The following 14 questions (Q78 to Q91) 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 (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.

Q78: The appropriate statistical procedure for this example would be a

A. z-test

B. t-test

Q79: Is this a one-tailed or a two-tailed test?

A. one-tailed

B. two-tailed

Q80: 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.

Q81: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

B. μSATprep = 1150

C. μSATprep ?<=1050

D. μSATprep ?>=1050

Q82: 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 of your 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 you

A. reject the null hypothesis

B. fail to reject the null hypothesis

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.

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.

Q87. Based on your evaluation of the null in Q25 and your conclusion is Q86, 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:

Q88: The mean you will use for this calculation is

A. 1050

B. 1150

Q89: 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:

__________ ?<= u<=? __________

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 _____.

The following 2 questions (Q92 to Q93) are based on the following situation:

The national average for the verbal section of the Graduate Record Exam (GRE) is 500 with a standard deviation of 100. A researcher uses a sampling distribution made up of samples of 100.

Q92: According to the Central Limit Theorem, what is the mean of the sampling distribution of means?

A. 10

B. 50

C. 100

D. 500

Q93: According to the Central Limit Theorem, what is the standard error of the mean?

a. 10

b. 50

c. 100

d. 500

Q94: As you increase the number of subjects in your sample, the calculated value of a t-test will

A.) increase

B.) decrease

C.) remain the same

Q95: As you decrease the true distance between the null and alternative hypotheses (μ1 - μ0), the likelihood of rejecting the null hypothesis

A. increases

B. decreases

C. remains the same

Q96: Keeping everything else the same, if you were to decrease your alpha level from .05 to .01, the likelihood of rejecting the null hypothesis

A. increases

B. decreases

C. remains the same

Answers;

Q78: The appropriate statistical procedure for this example would be a

B. z-test

Q79: Is this a one-tailed or a two-tailed test?

B. two-tailed

Q80: 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.

Q81: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

Q82: 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:

data

u = 1050 X??? = 1150 s = 300 n = 16

hypothesis

 μSATprep = 1050

 μSATprep is not equal to 1050

test statistic = z test = x??? -u / ?σ? /??? n

decision rule reject null hypothesis if t calculated

Q83: What is the numeric value of your standard error?

SE = 75

Q84: What is the z-value or t-value you obtained (your test statistic)?

z - value = 1.33

Q85: Based on your results (and comparing your Q84 and Q82 answers) would you

B. fail to reject the null hypothesis

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 your evaluation of the null in Q25 and your conclusion is Q86, as a researcher, you would be more concerned with a

B. Type II statistical error

Q88: The mean you will use for this calculation is

B. 1150

Q89: What is the new critical value you will use for this calculation?

2.58

Q90: As you know, two values will be required to complete the following equation:

956.5?<= u<=?1343.5

 Q91: Which of the following is a more accurate interpretation of the confidence interval you

just computed?

B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval 956.5 to 1343.5

Q92: According to the Central Limit Theorem, what is the mean of the sampling distribution of means?

D. 500

Q93: According to the Central Limit Theorem, what is the standard error of the mean?

a. 10

Q94: As you increase the number of subjects in your sample, the calculated value of a t-test will

A.) increase

Q95: As you decrease the true distance between the null and alternative hypotheses (μ1 - μ0), the likelihood of rejecting the null hypothesis

A. increases

Q96: Keeping everything else the same, if you were to decrease your alpha level from .05 to .01, the likelihood of rejecting the null hypothesis

B. decreases

Step-by-step explanation

The results are explained as shown;

Q78: The appropriate statistical procedure for this example would be a

B. z-test

This is because the standard deviation of the population is known and is not calculated even though our n is less than 30

also it is z test since it is comparing population mean and sample mean

Q79: Is this a one-tailed or a two-tailed test?

B. two-tailed

This is a two tailed test since we are determining if the difference between population mean and sample mean is significant that is it is not equals to zero or if it is not significant it might be equal to zero

Q80: 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.

Q81: The most appropriate null hypothesis (in symbols) would be

A. μSATprep = 1050

Q82: 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:

data

u = 1050 X??? = 1150 s = 300 n = 16

hypothesis

 μSATprep = 1050

 μSATprep is not equal to 1050

test statistic = z test = x??? -u / ?σ? /??? n

decision rule reject null hypothesis if z calculated is greater than z tabulated at alpha = 0.05

z = 1150-1050 / 100?? /??? 16 = 100/75 = 1.33

Q83: What is the numeric value of your standard error?

SE = 75

that is 300 /??? 16 = 75

Q84: What is the z-value or t-value you obtained (your test statistic)?

z - value = 1.33

Q85: Based on your results (and comparing your Q84 and Q82 answers) would you

B. fail to reject the null hypothesis

This is since Z tabulated at alpha = 0.05 is 1.96 which is greater than the z value of 1.33 hence we fail to reject

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 your evaluation of the null in Q25 and your conclusion is Q86, as a researcher, you would be more concerned with a

B. Type II statistical error

this is so because the error  occurs when one accepts a null hypothesis that is actually false . if the null hypothesis that we accepted is actually false which might be logically true then we would have committed type II error

Q88: The mean you will use for this calculation is

B. 1150

Q89: What is the new critical value you will use for this calculation?

2.58 the z score for 99% confidence interval

Q90: As you know, two values will be required to complete the following equation:

the calculation is;

X??? + or - z score x SE = X??? + or - 2.58 x 75 = X??? + or - 193.5

956.5?<= u<=?1343.5

Q91: Which of the following is a more accurate interpretation of the confidence interval you

just computed?

B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval 956.5 to 1343.5

Q92: According to the Central Limit Theorem, what is the mean of the sampling distribution of means?

According to the central limit theorem, the mean of a sampling distribution of means is an unbiased estimator of the population mean.

hence u_x??? = u or X??? = u hence mean of sampling distributions is equals to population mean

D. 500

Q93: According to the Central Limit Theorem, what is the standard error of the mean?

?σ?_x??? = ?σ? /??? n = 100?? /??? 100 = 100/10 = 10

a. 10

Q94: As you increase the number of subjects in your sample, the calculated value of a t-test will

A.) increase

this is because if you increase n the SE will reduce and hence increasing the t value calculated

Q95: As you decrease the true distance between the null and alternative hypotheses (μ1 - μ0), the likelihood of rejecting the null hypothesis

B. decreases

This is so because as the difference between the two means decreases the smaller the value to divide by the standard error for the test statistic value will decrease if the test statistic value reduce to be compared with values from the tables then chances of rejecting the null hypothesis will reduce since test values will be < test scores

Q96: Keeping everything else the same, if you were to decrease your alpha level from .05 to .01, the likelihood of rejecting the null hypothesis

B. decreases

alpha values are compared with p-values if a p-value is less than alpha then we reject null hypothesis if it is greater we fail to reject null hypothesis. if we reduce alpha level from 0.05 to 0.01 p-value of the test might exceed the new alpha(0.01) and hence decreasing chances of rejecting null hypothesis