Question

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.

Short answer

Part (a) Interpretation: We can be $95\%$positive that the average person watched $4.08$worth of television every day last year.

Variable	N	Mean	StDev	SE Mean	95% Lower Bound	T	F
CI	30	52.0403	2.8086	0.5124	51.1696	101.55	0.000

Part (b) because 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.

Step-by-step solution

Part (a) Step 1: Given information

55.02	50.76	52.08	57.03	52.13	53.51
51.31	51.46	46.35	55.29	45.52	54.10
55.29	50.34	47.18	53.79	50.68	51.52
50.45	51.75	53.61	51.97	51.91	54.32
48.04	53.34	53.50	55.98	49.06	53.92

Part (a) Step 2: Concept

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

Part (a) Step 3: Explanation

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.

Part (a) Step 4: Calculation

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.

Part (b) Step 1: Explanation

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.

Part (c) Step 1: Explanation

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 $95\%$