Question

In a sample of college seniors, responded positively when asked if they have spring fever. Based upon this, compute a confidence interval for the proportion of all college seniors who have spring fever. Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.

What is the lower limits 95% confidence interval ?
What is the higher limit 95% confidence interval ?

Answer 1

Answer on Question #41512 – Math - Statistics and Probability

$$\alpha = .05, 1 - \alpha = .95, \frac{\alpha}{2} = .025, z_{.025} = 1.960.$$

Lower limit for 95% confidence interval:

$$\bar{p} - z_{\frac{\alpha}{2}} \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \bar{p} - 1.960 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}}, \text{ where } \bar{p} = \frac{x}{n}, x = \text{number having the characteristic in a random sample of size } n.$$

Upper limit for 95% confidence interval:

$$\bar{p} + z_{\frac{\alpha}{2}} \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \bar{p} + 1.960 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}}, \text{ where } \bar{p} = \frac{x}{n}, x = \text{number having the characteristic in a random sample of size } n.$$

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