A university has 807 faculty members. For each faculty member, the number of refereed publications was recorded. This number is not directly available on the database, so requires the investigator to examine each record separately. A frequency table number of refereed publication is given below for an SRS of 50 faculty members.
\begin{array}{|c|c|c|c|} \hline \text{Refereed Publications}& 0 & 1 & 2& 3 & 4 & 5& 6 & 7 & 8& 9 & 10 \\ \hline \text{Faculty Members} &28 &4 &3& 4 & 4 & 2& 1 & 0 & 2& 1 & 1\\ \hline \end{array}
b) Estimate the mean number of publications per faculty member, and give the SE for your estimate: \begin{array}{|c|c|c|c|} \hline y& f & f*y & f*y^2 \\ \hline 0 &28 & 0&0\\ \hline 1 & 4& 4&4\\ \hline 2& 3& 6&12\\ \hline 3& 4& 12&36\\ \hline 4& 4& 16&64\\ \hline 5& 2& 10&50\\ \hline 6 & 1& 6&36\\ \hline 7 & 0& 0&0\\ \hline 8& 2& 16&128\\ \hline 9& 1& 9&81\\ \hline 10& 1& 10&100\\ \hline \text{total}& 50&89 &511\\ \hline \end{array}
\begin{align} \bar{y} &= \frac{\sum fy}{\sum f} &\rightarrow& &\frac{89}{50} &= 1.78 \\[10pt] SD &= \sqrt{ \frac{\sum fy^2}{\sum f}-(\bar{y})^2} &\rightarrow& &\sqrt{\frac{511}{50}-(1.78)^2} &= 2.66 \\[10pt] SE(\bar{y}) &= \frac{s}{\sqrt{n}}\sqrt{1-\frac{n}{N}} &\rightarrow& &\frac{2.66}{\sqrt{50}}\sqrt{1-\frac{50}{807}} &= 0.3643 \end{align}
Did I do it correctly?
d) Estimate the proportion of faculty members with no publications and give a $95\%$ CI.
\begin{align} p &= \frac{y}{n} &\rightarrow& &\frac{28}{50} &= 0.56 \\[10pt] SE(p) &= \sqrt{\frac{p(1-p)}{n-1}\bigg(1-\frac{n}{N}\bigg)} &\rightarrow& &\sqrt{\frac{0.56(0.44)}{49}\bigg(1-\frac{50}{807}\bigg)} &= 0.0687 \\[10pt] 95\%\ CI &= p\pm 1.96[SE(p)] &\rightarrow& &0.56 \pm1.96(0.0687) &= [0.425,0.695] \end{align}
Am I using the correct formulas?
