Chapter 10.6, Problem 26E

To determine

To find:

To perfom the test statistic for the following scenerio,

“A manufacturer of children’s vitamins claims that its vitamins are mixed so the each batch has exactly the following percentage of each color: 20% green, 40% yellow, 10% red, and 30% orange. To test the claim that these percentage are incorrect, 100 bottles of vitamins were pulled and the clors of the vitamins were tailled”.

Answer to Problem 26E

Solution:

(a) $H_0:p_1=0.02,p_2=0.40,p_3=0.10,and{p}_4=0.30$

$H_a$: There is some difference amongst the probabilities.

(b) Use the chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

(c) The test statistic value is ${\chi }^2=34.6485$.

(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.

Explanation of Solution

The following table represents the random sample of number of vitamins and the color

of vitamins is,

Children's Vitamins
	Green	Yellow	Red	orange
Number	1149	1948	552	1401

Step 1: To state the null and alternative hypotheses.

When starting the hypotheses to be used we take the null hypothesis to be that the proportion percentages stated by the vitamin manufacturer are not incorrect.

Null hypothesis: $H_0$: The proportion of vitamins are not incorrect.

Alternative hypothesis: $H_a$: The proportion of vitamins are incorrect.

Let $p_1,p_2,p_{3}and{p}_4$ be the probabilities for Green, Red, Yellow and Orange respectively.

Mathematically, we can write the null and alternative hypothesis for four different vitamins as follows,

$H_0:p_1=0.02,p_2=0.40,p_3=0.10,and{p}_4=0.30$

$H_a$: There is some difference amongst the probabilities.

Step 2: To determine whether the observed values proportion of color of vitamins is match with the expected values of proportion of color of vitamins. Since we assume that the necessary condition that have been met. So, use the chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

Step 3: To calculate the test statistic.

Let‘s calculate the expected value of color of vitamins. Since we are assuming that the proportion percentages stated by the vitamin manufacturer are not incorrect is calculated as follows,

$\begin{array}{rll}n& =& 1149+1948++552+1401\\ & =& 5050\end{array}$

And,

$\begin{array}{l}E\left(x_1\right)=n{p}_1=5050\times 0.20=1010\\ E\left(x_2\right)=n{p}_2=5050\times 0.40=2020\\ E\left(x_3\right)=n{p}_3=5050\times 0.10=505\\ E\left(x_4\right)=n{p}_4=5050\times 0.30=1515\end{array}$

Formula for calculating test statistic for a Chi-Square test for goodness of fit:

The test statistic for a chi-square test for goodness of fit is given below,

${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Where O_i, is the observed frequency for the i^th possible outcome and E_i is the expected frequency for the i^th possible outcome and n is the sample size.

From the above table represent the value of n is 5050.

The following table represents the test statistic, ${\chi }^2$, for a chi- square test for goodness of fit is given below,

Children's Vitamins	Observed Values	Expected Values	$O_i-E_i$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
Green	1149	1010	139	19321	19.1297
Yellow	1948	2020	-72	5184	2.56634
Red	552	505	47	2209	4.37426
Orange	1401	1515	-114	12996	8.57822
					${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}=34.6485}$

Step 4: Degrees of freedom in a Chi-square test for goodness of fit:

In a chi-square test for goodness of fit the number of degrees of freedom for the chi-square distribution of the test statistic is given by,

$df=k-1$

Where k is the number of possible outcomes for each trial.

Rejection Region for Chi-Square Test for Association:

Reject the null hypothesis, $H_0$, if ${\chi }^2\ge {\chi }_{\alpha }^2$.

From the given information k = 4.

Substitute the above values in the formula of degrees of freedom to get the following,

$\begin{array}{rll}df& =& k-1\\ & =& 4-1\\ & =& 3\end{array}$

Conclusion:

Use the level of significance of $\alpha =0.05$ to find the critical value ${\chi }_{0.05}^2$ of the chi-square distribution with df = 3 in order to make a decision about the null hypothesis.

Use the “Area to the right of the critical value ${\chi }^2$” table to get the following,

${\chi }_{0.05}^2=7.815$

The test statistic value is ${\chi }^2=34.6485$.

By using the above condition we reject the null hypothesis.

Since, $34.6485\ge 7.815\iff {\chi }^2\ge {\chi }_{0.05}^2$.

So, there is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.

Final statement:

(a) $H_0:p_1=0.02,p_2=0.40,p_3=0.10,and{p}_4=0.30$

$H_a$: There is some difference amongst the probabilities.

(b) Use the chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

(c) The test statistic value is ${\chi }^2=34.6485$.

(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.