A health spa has advertised a weight-reducing program and has claimed that the average participant in the program loses is about 17 pounds. A somewhat overweight executive is interested in the program but is sceptical about the claims and asks for some hard evidence. The spa allows him to select randomly the records of 10 participants and record their weights before and after the program. These data are recorded below: weight Before 189 202 220 207 194 177 193 202 208 233 After 170 179 203 192 172 161 174 187 186 204 Construct the 99% confidence interval on the mean difference of weight before and after the program. Does the claim from the health spa is true? Explain.
Question
A health spa has advertised a weight-reducing program and has claimed that the average participant in the program loses is about 17 pounds. A somewhat overweight executive is interested in the program but is sceptical about the claims and asks for some hard evidence. The spa allows him to select randomly the records of 10 participants and record their weights before and after the program. These data are recorded below: weight Before 189 202 220 207 194 177 193 202 208 233
After 170 179 203 192 172 161 174 187 186 204
Construct the 99% confidence interval on the mean difference of weight before and after the program. Does the claim from the health spa is true? Explain.
Solution
To construct the 99% confidence interval on the mean difference of weight before and after the program, we first need to calculate the mean difference and the standard deviation of the differences.
Step 1: Calculate the differences in weight before and after the program for each participant.
Differences: 19, 23, 17, 15, 22, 16, 19, 15, 22, 29
Step 2: Calculate the mean difference.
Mean difference = (19+23+17+15+22+16+19+15+22+29) / 10 = 20.7 pounds
Step 3: Calculate the standard deviation of the differences.
Standard deviation = sqrt[((19-20.7)^2 + (23-20.7)^2 + (17-20.7)^2 + (15-20.7)^2 + (22-20.7)^2 + (16-20.7)^2 + (19-20.7)^2 + (15-20.7)^2 + (22-20.7)^2 + (29-20.7)^2) / (10-1)] = 4.3 pounds
Step 4: Calculate the standard error of the mean difference.
Standard error = standard deviation / sqrt(number of participants) = 4.3 / sqrt(10) = 1.36 pounds
Step 5: Construct the 99% confidence interval.
The critical value for a 99% confidence interval is approximately 2.82 (from the t-distribution table with 9 degrees of freedom).
Lower limit = mean difference - (critical value * standard error) = 20.7 - (2.82 * 1.36) = 17.1 pounds
Upper limit = mean difference + (critical value * standard error) = 20.7 + (2.82 * 1.36) = 24.3 pounds
So, the 99% confidence interval on the mean difference of weight before and after the program is from 17.1 pounds to 24.3 pounds.
The health spa claimed that the average participant in the program loses about 17 pounds. This value is within our confidence interval, so we do not have enough evidence to reject the spa's claim. Therefore, it seems that the claim from the health spa could be true.
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