Answer:
(16.528, 17.512) is a 99% two-sided confidence interval for the true mean yield. The sample mean is 17.02 and the sample standard deviation is 0.299.
Step-by-step explanation:
We have a small sample of size n = 6, where the sample mean and standard deviation are [tex]\bar{x} = 17.02[/tex] and s = 0.299. The confidence interval is given by [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n-1=5 degrees of freedom, this because we are dealing with a small sample which comes from the normal distribution (the pivotal quantity is [tex]T=\frac{\bar{X}-\mu}{S/\sqrt{n}}[/tex]). As we want the 99% confidence interval, we have that [tex]\alpha = 0.01[/tex] and the confidence interval is [tex]17.02\pm t_{0.005}(\frac{0.299}{\sqrt{6}})[/tex] where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with 5 df, i.e., [tex]t_{0.005} = -4.0321[/tex]. Then, we have [tex]17.02\pm (-4.0321)(\frac{0.299}{\sqrt{6}})[/tex] and the 99% confidence interval is given by (16.528, 17.512)
