116k views
3 votes
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.

User BuffK
by
7.6k points

1 Answer

3 votes

Answer:


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


0.28 + 1.96\sqrt{(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

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 = (X)/(n)= (28)/(100)=0.28

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


\hat p \pm z_(\alpha/2)\sqrt{(\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{(0.28(1-0.28))/(100)}=0.192


0.28 + 1.96\sqrt{(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

User Micadelli
by
8.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories