

A. We are 99% confident that the scores fall in the interval _____ to _____.

B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval _____ to _____.

C. We are 99% confident that the example above has correct values.

D. We are 99% confident that the difference in SAT scores between the students who took the prep course and the students who did not falls in the interval _____ to _____.

Answer 1

Q1:

Since we do not know the population standard deviations, the appropriate statistical procedure for this example would be a

B. t-test

Q2:

The most appropriate null hypothesis (in words) would be

A. There is no 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.

Q3: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

Q4:

Type I statistical error is rejection of true null hypothesis.

Type I statistical error is we conclude that 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 but in reality there is no difference.

Type II statistical error is acceptance of false null hypothesis.

Type II statistical error is we conclude that there is no 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 but in reality there is a difference.

The researcher you would be more concerned with a

A. Type I statistical error

because in Type I error, we do preparation course based on conclusion but it leads to no change in SAT scores.

Calculate the 99% confidence interval. Steps:

Q5: The mean you will use for this calculation is sample mean

B. 1150

Q6:

Degree of freedom = 16-1 = 15

Critical value of t at df = 15 and 99% confidence interval is 2.95

Q7: As you know, two values will be required to complete the following equation:

where SE = s / = 300 / = 75

(1150 - 2.95 * 75 , 1150 + 2.95 * 75)

(928.75 , 1371.25)

Q8:

The accurate interpretation of the confidence interval is,

B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval 928.75 to 1371.25.