Question

Sixteen randomly chosen people suffering from excess weight are placed on a year-long exercise regimen and their weights before and after are compared. The claim is that this exercise regimen will allow participants to lose an average of at least 20 lbs in a year. After one year, the weight losses for each participant were measured. The sample data were plotted and found to follow an approximately Normal distribution with no outliers. The average weight loss in this group was 24.7 lbs. with a sample standard deviation of 12.4 lbs. (Please only answer one part at a time to give other students a chance to answer as well! Start with the first one!) 1. Explain the circumstances under which it is appropriate to use a z-procedure and under which circumstances one should use a t-procedure. What is the appropriate procedure to use in this case? 2. State the null hypothesis and the alternative hypothesis for this test. Is it one-sided or two-sided? 3. What is the degree of freedom of this statistical test? 4. Calculate the test statistic for this scenario. 5. Based on these results, can we conclude at the 5% significance level that the exercise regimen produces the claimed average weight loss? 6. What about the 10% significance level?

Answer 1

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 standard deviation i.e

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

So,in our case we have sample standard deviation, hence we going to 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 test is 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 of significance if t > t_0.05,15

t_0.05,15 = 1.753, So here 1.52 < 1.753m

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

Part 6). = 10% = 0.1 So, t_0.1,15 = 1.341

Compare  t_0.1,15 with the calculate value of t,  t > t_0.1,15 = 1.52 > 1.341

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