Questions
Questions

FA25_QTM_100_3 Midterm Exam 2 - Requires Respondus LockDown Browser

Single choice

A chemistry professor investigated the relationship between attending weekly mentoring sessions and the improvement on the exam performances. To test this, 5 students were randomly selected from a large group of students who never attended mentoring sessions before the midterm exam, then were asked to attend weekly mentoring sessions until the end of the semester. After the final exam, the professor used a summary table below to organize the exam scores for these 5 students. Assume all conditions for valid statistical inferences are satisfied.  Student Midterm Exam Score Final Exam Score Difference of Exam Scores 1 88.5 91.5 3 2 79 81 2 3 81 87 6 4 86 86 0 5 84 89 5 Mean 3.2 Standard Deviation 3.80 3.91   Which of the following is true?

Options
A.The correct degrees of freedom for this hypothesis test is 8.
B.Two sample z-test is the correct hypothesis test to answer the professor's research question.
C.In this dataset, students had a higher average exam score after attending weekly mentoring sessions.
D.The standard deviation for the difference of exam scores is 0.11.
View Explanation

View Explanation

Verified Answer
Please login to view
Step-by-Step Analysis
First, let's restate the problem and the options to ensure clarity. The professor compared final exam scores to midterm scores for 5 students after a mentoring intervention, and the given data include the differences in scores: 3, 2, 6, 0, and 5, with a mean difference of 3.2. The question asks which statement is true about this dataset. Option 1: The correct degrees of freedom for this hypothesis test is 8. In a paired dataset like this (each student has a midterm and a corresponding final score), the appropriate test ......Login to view full explanation

Log in for full answers

We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!

Similar Questions

Forty-five families are randomly selected and their daily water usage (in gallons) before and after viewing a water conservation video is measured. What is the appropriate null hypothesis to determine if the mean amount of water conserved between the before and after measurements significantly differs.

Question at position 27 The impact of Greta Thunberg’s documentary “A Year to Change the World” had on older U.S residents' attitudes towards environmental protection policies. Before watching the documentary, 25 participants ages 65-75-years answered a series of questions regarding their views on the implementation of stringent environmental regulations. After completing the questionnaire, they watched the documentary, and then they completed the questionnaire a second time. Scores ranged from 30 to 100, with higher scores indicating more favorable attitudes towards implementing environmental regulations. What kind of statistical test should you run? Printed Blank[input]Question Blank 1 of 6[input] The Printed Blank[input]Question Blank 2 of 6[input] hypothesis is: There is NO difference between population mean attitudes towards environmental regulations before and after watching the documentary. In symbols, the null hypothesis is: Printed Blank[input]Question Blank 3 of 6[input] Instead of having you run through the steps to calculate a t-test by hand, I'll give you the calculated t-test statistic: t = -2.93. Now, find your t-critical value: Printed Blank[input]Question Blank 4 of 6[input]. Should you reject or fail to reject the null? Printed Blank[input]Question Blank 5 of 6[input] In your interpretation, you'd say: We Printed Blank[input]Question Blank 6 of 6[input] enough evidence to reject the null. We can conclude that there is a difference between population mean attitudes towards environmental regulations before and after watching the documentary. independent samples t-test1-sample t-testdependent samples t-testcorrelationz-testalternativenullµ = 30µ ≠ 30µbefore = µafterµbefore ≠ µafter2.0642.0602.0522.056fail to rejectrejectdo not havehave

A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their competition scores were recorded before and after the training. The results are shown below.   Subject A B C D E F G Before 9.5 9.4 9.6 9.5 9.5 9.6 9.7 After 9.6 9.6 9.6 9.4 9.6 9.9 9.5 Using a 0.01 level of significance, test the claim that the training technique is effective in raising the gymnasts' scores. (Before-After)    Complete the following analysis. 1. This is a [ Select ] right two left -tailed test. 2. The test statistic is equal to t=-0.88 . 3. If the value of the test statistic is less than -3.14 then you will reject the null hypothesis.  4. If the P-value is less than [ Select ] .005 .01 .1 .05 , then you will reject the null hypothesis. 5. The P-value is equal to [ Select ] 0.21 0.054 .14 0.05 .  6. [ Select ] Fail to reject Reject the null hypothesis. 7. At the 1% significance level, there [ Select ] is is not evidence to support the claim that the technique is effective in raising the gymnasts' scores.   (You should technically be checking graphs to make sure that certain assumptions are met. Of course, this data doesn't meet these conditions, so in real life you wouldn't do this analysis. You would need another method than what I have given you. However, we are just going to check that we can do the computations, even if they are meaningless in this case.)          

When should we use a matched pairs method for testing two population means?

More Practical Tools for Students Powered by AI Study Helper

Join us and instantly unlock extensive past papers & exclusive solutions to get a head start on your studies!