Question

Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. 13.03 14.98 13.12 13.13 13.11 13.03 13.16 14.98 13.06 13.05 13.12 13.13 2. Sand data Part: 0/2 WS Part 1 of 2 (a) Construct a 95% confidence interval for the mean weight. Round the answers to at least three decimal places. A 95% confidence interval for the mean weight is ________

Answer 1

Answer:

95% Confidence interval:  (12.93, 13.87)      

Step-by-step explanation:

We are given the following in the question:

13.03, 14.98, 13.12, 13.13, 13.11, 13.03, 13.16, 14.98, 13.06, 13.05, 13.12, 13.13

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{160.9}{12} = 13.4$

Sum of squares of differences = 5.94

$S.D = \sqrt{\dfrac{5.94}{11}} = 0.74$

95% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 11 and}~\alpha_{0.05} = \pm 2.201$  

$13.4 \pm 2.201(\dfrac{0.74}{\sqrt{12}} )\\\\ = 13.4 \pm 0.4701 = (12.9299 ,13.8701)\\\approx (12.93, 13.87)$