M\&Ms. In the article "Sweetening Statistics-What M\&M's Can Teach Us" (Minitab Inc., August $2008$), M. Paret and E. Martz discussed several statistical analyses that they performed on bags of M\&Ms. The authors took a random sample of $30$ small bags of peanut M\&Ms and obtained the following weights, in grams (g).

a. Determine a $95\%$lower confidence bound for the mean weight of all small bags of peanut M\&Ms. (Note: The sample mean and sample standard deviation of the data are $52.040\mathrm{g}$and $2.807\mathrm{g}$respectively.)

b. Interpret your result in pant (a).

c. According to the package, each small bag of peanut M\&Ms should weigh $49.3\mathrm{g}$Comment on this specification in view of your answer to part (b) It provides equal confidence with a greater lower limit.

Part (c) Because the weight of $49.3\mathrm{g}$ is below the $95\%$ lower confidence bound.

The formula used: Lower confidence bound$\overline{x}-t_{\alpha }·\frac{s}{\sqrt{n}}$

Because the sample size is moderate $(n=30)$, we must first evaluate the issues of normalcy and outliers. To accomplish so, we create a normal probability plot in the picture using Minitab. There are no outliers in the plot, and it follows a fairly straight path. The normal probability map shown below gives strong evidence for the population being regularly distributed. Finally, for the mean weight of all small bags of peanut M&Ms, we find a $95\%$ lower confidence bound.

Step1 We want a $95\%$lower confidence bound, so $\alpha =1-0.95=0.05$For $n=30$, we have degrees of freedom, $df=n-1=30-1=29$From the $t-$distribution tables, $t_{\alpha }=1.699$

Step2 From step1, $t_{\alpha }=1.699$Applying the usual formulas for $\overline{x}$and $s$to the data given in table gives $\overline{x}=52.040\mathrm{g}$and $s=2.807\mathrm{g}$So a $95\%$lower confidence bound for $\mu$is from

$$\begin{gathered} \overline{x}-t_{\alpha }·\frac{s}{\sqrt{n}}=52.040-1.699·\frac{2.807}{\sqrt{30}} \\ =52.040-0.871 \\ =51.169 \end{gathered}$$

$95\%$ lower confidence bound is provided by the Minitab display.

Step3 Interpretation: We may be a $95\%$lower certain that the average person watched$4.08$the worth of television each day last year, according to the related Minitab display.

The one-sided interval does not bind $\mu$from above, but it still achieves $95\%$confidence with a lower bound that is higher. If only the lower bound $\mu$is of importance, the one-sided interval is favored since it provides equal confidence while having a larger lower limit.

Each small bag of peanut M&Ms should weigh $49.3\mathrm{g}$according to the packaging. In light of the solution in part (b), the comment on this weight specification of $49.3\mathrm{g}$ is not appropriate. Because the weight of $49.3\mathrm{g}$ is less than the lower confidence bound of $95\%$