Question

Suppose we wanted to estimate the mean number of hours that all PACollege students spend on their MAT260 homework. We don't know the actual mean, but from previous studies, we have concluded that the standard deviation of all students is 3.2 hours. A sample of 32 students found that the average number of hours spent on homework was 10.4. Create a 95% confidence interval to estimate the average for all students at PACollege.

Answer 1

Given the confidence interval value formula

$CI=\bar{x}\pm z\times\frac{s}{\sqrt[]{n}}$$\begin{gathered} \bar{x}\Rightarrow\operatorname{mean} \\ \bar{x}=10.4 \\ s\Rightarrow\text{standard deviation} \\ s=3.2 \\ \alpha=\frac{95}{100}=0.95 \\ z_{\alpha}=1.96 \\ n=32 \end{gathered}$

Substitute the values defined above and solve for confidence interval

$\begin{gathered} CI=10.4\pm1.96\times\frac{3.2}{\sqrt[]{32}} \\ CI=10.4\pm1.96\times\frac{3.2}{5.66} \\ CI=10.4\pm1.96\times0.57 \\ CI=10.4\pm1.12 \\ CI=10.4-1.12\text{ to 10.4+1.12} \\ CI=9.28\text{ to 11.52} \end{gathered}$

Hence,the average for all students fall in between confidence interval of 9.28 to 11.52