The brightness of a picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in ¯x = 317.2 and s = 15.7 measured in microamps. (a) Find a 99% CI on mean current required.

Answer:The 99% confidence interval for the mean would be (301.064;333.336) mA  Step-by-step explanation:1) Notation and some definitionsn=10 sample selected[tex]\bar x=317.2mA[/tex] sample mean for the sample tubes selected[tex]s=15.7mA[/tex] sample deviation for the sample selectedConfidence = 99% or 0.99[tex]\alpha=1-0.99=0.01[/tex] significance levelA confidence interval for the mean is used to "places boundaries around an estimated [tex]\bar X[/tex] so that the true population mean [tex]\mu[/tex] would be expected to lie within those boundaries with a confidence specified. If the uncertainty is large, then the interval between the boundaries must be wide; if the uncertainty is small, then the interval can be narrow"2) Formula to useFor this case the sample size is <30 and the population standard deviation [tex]\sigma[/tex] is not known, so for this case we can use the t distributon to calculate the critical value. The first step would be calculate alpha [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], then we can calculate the degrees of freedom given by:[tex]df=n-1=10-1=9[/tex]Now we can calculate the critical value [tex]t_{\alpha/2}=3.25[/tex]And then we can calculate the confidence interval with the following formula[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]  (1)3) Calculate the intervalUsing the formula (1) and replacing the values that we got we have:[tex]317.2 - 3.25\frac{15.7}{\sqrt{10}}=301.064[/tex]  [tex]317.2 + 3.25\frac{15.7}{\sqrt{10}}=333.336[/tex]So then the 99% confidence interval for the mean would be (301.064;333.336)mA  Interpretation: A point estimate for the true mean of brightness level for the tubes in the population is 317.2mA, and we are 99% confident that the true mean is between 301.064 mA and 333.336 mA.