Microsoft Word - Norm exam 4 sp 20.docx Q1)____ 20 pts Q2)____ 10 pts Q3)____ 5 pts Total_______ 35 pts #1) A company owning a national chain of health clubs is starting an ad campaign to encourage...

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Microsoft Word - Norm exam 4 sp 20.docx Q1)____ 20 pts Q2)____ 10 pts Q3)____ 5 pts Total_______ 35 pts #1) A company owning a national chain of health clubs is starting an ad campaign to encourage members to hire one of the clubs’ personal trainers. The ads contend that not only does using a personal trainer lead to more effective use of the clubs’ equipment, but also “increases the member’s motivation to spend more time using the clubs”. As evidence to support their claim, the company selected a simple random sample of 100 members who had contracted to use one of the clubs personal trainers, and recorded the total number of hours each such member has spent in the club since Sept. 1 of 2019. Similarly the company selected a simple random sample of 100 members who had not used any personal trainer, and recorded the total number of hours each of those members spent in the club since last Sept. 1 . Some illustrative data values and some summarized results are shown below: “YES” used trainer “NO” didn’t use trainer ------------------------------------- -------------------------------------------- ……….. ………………… Summary results: ………. …………………. Sample mean for ‘Yes’ = 80 ……… ………………… Sample mean for ‘No’ = 58 92 39 164 68 standard deviation for ‘Yes’ = 60 45 103 standard deviation for ‘No’ = 80 ----------- ------------------ ---------- ------------------- 2 pts. a) Explain clearly why we need to treat these as “independent samples”, rather than “paired”. 8 pts b) Construct a 95% confidence interval for the true difference in population means of “number of hours spent in the club since Sept. 1” for those “YES” and those “NO”. Be sure to summarize/interpret what your interval says verbally (including whether we are/aren’t able to be as much as 95% confident that the population mean for “YES” is greater than the population mean for “NO”). 4 pts c) To what extent does this study and its results “allow you to argue” AND/OR “not allow you to argue” that using a personal trainer is the reason for the “YES” group having a higher population mean of “hours spent in the club since Sept. 1” than the “NO” group ? Explain your reasoning. 6 pts d) Suggest a variable that you feel would be a good basis for creating “matched pairs” for this study (i.e. where within each pair, 1 person did have a personal trainer and 1 person did not). Explain the rationale for your choice. Suppose you followed through on your suggestion, formed 100 pairs of members , analyzed the data and constructed a 95% confidence interval for the “Difference in population means” based on those 100 pairs. What would you be able to look at, to see whether your idea (above) for the basis for creating the pairs was a good one (i.e. where could you see “how much benefit” you got from doing the extra effort to create the 100 paired samples) ? NOT RELATED TO QUESTION 1 2) A national health club chain has begun a study, which they hope can provide results touting the benefits provided by using one of their personal trainers. They have selected a random sample of 300 new members (i.e. who joined the club sometime between Jan. 1 and April 1, 2019). Of those 300, 190 did not make use of a personal trainer during the following year, while 110 did make use of one of the club’s personal trainers (for a moderate fee). On the 1-year anniversary of having joined the club, the change in Body Mass Index (from that recorded at their enrollment) was noted (BMI is a measure of body fat based on height weight). 2 pts each a) What is the ‘Response variable’ for this study? What is the Explanatory (Predictor) variable ? b) Is this an observational study, or an experiment (explain your reasoning) ? c) Address each of the following potential problems, and explain why you think each WOULD or WOULD NOT be a potential complication for this study. n Confounding variables and implication for causation n Ecological validity n Extending the results inappropriately 5 pts 3) A counselor in the Admissions Office of a local university has compiled a database of information on a random sample of 85 colleges and universities in the U.S. For purposes of this question the relevant variables are : n “Tuition” (in $, per semester, ranging from $4800 to $24,000) n “Percent of applicants offered admission to the school (in %, ranging from 12% to 89%) To investigate the possible relationship between “Acceptance %” and “Tuition”, the following results were obtained: Regression Equation Tuition = 21123 - 137.5 AccPct What are the values for b0 and b1 in this problem ? Explain the meaning of each of these coefficients IN THE CONTEXT OF THIS PROBLEM (be sure your explanation includes any relevant units). Are there any limitations/cautions on your interpretations ?
Answered Same DayApr 27, 2021

Answer To: Microsoft Word - Norm exam 4 sp 20.docx Q1)____ 20 pts Q2)____ 10 pts Q3)____ 5 pts Total_______ 35...

Pooja answered on Apr 27 2021
140 Votes
1)
a) The two groups of “YES” used trainer and “NO” didn’t use trainer are not related to each other (by same
participant or any other form). Thus, we treat these as “independent samples”.
b)    sample 1    sample 2
n=    100.000    100.000
mean=    80.0000000    58.0000000
s=    60.0000000    80.0000000
s^2/n    36.0000    64.0000
Sp^2 = {(n1-1)*s1^2 + (n2-1)*s2^2} / {n1+n2-2}
Sp^2 = ((100-1)*60^2+(100-1)*80^2)/(100+100-2)
Sp^2 = 5000.0000
df= n1+n2-2
100+100-2
198
critical value: t(a/2, df)
=t(0.05/2,198)
=t.INV.2T(0.05,198)
=1.9720
CI = (Xbar1 - Xbar2) +- t(a/2,df)*sqrt(Sp^2*(1/n1+1/n2))        
lower = (80-58)-1.97202*SQRT(5000*(1/100+1/100)) = 2.2798
upper = (80-58)+1.97202*SQRT(5000*(1/100+1/100)) = 41.7202
i am 95% confident that the estimated difference in population means of “number of hours spent in the club since Sept. 1” for those “YES” and those “NO” lie in the interval (2.28, 41.72).
Since the confidence interval is positive (contains all values greater than 0), i can say that the population mean for “YES” is greater than the population mean for “NO” at 5% level of significance (or...
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