Chapter 14.3, Problem 14.13E

To determine

The provided data is sufficient or not to support the percentage reported by Mars, incorporated.

Answer to Problem 14.13E

Yes, provided data can substantiate to support the percentage reported by Mars, incorporated.

Given:

Disease	Brown	Yellow	Red	Blue	Orange	Green
Deaths	$13\%$	$14\%$	$13\%$	$24\%$	$20\%$	$16\%$

  $\begin{array}{l}{n}_1=70\\ n_2=72\\ n_3=61\\ n_4=118\\ n_5=108\\ n_5=85\\ k=6\end{array}$

Formula Used:

The degrees of freedom are expressed as:

  $df=k-1$

The observed value is expressed as:

  $X^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Calculation:

Consider:

Probability of M&M’s containing brown candies $p_1$ .

Probability of M&M’s containing yellow candies $p_2$ .

Probability of M&M’s containing red candies $p_3$ .

Probability of M&M’s containing blue candies $p_4$ .

Probability of M&M’s containing orange candies $p_5$ .

Probability of M&M’s containing green candies $p_6$ .

The null hypothesis $\left(H_0\right)$ is

  $\begin{array}{l}{p}_1=0.13\\ p_2=0.14\\ p_3=0.13\\ p_4=0.24\\ p_5=0.20\\ p_6=0.16\end{array}$

The alternative hypothesis $H_a:$ at least one probability of above six probabilities which may vary from the certain probability.

Total number count:

  $\begin{array}{l}n=70+72+61+118+108+85\\ n=514\end{array}$

The expected count is calculated as:

For $i=1$ ,

  $\begin{array}{l}{E}_1=n{p}_1\\ E_1=514\times 0.13\\ E_1=66.82\end{array}$

For $i=2$ ,

  $\begin{array}{l}{E}_2=n{p}_2\\ E_2=514\times 0.14\\ E_2=71.96\end{array}$

For $i=3$ ,

  $\begin{array}{l}{E}_3=n{p}_3\\ E_3=514\times 0.13\\ E_3=66.82\end{array}$

For $i=4$ ,

  $\begin{array}{l}{E}_4=n{p}_4\\ E_4=514\times 0.24\\ E_4=123.36\end{array}$

For $i=5$ ,

  $\begin{array}{l}{E}_5=n{p}_5\\ E_5=514\times 0.20\\ E_5=102.8\end{array}$

For $i=6$ ,

  $\begin{array}{l}{E}_6=n{p}_6\\ E_6=514\times 0.16\\ E_6=82.24\end{array}$

Now, the observed value is calculated as: $\begin{array}{l}{X}^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ X^2=\frac{{\left(70-66.82\right)}^2}{66.82}+\frac{{\left(72-71.96\right)}^2}{71.96}+\frac{{\left(21-43.12\right)}^2}{43.12}+\frac{{\left(61-66.82\right)}^2}{66.82}+\frac{{\left(118-123.36\right)}^2}{123.36}+\frac{{\left(85-82.24\right)}^2}{82.24}\\ X^2=1.2468\end{array}$

Calculate the degrees of freedom:

  $\begin{array}{l}df=k-1\\ df=6-1\\ df=5\end{array}$

Therefore, the value of degrees of freedom is $df=5$ .

Need to estimate the a chi-square test statistics of $X^2=1.2468$ with degree of freedom $df$ .

The test statistics value lies between $X_{0.95}{}^2$ and $X_{0.90}{}^2$ .

Here,

  $X_{0.95}{}^2=1.145476$

  $X_{0.90}{}^2=1.11031$

Therefore, the $p$ -value lies between $0.95$ and $0.90$ .

That means: $0.90\le p\le 0.95$ .

Here the $p$ -value is very large. So, one can not reject the null hypothesis $H_0$ .

So, one can conclude that the given data can substantiate for the percentages reported by Mars, incorporated.

Conclusion:

Therefore, given data can substantiate for the percentages reported by Mars, incorporated.

Explanation of Solution

Given:

Disease	Brown	Yellow	Red	Blue	Orange	Green
Deaths	$13\%$	$14\%$	$13\%$	$24\%$	$20\%$	$16\%$

  $\begin{array}{l}{n}_1=70\\ n_2=72\\ n_3=61\\ n_4=118\\ n_5=108\\ n_5=85\\ k=6\end{array}$

Formula Used:

The degrees of freedom are expressed as:

  $df=k-1$

The observed value is expressed as:

  $X^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Calculation:

Consider:

Probability of M&M’s containing brown candies $p_1$ .

Probability of M&M’s containing yellow candies $p_2$ .

Probability of M&M’s containing red candies $p_3$ .

Probability of M&M’s containing blue candies $p_4$ .

Probability of M&M’s containing orange candies $p_5$ .

Probability of M&M’s containing green candies $p_6$ .

The null hypothesis $\left(H_0\right)$ is

  $\begin{array}{l}{p}_1=0.13\\ p_2=0.14\\ p_3=0.13\\ p_4=0.24\\ p_5=0.20\\ p_6=0.16\end{array}$

The alternative hypothesis $H_a:$ at least one probability of above six probabilities which may vary from the certain probability.

Total number count:

  $\begin{array}{l}n=70+72+61+118+108+85\\ n=514\end{array}$

The expected count is calculated as:

For $i=1$ ,

  $\begin{array}{l}{E}_1=n{p}_1\\ E_1=514\times 0.13\\ E_1=66.82\end{array}$

For $i=2$ ,

  $\begin{array}{l}{E}_2=n{p}_2\\ E_2=514\times 0.14\\ E_2=71.96\end{array}$

For $i=3$ ,

  $\begin{array}{l}{E}_3=n{p}_3\\ E_3=514\times 0.13\\ E_3=66.82\end{array}$

For $i=4$ ,

  $\begin{array}{l}{E}_4=n{p}_4\\ E_4=514\times 0.24\\ E_4=123.36\end{array}$

For $i=5$ ,

  $\begin{array}{l}{E}_5=n{p}_5\\ E_5=514\times 0.20\\ E_5=102.8\end{array}$

For $i=6$ ,

  $\begin{array}{l}{E}_6=n{p}_6\\ E_6=514\times 0.16\\ E_6=82.24\end{array}$

Now, the observed value is calculated as: $\begin{array}{l}{X}^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ X^2=\frac{{\left(70-66.82\right)}^2}{66.82}+\frac{{\left(72-71.96\right)}^2}{71.96}+\frac{{\left(21-43.12\right)}^2}{43.12}+\frac{{\left(61-66.82\right)}^2}{66.82}+\frac{{\left(118-123.36\right)}^2}{123.36}+\frac{{\left(85-82.24\right)}^2}{82.24}\\ X^2=1.2468\end{array}$

Calculate the degrees of freedom:

  $\begin{array}{l}df=k-1\\ df=6-1\\ df=5\end{array}$

Therefore, the value of degrees of freedom is $df=5$ .

Need to estimate the a chi-square test statistics of $X^2=1.2468$ with degree of freedom $df$ .

The test statistics value lies between $X_{0.95}{}^2$ and $X_{0.90}{}^2$ .

Here,

  $X_{0.95}{}^2=1.145476$

  $X_{0.90}{}^2=1.11031$

Therefore, the $p$ -value lies between $0.95$ and $0.90$ .

That means: $0.90\le p\le 0.95$ .

Here the $p$ -value is very large. So, one can not reject the null hypothesis $H_0$ .

So, one can conclude that the given data can substantiate for the percentages reported by Mars, incorporated.

Conclusion:

Therefore, given data can substantiate for the percentages reported by Mars, incorporated.