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For patients with a history of osteomyelitis, it is desired to study a tannium implant method. Suppose the The effectiveness of the tannium implant is determined by the amount of bone growth 1 year after place the implant. The mean bone growth for patients with no history of previous infection is about 3.2 mm. Based on her clinical experience, Dr. Calderón believes that bone growth in patients with a history of previous infection is slower than that of those without a medical history. To confirm her belief, Dr. Calderón wishes to test her hypotheses with a level of significance! = 0.05. In fact, You want to be able to detect a significant difference of 0.9 mm with a probability of 0.90. Suppose the The amount of bone growth is normally distributed with a variance &2 = 2.68. How many patients with history of previous bone infection required for investigation

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Answer to a math question For patients with a history of osteomyelitis, it is desired to study a tannium implant method. Suppose the The effectiveness of the tannium implant is determined by the amount of bone growth 1 year after place the implant. The mean bone growth for patients with no history of previous infection is about 3.2 mm. Based on her clinical experience, Dr. Calderón believes that bone growth in patients with a history of previous infection is slower than that of those without a medical history. To confirm her belief, Dr. Calderón wishes to test her hypotheses with a level of significance! = 0.05. In fact, You want to be able to detect a significant difference of 0.9 mm with a probability of 0.90. Suppose the The amount of bone growth is normally distributed with a variance &2 = 2.68. How many patients with history of previous bone infection required for investigation

Expert avatar
Murray
4.5
66 Answers
To calculate the sample size, we use the formula for detecting a difference in means:

n = \left( \frac{(Z_{\alpha} + Z_{\beta})^2 \cdot \sigma^2}{\Delta^2} \right)

Where:
- Z_{\alpha} is the z-value corresponding to the level of significance, \( \alpha = 0.05 \)
- Z_{\beta} is the z-value corresponding to the power, \( \beta = 1 - 0.90 = 0.10 \)
- \sigma^2 is the variance (\(2.68\))
- \Delta is the detectable difference (\(0.9\))

First, find the critical values:

Z_{\alpha} = 1.645 \text{ (one-tailed test at } \alpha = 0.05)

Z_{\beta} = 1.28 \text{ (power of 0.90)}

Next, substitute the values into the formula:

n = \left( \frac{(1.645 + 1.28)^2 \cdot 2.68}{0.9^2} \right)

Simplify the expression inside the parentheses:

(1.645 + 1.28)^2 = 2.925^2 = 8.55625

Now, calculate the sample size:

n = \left( \frac{8.55625 \cdot 2.68}{0.9^2} \right)

n = \left( \frac{22.9488}{0.81} \right)

n \approx 28.3368

Since the sample size must be an integer, round up to the next whole number:

n = 39

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