Questions
An SAT prep course claims to improve the test score of students. The table below shows...

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.01 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1 2 3 4 5 6 7
Score on first SAT 560 490 400 350 360 470 560
Score on second SAT 590 540 470 460 380 510 620

a) State the null and alternative hypotheses for the test.

b) Find the value of the standard deviation of the paired differences. Round your answer to three decimal place.

c) Compute the value of the test statistic. Round your answer to three decimal places.

d) Determine the decision rule for rejecting the null hypothesis. Round the numerical portion of your answer to three decimal places.

e) Make the decision for the hypothesis test.

In: Statistics and Probability

An SAT prep course claims to improve the test score of students. The table below shows...

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.1 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1 2 3 4 5 6 7
Score on first SAT 380 410 400 410 360 550 550
Score on second SAT 410 510 430 480 390 590 600

Step 1: State the null and alternative hypotheses for the test.

Step 2: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5: Make the decision for the hypothesis test.

In: Statistics and Probability

A counselor hypothesizes that a popular new cognitive therapy increases depression. The counselor collects a sample...

A counselor hypothesizes that a popular new cognitive therapy increases depression. The counselor collects a sample of 28 students and gives them the cognitive therapy once a week for two months. Afterwards the students fill out a depression inventory in which their mean score was 50.73. Normal individuals in the population have a depression inventory mean of 50 with a variance of 12.96. What can be concluded with α = 0.01?

- c) Obtain/compute the appropriate values to make a decision about H0.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value = __________; test statistic = ____________
Decision: (choose one) Reject H0 or Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[ ________ , _______ ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d = ___________ ;   -(choose one) 1. na 2. trivial effect 3. small effect 4. medium effect 5. large effect
r2 = ____________ ;  -(choose one) 1. na 2. trivial effect 3. small effect 4. medium effect 5. large effect

f) Make an interpretation based on the results. (choose one)

1.) The depression of students that underwent cognitive therapy is significantly higher than the population.

2.) The depression of students that underwent cognitive therapy is significantly lower than the population.  

3.) The new cognitive therapy technique does not significantly impact depression.

In: Statistics and Probability

The Gourmand Cooking School runs short cooking courses at its small campus. Management has identified two...

The Gourmand Cooking School runs short cooking courses at its small campus. Management has identified two cost drivers it uses in its budgeting and performance reports—the number of courses and the total number of students. For example, the school might run two courses in a month and have a total of 64 students enrolled in those two courses. Data concerning the company’s cost formulas appear below:

Fixed Cost per Month Cost per Course Cost per
Student
Instructor wages $ 2,910
Classroom supplies $ 310
Utilities $ 1,240 $ 50
Campus rent $ 5,200
Insurance $ 2,200
Administrative expenses $ 3,700 $ 42 $ 7

For example, administrative expenses should be $3,700 per month plus $42 per course plus $7 per student. The company’s sales should average $890 per student.

The company planned to run four courses with a total of 64 students; however, it actually ran four courses with a total of only 56 students. The actual operating results for September appear below:

Actual
Revenue $ 54,060
Instructor wages $ 10,920
Classroom supplies $ 19,690
Utilities $ 1,850
Campus rent $ 5,200
Insurance $ 2,340
Administrative expenses $ 3,742

Required:

Prepare a flexible budget performance report that shows both revenue and spending variances and activity variances for September. (Indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values.)

In: Accounting

3. a. A survey of 1000 U.S. adults found that 34% of people said that they...

3.

a. A survey of 1000 U.S. adults found that 34% of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the 90% confidence interval of the true proportion. ROUND TO FIVE DECIMAL PLACES

b. In a recent year, 6% of cars sold had a manual transmission. A random sample of college students who owned cars revealed the following: out of 129 cars, 23 had manual transmissions. Estimate the proportion of college students who drive cars with manual transmissions with 99% confidence. ROUND TO THREE DECIMAL PLACES

c. A random sample of 325 college students were asked if they believed that places could be haunted, and 147 responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with 95% confidence. According toTime magazine, 37% of Americans believe that places can be haunted. ROUND TO THREE DECIMAL PLACES

___ <p<____

d. A CBS News/New York times poll found that 329 out of 763 randomly selected adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 94% accuracy. ROUND TO THREE DECIMAL PLACES

___ <p < ____

e. Took a random sample of 200 people, 142 said that they watched educational television. Find the 95% confidence interval of the true proportion of people who watched educational television. ROUND TO THREE DECIMAL PLACES

____ < p < ____

In: Statistics and Probability

An SAT prep course claims to improve the test score of students. The table below shows...

