Question

Refer to Exercise 80. Construct and interpret a $95\%$ confidence interval for the population mean $M$. What additional information does the confidence interval provide?

Short answer

The $95\%$confldence interval is $(297.46,300.61)$. It intends that there are $95\%$chances that the mean amount of bottle fllled is somewhere in the range of $297.46$and $300.61\mathrm{ml}$.

Step-by-step solution

Given information

From Question 80, Bottles of a popular cola are supposed to contain $300$milliliters (ml) of cola. Because the filling technology is not entirely precise, there is some fluctuation from bottle to bottle. The distribution of the information, in my experience, is about Normal. An inspector measures the contents of six randomly selected bottles from a single day’s production. The results are$299.4297.7301.0298.9300.2297.0$.

We need to construct and interpret a $95\%$confidence interval for the population mean$M$.

Explanation

The dataset is

$299.4$

$297.7$

$301$

$298.9$

$300.2$

$297$

The formula to compute the t-interval is:

$CI=\overline{x}\pm t_{\alpha /2}\times \frac{s}{\sqrt{n}}$

The obtained output using Ti-83 plus calculator is:

From the above output, the $95\%$confidence interval is $(297.46,300.61)$. It intends that there are $95\%$chances that the mean amount of bottle filled is somewhere in the range of $297.46$and $300.61\mathrm{ml}$. Here, 300 lies in the registered confidence interval. Subsequently, there is sufficient evidence to conclude that the mean amount of cola filled in all bottles isn't unique at about $300\mathrm{ml}$.