Answer to Question #346700 in Statistics and Probability for Pearl

Question #346700

A sample of 390 senior high school students applied to different state universities, wherein 117 are female. Find the 99% confidence interval of the true population proportion who applies for different state universities

1
Expert's answer
2022-06-01T13:23:27-0400

The sample proportion is computed as follows, based on the sample size N=390N = 390 and the number of favorable cases X=117X = 117


p^=XN=117390=0.3\hat{p}=\dfrac{X}{N}=\dfrac{117}{390}=0.3

The critical value for α=0.01\alpha = 0.01 is zc=z1α/2=2.5758.z_c = z_{1-\alpha/2} = 2.5758.

The corresponding confidence interval is computed as shown below:


CI(Proportion)=(p^zc×p^(1p^)N,CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}},

p^+zc×p^(1p^)N)\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}})

=(0.32.5758×0.3(10.3)390,=(0.3-2.5758\times \sqrt{\dfrac{0.3(1-0.3)}{390}},

0.3+2.5758×0.3(10.3)390)0.3+2.5758\times \sqrt{\dfrac{0.3(1-0.3)}{390}})

=(0.24023,0.35977)=(0.24023, 0.35977)

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.24023<p<0.35977,0.24023 < p < 0.35977, which indicates that we are 99% confident that the true population proportion pp is contained by the interval (0.24023,0.35977).(0.24023, 0.35977).


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