Question

A survey of n = 100 students found that 28 of them were majoring in a STEM field. Give a 95% confidence interval estimate for the true proportion, p, of students who major in a STEM field.

Answer 1

Answer:

$0.28 - 1.96\sqrt{\frac{0.28(1-0.28)}{100}}=0.192$

$0.28 + 1.96\sqrt{\frac{0.28(1-0.28)}{100}}=0.368$

We are 95% that the true proportion of interest for the students were majoring in a STEM field is between 0.192 and 0.368

Step-by-step explanation:

We want to calculate a confidence interval for the true proportion p who represent the % of students were majoring in a STEM field. The best point of estimate for this proportion is given by:

$\hat p = \frac{X}{n}= \frac{28}{100}=0.28$

The confidence interval for the true proportion of interest is given by:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Since the confidence level is 95%, the significance level would be $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. The critical values for this case are:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

And after replace into the formula we got:

$0.28 - 1.96\sqrt{\frac{0.28(1-0.28)}{100}}=0.192$

$0.28 + 1.96\sqrt{\frac{0.28(1-0.28)}{100}}=0.368$

We are 95% that the true proportion of interest for the students were majoring in a STEM field is between 0.192 and 0.368