Question

for the sample of 39 individuals who had taken olanzapine, the article reported (7.595, 9.915) as a 95% ci for true average weight gain (kg). what is a 99% ci? (round your answers to three decimal places.)

Answer 1

The 99% confidence interval for the true average weight gain based on the provided 95% CI (7.595, 9.915) is approximately (7.380, 10.130) kg. This calculation involves a point estimate of 8.755 kg, a standard error of 0.645 kg, and a critical value of 2.576.

Explanation:

To calculate the 99% confidence interval (CI) for the true average weight gain (kg) based on the provided 95% CI (7.595, 9.915), you can use the formula:

`\[\text{{99\% CI}} = \text{{Point Estimate}} \pm \left( \text{{Critical Value}} * \text{{Standard Error}} \right)\]`

The point estimate is the midpoint of the 95% CI, and the critical value corresponds to the 99% confidence level.

1. Point Estimate:

`\[ \text{{Point Estimate}} = \frac{{\text{{Lower Limit}} + \text{{Upper Limit}}}}{2} \]`

`\[ = \frac{{7.595 + 9.915}}{2} \]`

`\[ \approx 8.755 \text{{ (rounded to three decimal places)}} \]`

2. Standard Error (SE):

The standard error is calculated using the formula:

`\[ \text{{SE}} = \frac{{\text{{Upper Limit}} - \text{{Lower Limit}}}}{{2 * \text{{Z-score}}}} \]`

Since we're moving from a 95% CI to a 99% CI, the Z-score changes. For a 99% confidence level, the Z-score is larger than that for a 95% confidence level. You need to find the Z-score for a 99% confidence interval.

You can use a Z-table or a statistical calculator to find the Z-score for a 99% confidence interval. For a two-tailed 99% confidence interval, the Z-score is approximately 2.576.

`\[ \text{{SE}} = \frac{{9.915 - 7.595}}{{2 * 2.576}} \]`

`\[ \approx 0.645 \text{{ (rounded to three decimal places)}} \]`

3. Critical Value (CV):

The critical value for a 99% confidence interval is 2.576.

4. 99% CI Calculation:

`\[ \text{{99\% CI}} = 8.755 \pm (2.576 * 0.645) \]`

`\[ \text{{99\% CI}} \approx (7.380, 10.130) \text{{ (rounded to three decimal places)}} \]`

Therefore, the 99% confidence interval for the true average weight gain (kg) is approximately (7.380, 10.130).

Answer 2

The 99% CI for the true average weight gain is (5.766, 11.744) when the 95% CI is (7.595, 9.915).

To find the 99% confidence interval (CI) for the true average weight gain given the reported 95% CI (7.595, 9.915), you can use a formula to adjust it.

The formula to adjust a CI when changing the confidence level is:

`\[ \text{New CI} = \text{Old CI} + \text{Z-value} * \text{Standard Error} \]`

For a 95% CI, the Z-value is 1.96, and for a 99% CI, the Z-value is 2.576.

Given the 95% CI (7.595, 9.915), let's first find the mean of the interval:

`\[ \text{Mean} = \frac{\text{Upper Bound} + \text{Lower Bound}}{2} \]`

`\[ \text{Mean} = (7.595 + 9.915)/(2) = (17.51)/(2) = 8.755 \]`

Now, calculate the half-width of the interval:

`\[ \text{Half-width} = \frac{\text{Upper Bound} - \text{Lower Bound}}{2} \]`

`\[ \text{Half-width} = (9.915 - 7.595)/(2) = (2.32)/(2) = 1.16 \]`

For a 99% CI, the Z-value is 2.576. Multiply this by the standard error (half-width):

`\[ \text{New Half-width} = 2.576 * 1.16 = 2.989 \]`

Now, construct the new 99% CI using the mean and the new half-width:

`\[ \text{New Lower Bound} = \text{Mean} - \text{New Half-width} \]`

`\[ \text{New Lower Bound} = 8.755 - 2.989 = 5.766 \]`

`\[ \text{New Upper Bound} = \text{Mean} + \text{New Half-width} \]`

`\[ \text{New Upper Bound} = 8.755 + 2.989 = 11.744 \]`

Therefore, the answer is (5.766, 11.744)