(iii) How large would the sample size need to be so that the margin of error for your 95% confidence interval is at most 5? (iv) Unfortunately, Juanita has not taken statistics in college, and treats her sample variance as the population variance, and carries out a test accordingly. Find out what her p-value is, and whether she accepted or rejected Ho at the same level of significance as in part (ii) above,

Given: Sample size, n = 9

          Sample mean, m = 94.619

          SAmple variance, s² = 49.0001  

Part (iv):  Since the population variance is unknown, the margin of error "E" to be at most 5 implies

               $\begin{array}{rcl} E & =& 5\\ & \Rightarrow & t_{(n-1, \alpha /2)}\times \sqrt{\raisebox{1ex}{$s^2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.} = 5\\ & \Rightarrow & 2.306\times \sqrt{\frac{49.0001}{n}} =5\\ & \Rightarrow & n = {\left(\frac{ 2.306\times \sqrt{49.0001}}{5}\right)}^2\\ & \Rightarrow & n = 10.4 ~ 10 \end{array}$