Question #41511

In a sample of 350 college seniors, 290 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 limit of 95% confidence interval ?
What is the higher limit of 95% confidence interval ?
1

Expert's answer

2014-04-18T09:52:25-0400

Answer on Question #41511 – Math - Statistics and Probability

α=.05,1α=.95,α2=.025,z.025=1.960,n=350,x=290,\alpha = .05, 1 - \alpha = .95, \frac{\alpha}{2} = .025, z_{.025} = 1.960, n = 350, x = 290,

xx = number having the characteristic in a random sample of size nn.


pˉ=xn=290350\bar{p} = \frac{x}{n} = \frac{290}{350}


Lower limit for 95% confidence interval:


pˉzα2pˉ(1pˉ)n=2903501.960290350(1290350)3500.79.\bar{p} - z_{\frac{\alpha}{2}} \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \frac{290}{350} - 1.960 \sqrt{\frac{\frac{290}{350} \left(1 - \frac{290}{350}\right)}{350}} \approx 0.79.


Upper limit for 95% confidence interval:


pˉ+zα2pˉ(1pˉ)n=290350+1.960290350(1290350)3500.87.\bar{p} + z_{\frac{\alpha}{2}} \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \frac{290}{350} + 1.960 \sqrt{\frac{\frac{290}{350} \left(1 - \frac{290}{350}\right)}{350}} \approx 0.87.


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