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 :

(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

1. t= with df=

2. 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

1. t= with df=

2. 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

1. t= with df=

2. 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.
   • substantial/strong evidence against the null hypothesis. • overwhelming evidence against the null hypothesis.

3. Pizza! A national pizza chain advertises that its large pizzas are 14 inches in diameter. The CEO wants to determine whether the average size of all of its pizzas adheres to that claim. She collects a random sample of 31 pizzas and measures each one. The sample results in an average diameter of 14.9 inches and a standard deviation of s = 2.10.

(a) Identify μ in this scenario. (Choose One)

• Average diameter of all pizzas in the sample.

• 14 inches.

• True average diameter of all pizzas sold by the chain.

• We do not have enough information to answer this question.

(b) Conduct a hypothesis test for the above situation.

1. Identify the hypotheses H0 and Ha.

2. Find the test statistic (Round your answer to 2 decimal places).

3. Using the table below, select the most appropriate p-value.

   P(T <t) 0.99 P(T >t) 0.01 P(T>|t|) 0.02

4. Provide an interpretation of the p-value you chose from the table within the context of the problem by appropriately filling in the blanks:
   This p-value is the probability, assuming the is , of observing a test statistic at least as unusual as the one observed.

5. Based on the p-value, provide a conclusion in the context of the problem by appropriately filling in the blanks:
   There is evidence indicating that the _____is________ .

Answer 1

SOLUTION 1A:  True average increase in score of all people in the population

SOLUTION 1B] H0:

HA:

C] t= 3/5.4/sqrt(9)

t=1.67

degreesof freedom= n-1=9-1=8

little to no evidence against the null hypothesis.

ii) a sample of size n = 24

1. t=2.72 with df= 23

2. The p-value corresponding to the correct test statistic is 0.00611. Based on this

   p-value, there is... moderate evidence against the null hypothesis.

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

t=11.71 with df=443

overwhelming evidence against the null hypothesis.

AS PER THE q& a GUIDELINES I HAVE DONE THE FIRST QUESTION PLEASE RE POST THE REST THANK YOU.