Question #336179

X=842, n=1200, 95% confidence

1
Expert's answer
2022-05-02T13:30:28-0400

The sample proportion is computed as follows, based on the sample size N=1200N = 1200 and the number of favorable cases X=842:X = 842:


p^=XN=84212000.701667\hat{p}=\dfrac{X}{N}=\dfrac{842}{1200}\approx0.701667

The critical value for α=0.05\alpha = 0.05 is zc=z1α/2=1.96.z_c = z_{1-\alpha/2} = 1.96.

The corresponding confidence interval is computed as shown below:


CI(Proportion)=(p^zcp^(1p^)n,CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},

p^+zcp^(1p^)n)\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})

=(0.7016671.960.701667(10.701667)1200,=(0.701667-1.96\sqrt{\dfrac{0.701667(1-0.701667)}{1200}},

0.701667+1.960.701667(10.701667)1200)0.701667+1.96\sqrt{\dfrac{0.701667(1-0.701667)}{1200}})

=(0.615,0.788)=(0.615,0.788)

Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.615<p<0.788,0.615 < p < 0.788, which indicates that we are 95% confident that the true population proportion pp is contained by the interval (0.615,0.788).(0.615, 0.788).



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