Answer to Question #199878 in Statistics and Probability for Delmundo

Question

M.Y University claims that some of their BS accountancy first year students shifted to online classes because of the difficulties they’d encountered, which is normally distributed. A random sample of 200 students revealed that 55 of them shifted.

Compute for the proportion estimate of the parameter.

Compute for the margin of error of the proportion at 90% CL.

Find 95% CI for all the students who shifted.

Interpret the result.

can someone solve this please

Answer 1

a) The sample proportion is computed as follows, based on the sample size $N=200$

and the number of favorable cases $X=55:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{55}{200}=0.275$$

b) The critical value for $\alpha=0.1$ is $z_{1-\alpha/2}=1.6449.$

The margin of error of the proportion at 90% CL

$$ME=z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$$

$$=1.6449\times\sqrt{\dfrac{0.275(1-0.275)}{200}}$$

$$\approx 0.051935$$

c) The sample proportion is computed as follows, based on the sample size $N=200$

and the number of favorable cases $X=55:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{55}{200}=0.275$$

The critical value for $\alpha=0.05$ is $z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)$$

$$=(\hat{p}-z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=\bigg(0.275-1.96\times\sqrt{\dfrac{0.275(1-0.275)}{200}},$$

$$0.275+1.96\times\sqrt{\dfrac{0.275(1-0.275)}{200}}\bigg)$$

$$=(0.213116, 0.336884)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.213116<p<0.336884),$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.213116, 0.336884).$