Question

ACT Prep Course. ACT prep courses like to market that you can increase your ACT score by taking their courses. Some statisticians were curious how effective these courses really were. They decided to investigate the truth of the claim by measuring the average score increase for a random sample of students selected to take an ACT prep course. These students took the ACT twice, once before and once after taking the course. The variable of interest was the increase in scores between first and second attempts. NOTE: The population of interest is all students who took the ACT prep course.

(a) Identify µ in this scenario.

(Choose One)

• Average score of all people in the sample.

• True average increase in score of all people in the population

. • True average score of all people in the population.

• Average increase in score by all people in the sample

. (b) In order to test the researcher’s claim, identify the appropriate: i. null hypothesis – H0 : ii. alternative hypotheses – Ha :

1 (c) Suppose that an average increase in score of 3 points (x = 3) and standard deviation of s = 5.4 points were found in each of the following scenarios. Further, assume the difference in score is normally distributed. For each case, compute the degrees of freedom and test statistic and assess the strength of evidence against the null hypothesis. Round t-statistics to two decimals. i.

Scenario 1: a sample of size n = 9 A. t = with df = B. The p-value corresponding to the correct test statistic is 0.06673. Based on this p-value, there is... • little to no evidence against the null hypothesis.

• borderline/weak evidence against the null hypothesis.

• moderate evidence against the null hypothesis. •

substantial/strong evidence against the null hypothesis.

• overwhelming evidence against the null hypothesis.

ii. Scenario 2: a sample of size n = 24 A. t = with df = B. The p-value corresponding to the correct test statistic is 0.00611. Based on this p-value, there is... • little to no evidence against the null hypothesis.

• borderline/weak evidence against the null hypothesis.

• moderate evidence against the null hypothesis.

• substantial/strong evidence against the null hypothesis.

• overwhelming evidence against the null hypothesis.

iii. Scenario 3: a sample of size n = 444 A. t = with df = B. The p-value corresponding to the correct test statistic is <0.0001. Based on this p-value, there is...

• little to no evidence against the null hypothesis.

• borderline/weak evidence against the null hypothesis. •

moderate evidence against the null hypothesis. • s

substantial/strong evidence against the null hypothesis. •

overwhelming evidence against the null hypothesis.

Answer 1

(a) Identify µ in this scenario.

• True average increase in score of all people in the population

. (b) In order to test the researcher’s claim, identify the appropriate: i. null hypothesis – H0 : ii. alternative hypotheses – Ha :

1 (c) Suppose that an average increase in score of 3 points (x = 3) and standard deviation of s = 5.4 points were found in each of the following scenarios. Further, assume the difference in score is normally distributed. For each case, compute the degrees of freedom and test statistic and assess the strength of evidence against the null hypothesis. Round t-statistics to two decimals. i.

Scenario 1: a sample of size n = 9

A.

Standard error of mean , SE = 5.4 / = 1.8

t = (x - ) SE = (3 - 0) / 1.8 = 1.67

with df = n-1 = 9-1 = 8

B. The p-value corresponding to the correct test statistic is 0.06673. Based on this p-value, there is... • little to no evidence against the null hypothesis.

• borderline/weak evidence against the null hypothesis.

because p-value is greater than 0.05 significance level but still close to it.

ii. Scenario 2: a sample of size n = 24

A.

Standard error of mean , SE = 5.4 / = 1.10227

t = (x - ) SE = (3 - 0) / 1.10227 = 2.72

with df = n-1 = 24-1 = 23

B. The p-value corresponding to the correct test statistic is 0.00611. Based on this p-value, there is... • little to no evidence against the null hypothesis.

• substantial/strong evidence against the null hypothesis.

because p-value is less than 0.05 significance level

iii. Scenario 3: a sample of size n = 444

Standard error of mean , SE = 5.4 / = 0.2562727

t = (x - ) SE = (3 - 0) / 0.2562727 = 11.71

with df = n-1 = 444-1 = 443

B. The p-value corresponding to the correct test statistic is <0.0001. Based on this p-value, there is...

overwhelming evidence against the null hypothesis.

because p-value is very low as comared with 0.05 significance level