confint3: 2-Sided Confidence Interval (Extended Moodle Version)
confint3It is suspected that a supplier systematically underfills 5 l canisters of detergent. The filled volumes are assumed to be normally distributed. A small sample of \(13\) canisters is measured exactly. This shows that the canisters contain on average \(4948.1\) ml. The sample variance \(s^2_{n-1}\) is equal to \(352.1\).
Determine a \(95\%\) confidence interval for the average content of a canister (in ml).
What is the lower confidence bound?
What is the upper confidence bound?
The \(95\%\) confidence interval for the average content \(\mu\) in ml is given by: \[ \begin{aligned} & \left[\bar{y} \, - \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ &= \left[ 4948.1 \, - \, 2.1788\sqrt{\frac{352.1}{13}}, \; 4948.1 \, + \, 2.1788\sqrt{\frac{352.1}{13}}\right] \\ &= \left[4936.761, \, 4959.439\right]. \end{aligned} \]
- The lower confidence bound is \(4936.761\).
- The upper confidence bound is \(4959.439\).
It is suspected that a supplier systematically underfills 5 l canisters of detergent. The filled volumes are assumed to be normally distributed. A small sample of \(13\) canisters is measured exactly. This shows that the canisters contain on average \(4956.6\) ml. The sample variance \(s^2_{n-1}\) is equal to \(1591.7\).
Determine a \(99\%\) confidence interval for the average content of a canister (in ml).
What is the lower confidence bound?
What is the upper confidence bound?
The \(99\%\) confidence interval for the average content \(\mu\) in ml is given by: \[ \begin{aligned} & \left[\bar{y} \, - \, t_{n-1;0.995}\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, t_{n-1;0.995}\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ &= \left[ 4956.6 \, - \, 3.0545\sqrt{\frac{1591.7}{13}}, \; 4956.6 \, + \, 3.0545\sqrt{\frac{1591.7}{13}}\right] \\ &= \left[4922.801, \, 4990.399\right]. \end{aligned} \]
- The lower confidence bound is \(4922.801\).
- The upper confidence bound is \(4990.399\).
It is suspected that a supplier systematically underfills 5 l canisters of detergent. The filled volumes are assumed to be normally distributed. A small sample of \(13\) canisters is measured exactly. This shows that the canisters contain on average \(4987.4\) ml. The sample variance \(s^2_{n-1}\) is equal to \(2763.1\).
Determine a \(95\%\) confidence interval for the average content of a canister (in ml).
What is the lower confidence bound?
What is the upper confidence bound?
The \(95\%\) confidence interval for the average content \(\mu\) in ml is given by: \[ \begin{aligned} & \left[\bar{y} \, - \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ &= \left[ 4987.4 \, - \, 2.1788\sqrt{\frac{2763.1}{13}}, \; 4987.4 \, + \, 2.1788\sqrt{\frac{2763.1}{13}}\right] \\ &= \left[4955.635, \, 5019.165\right]. \end{aligned} \]
- The lower confidence bound is \(4955.635\).
- The upper confidence bound is \(5019.165\).
(Note that the HTML output contains mathematical equations in MathML, rendered by MathJax using ‘mathjax = TRUE’. Instead it is also possible to use ‘converter = “pandoc-mathjax”’ so that LaTeX equations are rendered by MathJax directly.)
Demo code:
library("exams")
set.seed(403)
exams2html("confint3.Rmd", mathjax = TRUE)
set.seed(403)
exams2pdf("confint3.Rmd")
set.seed(403)
exams2html("confint3.Rnw", mathjax = TRUE)
set.seed(403)
exams2pdf("confint3.Rnw")