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores? Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.05 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1 2 3 4 5 6 7

Score on first SAT 600 380 460 520 470 470 380

Score on second SAT 620 430 480 570 490 550 400

Step 1 of 5:

State the null and alternative hypotheses for the test.

Step 2 of 5:

Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:

Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5:

Make the decision for the hypothesis test.

In: Statistics and Probability

An SAT prep course claims to improve the test score of students. The table below shows...

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.1 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1 2 3 4 5 6 7
Score on first SAT 510 490 540 410 520 550 510
Score on second SAT 530 540 560 490 550 590 530

Step 1 of 5:

State the null and alternative hypotheses for the test

Step 2 of 5:

Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:

Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5:

Make the decision for the hypothesis test.

In: Statistics and Probability

Use this scenario to answer the following questions. Ms. Fitness-Buff, a high school gym teacher, wants...

  1. Use this scenario to answer the following questions.

Ms. Fitness-Buff, a high school gym teacher, wants to propose an after-school fitness program. To get an idea of the fitness level of the students at her school, she takes a random sample of 75 students and records the number of hours the students exercised in the past week. Her sample mean is 2.25 hours and she knows from past research that the population standard deviation is 2 hours. She wants to know if this varies from a population mean of 3 hours/week.

  1. Construct a 95% confidence interval. (2 points)
  1. Draw a conclusion for a two-sided test at a = .05. (1 point)
  1. Ms. Fitness-Buff is told that if students exercise less than 3 hours per week, she can start an after-school fitness program. Test this one-sided hypothesis and draw a conclusion at a = .05. (2 points)
  1. What would the consequences of a Type I error be in the test from part C? (1 point)
  1. What's the probability of a Type I error for the test in part C? (1 point)
  1. What would the consequences of a Type II error be for the test from part C? (1 point)
  1. What is the rejection region for Ho: m = 3 for the test from part C? (1 point)
  1. Calculate the probability of making a Type II error if the true population mean is 2.75. (2 points)
  1. What's the power of the test if the true population mean is 2.75? (1 point)

In: Statistics and Probability

The Gourmand Cooking School runs short cooking courses at its small campus. Management has identified two...

The Gourmand Cooking School runs short cooking courses at its small campus. Management has identified two cost drivers it uses in its budgeting and performance reports—the number of courses and the total number of students. For example, the school might run two courses in a month and have a total of 63 students enrolled in those two courses. Data concerning the company’s cost formulas appear below:

Fixed Cost per Month Cost per Course Cost per
Student
Instructor wages $ 2,950
Classroom supplies $ 290
Utilities $ 1,220 $ 65
Campus rent $ 4,600
Insurance $ 2,400
Administrative expenses $ 3,800 $ 40 $ 4

For example, administrative expenses should be $3,800 per month plus $40 per course plus $4 per student. The company’s sales should average $860 per student.

The company planned to run four courses with a total of 63 students; however, it actually ran four courses with a total of only 61 students. The actual operating results for September were as follows:

Actual
Revenue $ 51,280
Instructor wages $ 11,080
Classroom supplies $ 18,120
Utilities $ 1,890
Campus rent $ 4,600
Insurance $ 2,540
Administrative expenses $ 3,638

Required:

Prepare a flexible budget performance report that shows both revenue and spending variances and activity variances for September. (Indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values.)

In: Accounting

10. A study was designed to compare the attitudes of two groups of nursing students towards...

10. A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 17 nursing students from Group 1 resulted in a mean score of 55.3 with a standard deviation of 6.4. A random sample of 11 nursing students from Group 2 resulted in a mean score of 65.4 with a standard deviation of 2.8. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.01 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Ho: μ1 (=,≠,<,>,≤,≥) μ2

Ha: μ1 (=,≠,<,>,≤,≥) μ2

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Reject Ho if (t, I t I) (<,>) ____

Step 4 of 4: State the test's conclusion.

A. Reject Null Hypothesis

B. Fail to Reject Null Hypothesis

In: Statistics and Probability