SOLUTION: The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers.
52 30 44 35 39 26
47 37 56 26 39 28
Note: a normal probability plo
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Question 1181344: The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers.
52 30 44 35 39 26
47 37 56 26 39 28
Note: a normal probability plot suggests that the data could come from a population that is normally distributed. A boxplot indicates there are no outliers.
a) (15 points) Construct a 99% confidence interval for the mean age at which a baby first crawls.
b) (Explain what the confidence interval you constructed means.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to construct the confidence interval and interpret it:
**a) Constructing a 99% Confidence Interval:**
1. **Calculate the sample mean (x̄):**
x̄ = (52 + 30 + 44 + 35 + 39 + 26 + 47 + 37 + 56 + 26 + 39 + 28) / 12
x̄ = 429 / 12
x̄ = 35.75 weeks
2. **Calculate the sample standard deviation (s):**
First, find the squared differences from the mean:
| Age (x) | x - x̄ | (x - x̄)² |
|---|---|---|
| 52 | 16.25 | 264.0625 |
| 30 | -5.75 | 33.0625 |
| 44 | 8.25 | 68.0625 |
| 35 | -0.75 | 0.5625 |
| 39 | 3.25 | 10.5625 |
| 26 | -9.75 | 95.0625 |
| 47 | 11.25 | 126.5625 |
| 37 | 1.25 | 1.5625 |
| 56 | 20.25 | 410.0625 |
| 26 | -9.75 | 95.0625 |
| 39 | 3.25 | 10.5625 |
| 28 | -7.75 | 60.0625 |
| **Total** | | **1175.5** |
s = √[Σ(x - x̄)² / (n - 1)]
s = √(1175.5 / 11)
s ≈ √106.86
s ≈ 10.34 weeks
3. **Find the critical t-value:**
Since the sample size is small (n < 30) and the population standard deviation is unknown, we use the t-distribution. For a 99% confidence interval with 11 degrees of freedom (n - 1 = 12 - 1 = 11), the critical t-value (t*) is approximately 3.106 (you can find this using a t-table or calculator).
4. **Calculate the margin of error (E):**
E = t* * (s / √n)
E = 3.106 * (10.34 / √12)
E ≈ 3.106 * 2.98
E ≈ 9.26 weeks
5. **Construct the confidence interval:**
Lower Bound = x̄ - E = 35.75 - 9.26 ≈ 26.49 weeks
Upper Bound = x̄ + E = 35.75 + 9.26 ≈ 45.01 weeks
Therefore, the 99% confidence interval is approximately (26.49, 45.01).
**b) Interpretation of the Confidence Interval:**
We are 99% confident that the true mean age at which babies first crawl in the population is between 26.49 weeks and 45.01 weeks. This means that if we were to repeat this study many times and calculate a 99% confidence interval each time, 99% of those intervals would contain the true population mean.
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