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The lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
How to find the confidence interval for population proportion from large sample?
Suppose we're given that:
- Favorable Cases X (in count, in sample)
- Sample Size N
- Level of significance = [tex]\alpha[/tex]
Then, the sample proportion of favorable cases is:
[tex]\hat{p} = \dfrac{X}{N}[/tex]
The critical value at the level of significance [tex]\alpha[/tex] is [tex]Z_{1- \alpha/2}[/tex]
The corresponding confidence interval is:
[tex]CI = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)[/tex]
We need to construct the 99% confidence interval for the population proportion.
We have been provided with the following information about the number of favorable cases(the number of tears located between 3 and 6 mm from the meniscus rim were healed):
- Favorable Cases X =25
- Sample Size N =37
The sample proportion is computed as follows, based on the sample size N = 37 and the number of favorable cases X = 25
[tex]\hat p = \displaystyle \frac{X}{N} = \displaystyle \frac{25}{37} = 0.676[/tex]
The critical value for [tex]\alpha = 0.01[/tex] is [tex]z_c = z_{1-\alpha/2} = 2.576[/tex]
The corresponding confidence interval is computed as shown below:
[tex]\begin{array}{ccl} CI & = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) \\\\ \\\\ & = & \displaystyle \left( 0.676 - 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}}, 0.676 + 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}} \right) \\\\ \\\\ & = & (0.477, 0.874) \end{array}[/tex]
Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.477<p<0.874, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.477, 0.874).
Thus, the lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
Learn more about confidence interval for population proportion here:
https://brainly.com/question/16427118
