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 $(0.213116, 0.336884).$