# Sixteen randomly chosen people

Sixteen randomly chosen people suffering from excess weight areplaced on a year-long exercise regimen and their weights before andafter are compared. The claim is that this exercise regimen willallow participants to lose an average of at least 20 lbs in a year.After one year, the weight losses for each participant weremeasured. The sample data were plotted and found to follow anapproximately Normal distribution with no outliers. The averageweight loss in this group was 24.7 lbs. with a sample standarddeviation of 12.4 lbs. (Please only answer one part at a time togive other students a chance to answer as well! Start with thefirst one!) 1. Explain the circumstances under which it isappropriate to use a z-procedure and under which circumstances oneshould use a t-procedure. What is the appropriate procedure to usein this case? 2. State the null hypothesis and the alternativehypothesis for this test. Is it one-sided or two-sided? 3. What isthe degree of freedom of this statistical test? 4. Calculate thetest statistic for this scenario. 5. Based on these results, can weconclude at the 5% significance level that the exercise regimenproduces the claimed average weight loss? 6. What about the 10%significance level?

Given :-

Sample size n = 16

= 24.7 lbs

S = 12.4 lbs

= 20 lbs

Part 1). We use Z procedure when we know the population standarddeviation i.e

t procedure is used when we are given sample standard deviationi.e S

So,in our case we have sample standard deviation, hence we goingto use t test

Part 2). H0 = = 20 Vs. H1 = > 20

In Alternative hypothesis we have used greater than (>) sign,hence it is one sided test

Part 3). sample size is 16, hence degree of freedom for t testis n-1 = 16-1 = 15

So, degree of freedom is 15.

Part 4) Test Statistic t = ( ) / (S / )

Putting the values t = ( 24.7 – 20 ) / (12.4 / ) = 1.52

Part 5). Test Criteria :- Reject H0 at 5% level ofsignificance if t > t0.05,15

t0.05,15 = 1.753, So here 1.52 < 1.753m

Result :- Do not reject null hypothesis at 5% level ofsignificance i.e  = 20, exercise regimen will allow participants to lose an averagenot more than 20 lbs.

Part 6). = 10% = 0.1 So, t0.1,15 = 1.341

Compare  t0.1,15 with the calculate valueof t,  t > t0.1,15 = 1.52 > 1.341

Result :- We accept the alternative hypothesis at 10% level ofsignificance i.e > 20,  exercise regimen will allow participants tolose an average of at least 20 lbs in a year.

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