Question 1199898: 1.A brochure inviting subscriptions for a new diet program states that the participants are expected to lose their weight over 11Kg in five weeks. Suppose that, from the data of the five-week weight losses of 56 participants, the sample mean and sample standard deviation are found to be 13Kg and 5.5 Kg, respectively.
A.Construct the 95% confidence interval for the average weight lose
B.Test whether this finding is in line with the expected at α = .05 level of significance?
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **A. Constructing the 95% Confidence Interval**
1. **Identify the given values:**
- Sample mean (x̄) = 13 kg
- Sample standard deviation (s) = 5.5 kg
- Sample size (n) = 56
- Confidence level = 95%
2. **Find the critical t-value:**
- Degrees of freedom (df) = n - 1 = 56 - 1 = 55
- Using a t-distribution table or a calculator, find the t-critical value for a 95% confidence level and 55 degrees of freedom. The t-critical value is approximately 2.004.
3. **Calculate the standard error:**
- Standard Error (SE) = s / √n = 5.5 / √56 ≈ 0.735
4. **Calculate the margin of error:**
- Margin of Error = t-critical * SE = 2.004 * 0.735 ≈ 1.47
5. **Construct the confidence interval:**
- Lower bound = x̄ - Margin of Error = 13 - 1.47 = 11.53 kg
- Upper bound = x̄ + Margin of Error = 13 + 1.47 = 14.47 kg
**Therefore, the 95% confidence interval for the average weight loss is (11.53 kg, 14.47 kg).**
**B. Hypothesis Testing**
1. **Set up hypotheses:**
- Null hypothesis (H0): μ = 11 kg (The average weight loss is 11 kg)
- Alternative hypothesis (H1): μ ≠ 11 kg (The average weight loss is not 11 kg)
2. **Calculate the t-statistic:**
- t = (x̄ - μ) / (s / √n) = (13 - 11) / (5.5 / √56) ≈ 2.72
3. **Find the p-value:**
- Using a t-distribution table or a calculator, find the p-value associated with the calculated t-statistic (2.72) and 55 degrees of freedom.
- The p-value is approximately 0.0086.
4. **Compare p-value with significance level:**
- p-value (0.0086) < α (0.05)
5. **Make a decision:**
- Since the p-value is less than the significance level (α = 0.05), we reject the null hypothesis.
**Conclusion:**
At the 0.05 level of significance, there is sufficient evidence to conclude that the average weight loss of participants in the diet program is significantly different from the expected 11 kg. The 95% confidence interval for the average weight loss, (11.53 kg, 14.47 kg), also supports this conclusion as it does not include the expected value of 11 kg.
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