Question

To estimate a proportion of students with a grade point average of 3.7 and higher in a college, a random sample of 200 students was taken and the number of successes counted to be 35. The 95% confidence interval for a population proportion is:

Answer 1

Answer: $(0.1481323,\ 0.2018677)$

Step-by-step explanation:

The confidence interval for population proportion :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , where n= sample size, $\hat{p}$ = sample proportion , z*= Critical  z-value.

Let p = population proportion of successes.

Given: n= 200 , $\hat{p}=\dfrac{35}{200}=0.175$

Critical z value for 95% confidence = 1.96

The 95% confidence interval for a population proportion is:

$0.175\pm (1.96)\sqrt{\dfrac{(0.175)(1-0.175)}{200}}\\\\=\ 0.175\pm (1.96)\sqrt{0.000721875}\\\\= 0.175\pm (1.96)(0.0268677)\\\\=(0.175-0.0268677,0.175+0.0268677)\\\\= (0.1481323,\ 0.2018677)$

Required confidence interval: $(0.1481323,\ 0.2018677)$