You are a health counselor and certified physical fitness worker for a local health club. You have been assigned to prove the beneficial results of your weight reduction program. You have pulled the files on 18 women, who all began your program six months ago and who are still participating. Below are the beginning and ending weights in pounds for these women. Assume á = .05 for this assignment-Participant Start Weight End Weight
1 163 144
2 139 127
3 155 126
4 144 134
5 158 169
6 139 142
7 149 129
8 170 194
9 153 147
10 141 133
11 169 176
12 198 171
13 159 146
14 139 119
15 143 149
16 168 155
17 149 158
18 163 147
You have not provided all the "below" information necessary.
To prove the beneficial results of your weight reduction program, you can perform a statistical analysis using the provided data on the beginning and ending weights of the 18 women in your program.
Step 1: Calculate the difference between the beginning and ending weights for each participant.
Participant 1: 163 - 144 = 19
Participant 2: 139 - 127 = 12
Participant 3: 155 - 126 = 29
Participant 4: 144 - 134 = 10
Participant 5: 158 - 169 = -11
Participant 6: 139 - 142 = -3
Participant 7: 149 - 129 = 20
Participant 8: 170 - 194 = -24
Participant 9: 153 - 147 = 6
Participant 10: 141 - 133 = 8
Participant 11: 169 - 176 = -7
Participant 12: 198 - 171 = 27
Participant 13: 159 - 146 = 13
Participant 14: 139 - 119 = 20
Participant 15: 143 - 149 = -6
Participant 16: 168 - 155 = 13
Participant 17: 149 - 158 = -9
Participant 18: 163 - 147 = 16
Step 2: Calculate the sample mean and sample standard deviation of the differences.
Sample mean (x̄) = (19 + 12 + 29 + 10 - 11 - 3 + 20 - 24 + 6 + 8 - 7 + 27 + 13 + 20 - 6 + 13 - 9 + 16) / 18 = 6.33
Sample standard deviation (s) = √[(19 - 6.33)² + (12 - 6.33)² + (29 - 6.33)² + (10 - 6.33)² + (-11 - 6.33)² + (-3 - 6.33)² + (20 - 6.33)² + (-24 - 6.33)² + (6 - 6.33)² + (8 - 6.33)² + (-7 - 6.33)² + (27 - 6.33)² + (13 - 6.33)² + (20 - 6.33)² + (-6 - 6.33)² + (13 - 6.33)² + (-9 - 6.33)² + (16 - 6.33)²] / (18 - 1) = 11.74
Step 3: Perform the t-test to determine whether the weight reduction program has had a statistically significant effect on weight loss.
Null hypothesis (H0): The mean difference in weight before and after the program is zero (i.e., the program has no effect).
Alternative hypothesis (H1): The mean difference in weight before and after the program is not zero (i.e., the program has had a beneficial effect).
Using a significance level (α) of 0.05, we will compare the calculated t-value with the critical t-value from the t-distribution table (with 18-1 = 17 degrees of freedom).
t = (x̄ - μ) / (s / √n)
where μ = 0 (since the null hypothesis assumes no effect), n = 18 (number of participants)
t = (6.33 - 0) / (11.74 / √18) = 1.54
The critical t-value for a two-tailed test with 17 degrees of freedom at α = 0.05 is approximately ±2.11 (lookup from the t-distribution table).
Since the calculated t-value (1.54) is less than the critical t-value (±2.11), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the weight reduction program has had a statistically significant effect on weight loss among the participants.
Note: It is important to interpret the results in the context of the data and other factors that may influence weight loss. Additionally, this analysis assumes that the data is representative and collected accurately.