Question

1. The "Spring-like effect' of a golf club driver is important for driving a golf ball. The spring-like effect' for drivers is the ratio of the outgoing velocity of the ball to the incoming velocity, which is called the coefficient of restitution of the drive. Coefficient of restitution measurements have been made and the results are below. 0.8411 0.8191 0.8182 0.8125 0.8750 0.8580 0.8532 0.8483 0.8276 0.7983 0.8042 0.8730 0.8282 0.8359 0.8660 1a. Calculate sample mean 1b. Calculate sample variance, by hand (ie make a table with columns showing (ci-캬 i x as shown in Lecture 3) for the first row of 5 data points above) 1c. Calculate sample variance for the entire data set using calculator (or other) 2. Two catalysts are tested for yield on a chemical reaction. Data is gathered and shown below. What are the sample mean and sample variance of samples 1 and 2? You may use calculator (or other) Observation Number Catalyst 1 91.50 94.18 92.18 95.39 91.79 89.07 94.72 89.21 Catalyst 2 89.19 90.95 90.46 93.21 97.19 97.04 91.07 92.75

Answer 1

Sample Mean =  

and

Sample Variance =

1a. Sample Mean = 0.83724

1b. Sample Variance calculation for the first row of 5 data points: where $Q8AsN3sAAAAASUVORK5CYII=$ = 0.83318

0.8411	0.00792	6.27264E-05
0.8191	0.01408	0.000198246
0.8182	0.01498	0.0002244
0.8125	0.02068	0.000427662
0.875	0.04182	0.001748912

Sample Variance for first 5 data points is 0.000665

1c. Sample Variance for entire data set is = 0.000603

2. For Sample 1 (catalyst 1): Sample Mean = 92.255 and sample variance = 5.688314

For Sample 2 (catalyst 2): Sample Mean = 92.7325 and sample variance = 8.900993