(Solved): Question 2: Construct a \( 99 \% \) confidence interval ( \( 1 \% \) level) whether the average GPA ...

Question 2: Construct a \( 99 \% \) confidence interval ( \( 1 \% \) level) whether the average GPA is different from \( 3.00 \). INTRODUCTION In Question 2, the aim is to find the \( 99 \% \) confidence interval, which contains the population mean. We are also required to determine whether the mean GPA is different from 3.00. In other words, is the mean GPA higher or lower than \( 3.00 \). STATISTICAL WORKING Stating the Hypothesis Null hypothesis \( \mathrm{H}_{0}: \mu(\mathrm{GPA})=3.0 \) Alternative Hypothesis \( \mathrm{H}_{1}: \mu(\mathrm{GPA}) \neq 3.0 \) Level of significance \( \alpha=1 \% \), i.e., \( 0.01 \) This is a two tailed test. Therefore \( \alpha / 2=0.005 \) Determining the appropriate test \( 99 \% \) confidence interval with a large sample \( n=50 \) and population standard deviation is unknown associates with a z- critical value Determining the critical value As it is a Z distribution with the critical value at \( 1 \% \) level of signifance and is a two tailed test, this is associated with the \( z \) value \( 2.57 \) with \( .5 \% \) in each tail with reference to the critical value table posted earlier. For a \( 99 \% \mathrm{Cl}, \mathrm{z}=2.57 \), Thus \( \mathrm{Cl}=\overline{\mathrm{x}} \pm \mathrm{zs} / \mathrm{Vn} \) Thus \( \mathrm{Cl}=3.117 \pm 2.57^{*} 0.603 / \mathrm{V} 50 \) Lower \( \mathrm{Cl}=2.898 \), Upper \( \mathrm{Cl}=3.3363 \). Thus, the GPA of \( 3.00 \) lie within the \( 99 \% \) confidence interval.