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 deviationof 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.Q78:The appropriate statistical procedure for this example would be aA.z-testB.t-testQ79:Is this a one-tailed or a two-tailed test?A.one-tailedB.two-tailed 11Q80:The most appropriate null hypothesis (in words) would beA.There is no statistical difference in SAT scores when comparing students who took theSAT 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 theSAT 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.Q81:The most appropriate null hypothesis (in symbols) would beA.μSATprep= 1050B.μSATprep= 1150C.μSATprep1050D.μSATprep1050Q82:Set up the criteria for making a decision. That is, find the critical value using an alpha = .05. (Make sure you are sign specific: + ; -; or ) (Use your tables)Summarize the data into the appropriate test statistic.Steps:Q83:What is the numeric value ofyour standard error?Q84:What is the z-value or t-value you obtained (your test statistic)?Q85:Based on your results (and comparing your Q84 and Q82 answers) would youA.reject the null hypothesisB.fail to reject the null hypothesisQ86:The best conclusion for this example would beA.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 generalpopulation 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. 12Q87:Based on your evaluation of the null in Q85 and your conclusion is Q86, as a researcher you would be more concerned with aA.Type I statistical errorB.Type II statistical errorCalculate the 99%confidence interval.Steps:Q88:The mean you will use for this calculation isA.1050B.1150Q89:What is the new critical value you will use for this calculation?Q90:As you know, two values will be required to complete the following equation:__________ __________Q91: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

Here 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 deviationof 300.

Here we consider the following hypothesis:

Q78: The appropriate statistical procedure for this example would be B.t-test .

Q79: B This is a two tailed test.

Q80: The most appropriate null hypothesis (in words) would be A.There is no statistical difference in SAT scores when comparing students who took theSAT prep course with the general population of students who did not take the SAT prep course.

Q81: The most appropriate null hypothesis (in symbols) would be A.   μSATprep= 1050.

Q82: The critical value of the above test is t = 2.602 ( Obtained from T distribution table).

        The test statistic for the given test is:

        

            where n= 16, = sample mean=1150, s=sample standard deviation=300

Q83: The standard error is given by:

      

               =300/4   =75

Q84: The test statictic value T= {4 x ( 1150 - 1050)} / 300

                                           = 1.333

Q85: Based on the result above we would B. Fail to reject the null hypothesis since T< t

Q86: The best conclusion for this example 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.

Q87: Based on our evaluation of the null in Q85 and our conclusion is Q86, as a researcher you would be more concerned with B.Type II statistical error

Q88: The mean we would use here is B. 1150

Q89: The new critical value we will use here is t= 3.386 which is obtained from the T distribution table at level 0.005

Q90: The required equation to calculate the 99% confidence interval is

         

Q91: The more accurate interpretation of the confidence interval we just computed 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 903.55 to 1396.45 .