Question #330101

1.     In a Harris poll of 630 human resource professionals, 38.4 % said that they had at least one child by the age of 30 years.

a. Among the 630 professionals who were surveyed, how many of them said they had at least one child before 30 years of age.

b. Construct a 99% confidence interval estimate for the proportion of all human resource professionals with at least one child before 30 years of age.

c. Repeat part (b) using a confidence interval level of 90%.

d. Compare the confidence intervals from part (b) and (c) and identify the interval that is wider. Why is it wider?


Expert's answer

a:630⋅0.384=241.92≈242b:LB=p^−p^(1−p^)nz1+γ2=0.384−0.384(1−0.384)630⋅2.5758=0.334089UB=p^+p^(1−p^)nz1+γ2=0.384+0.384(1−0.384)630⋅2.5758=0.433911c:LB=p^−p^(1−p^)nz1+γ2=0.384−0.384(1−0.384)630⋅1.6449=0.352127UB=p^+p^(1−p^)nz1+γ2=0.384+0.384(1−0.384)630⋅1.6449=0.415873d:The  99%interval  is  wider  because  the  wider  interval  the  higher  probability.a:\\630\cdot 0.384=241.92\approx 242\\b:\\LB=\hat{p}-\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}}=0.384-\sqrt{\frac{0.384\left( 1-0.384 \right)}{630}}\cdot 2.5758=0.334089\\UB=\hat{p}+\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}}=0.384+\sqrt{\frac{0.384\left( 1-0.384 \right)}{630}}\cdot 2.5758=0.433911\\c:\\LB=\hat{p}-\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}}=0.384-\sqrt{\frac{0.384\left( 1-0.384 \right)}{630}}\cdot 1.6449=0.352127\\UB=\hat{p}+\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}}=0.384+\sqrt{\frac{0.384\left( 1-0.384 \right)}{630}}\cdot 1.6449=0.415873\\d:\\The\,\,99\% interval\,\,is\,\,wider\,\,because\,\,the\,\,wider\,\,interval\,\,the\,\,higher\,\,probability.


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