The college Physical Education Department offered an Advanced First Aid course last summer. The scores on the comprehensive final exam were normally distributed, and the z scores for some of the students are shown below.

A clinical psychology student wanted to determine if there is a significant difference in the Picture Arrangement scores (a subtest of the WAIS-IV that some feel might tap right-brain processing powers) between groups of right- and left-handed college students. The scores were as follows:

Picture Arrangement Scores

Left-Handed students Right- Handed students

12 8

10 10

12 10

14 12

12 11

10 6

8 7

13 9

7 11

(a). Is there a significant difference in the Picture Arrangement scores between the right- and left-handed students? Use α = .05 in making your decision. Be sure to state your hypotheses and include the following, if necessary – test statistic, degrees of freedom, computations, critical value(s)…

(b) What is the 95% confidence interval for the difference between the means?

1. A high school principal is interested in the amount of time her students spend per week working at an after school job. 37 students are randomly selected and their working hours are recorded. The sample had a mean of 12.3 hours and a standard deviation of 11.2 hours. We would like to construct an 80% confidence interval for the population mean hours worked weekly by high school students.
   1. What critical value will you use for an 80% confidence interval? Give the value and specify whether it comes from the standard normal distribution or the t distribution ( or ).

_______________

1. Calculate the margin of error, E. (You must show the setup to receive credit. You may round to three decimal places, if needed.)

E = _______________

2. Construct the 80% confidence interval for the population mean hours worked weekly by high school students. (Round limits to one decimal place.)

_______________<  < _______________

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 54 students using Method 1 produces a testing average of 51.7. A sample of 90 students using Method 2 produces a testing average of 56.8. Assume that the population standard deviation for Method 1 is 7.35, while the population standard deviation for Method 2 is 16.72. Determine the 80% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 1 of 3: Find the point estimate for the true difference between the population means.
Step 2 of 3: Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.
Step 3 of 3: Construct the 80% confidence interval. Round your answers to one decimal place.

The BioPharm company has developed a stress test for college-aged students. Scores on the stress test are approximately normally distributed with a mean of 55 points and a standard deviation of 2 points. Scores of 58 points or higher indicate a high level of stress and are of concern to doctors. A random sample of college-aged students will be selected and each will be given this stress test. Answer the following questions based on the information given.

1. What is the probability that the first randomly selected student has a high level of stress?

2. Given that the first randomly selected student is highly stressed, what is the probability that the next randomly selected doesn’t have a high level of stress?

3. Suppose the stress test is given to six randomly selected college students. What is the probability that exactly two of the six students will have a high level of stress (a score of at least 58 points)?

Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.2 significance level.

The null and alternative hypothesis would be:

H0:μN=μD
H1:μN≠μD

H0:pN=pD
H1:pN≠pD

H0:μN≤μD
H1:μN>μD

H0:pN≤pD
H1:pN>pD

H0:μN≥μD
H1:μN<μD

H0:pN≥pD
H1:pN<pD

The test is:

two-tailed

right-tailed

left-tailed

The sample consisted of 70 night students, with a sample mean GPA of 3.41 and a standard deviation of 0.03, and 70 day students, with a sample mean GPA of 3.36 and a standard deviation of 0.06.

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

• Reject the null hypothesis
• Fail to reject the null hypothesis

1. A random sample of 10 Harper College students was asked the question, “How much did you spend on textbooks this semester?” Below are the data of their responses:

Amount Spent ($) 292 240 316 361 449 428 402 286 349 250

1. Construct and interpret a 95% confidence interval for the mean amount spent for textbooks by all Harper College students.

2. What was the margin of error associated with your estimate? Explain its meaning.

3. Follett claims the mean amount spent by students for textbooks is only $280. What does the interval suggest about Follet’s

   claim?

4. Suppose the true population mean is $268 with a standard deviation of $46. If we assume all amounts follow a normal

   distribution, what is the probability the mean amount spent on textbooks by a random sample of 16 students exceeds $290?

1. On a course pre-test at a large university, 102 out of 600 randomly selected students taking Accounting II answered every question correctly. At the end of the course, 144 out of 450 randomly selected students answered every question correctly. (The pre and post-tests were identical.)

1. What is considered “success” for the binomial variable being tested?

2. Fill in the blanks to complete the hypotheses for a test to determine if the course instruction was effective in improving students’ knowledge about accounting? (The pre-test population is population #1.)

H₀: p₁ ___ p₂

H₁: p₁ ___ p₂

3. Calculate a 98% confidence interval for the difference in proportions of success. Write a sentence reporting your confidence interval as percentage point differences rounded to the nearest tenth of a percent.

4. Based on the confidence interval, was the course instruction effective in improving students’ knowledge about accounting? Why or why not?

1. On a course pre-test at a large university, 102 out of 600 randomly selected students taking Accounting II answered every question correctly. At the end of the course, 144 out of 450 randomly selected students answered every question correctly. (The pre and post-tests were identical.)

1. What is considered “success” for the binomial variable being tested?

2. Fill in the blanks to complete the hypotheses for a test to determine if the course instruction was effective in improving students’ knowledge about accounting? (The pre-test population is population #1.)

H₀: p₁ ___ p₂

H₁: p₁ ___ p₂

3. Calculate a 98% confidence interval for the difference in proportions of success. Write a sentence reporting your confidence interval as percentage point differences rounded to the nearest tenth of a percent.

4. Based on the confidence interval, was the course instruction effective in improving students’ knowledge about accounting? Why or why not?