Answer to Question #318077 in Statistics and Probability for Owen

Question

3. Based from record, the population standard deviation of the monthly income of all newly hired IT graduates is Php500.00. A group of researchers surveyed 20 newly hired IT graduates and obtained a mean monthly income of Php33, 500.00. Assuming that their monthly income is approximately normally distributed, Find:

a. the point estimate of population

b. the margin of error if the confidence level is at 90%.

c. the 98% confidence interval of the monthly income of all the newly hired IT graduates.

4. XYZ University claims that some of their BS Accountancy first year students shifted to another course because of their retention policy, which is normally distributed. A random sample of 200 students revealed that 55 of them shifted to another course after their first year.

a. Compute the proportion estimate of the parameter.

b. Find the 95% confidence interval of the population proportion.

Answer 1

$a:\\\hat{\mu}=\bar{x}=33500\\b:\\E=\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}}=\frac{500}{\sqrt{20}}\cdot 1.645=183.917\\c:\\\left( \bar{x}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\bar{x}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 33500-\frac{500}{\sqrt{20}}\cdot 2.326,33500+\frac{500}{\sqrt{20}}\cdot 2.326 \right) =\\=\left( 33239.9,33760.1 \right) \\4.a:\\\hat{p}=\frac{55}{200}=0.275\\b:\\\left( \hat{p}-\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 0.275-\sqrt{\frac{0.275\cdot 0.725}{200}}\cdot 1.96,0.275+\sqrt{\frac{0.275\cdot 0.725}{200}}\cdot 1.96 \right) =\\=\left( 0.213116,0.336884 \right)$