Chi-Square Goodness of Fit Test – Question 16 ( A newly developed muesli…)

Question 16

A newly developed muesli contains five types of seeds (A, B, C, D and E). The percentage of which is 35%, 25%, 20%, 10% and 10% according to the product information. In a randomly selected saa, the following volume distribution was found.

Component A B C D E
Number of Pieces 184 145 100 63 63

Lets us decide about the null hypothesis whether the composition of the sample corresponds to the distribution indicated on the packaging at alpha = 0.1 significance level.


Solution Steps 

Read the question again and made user you understand it. Think of what values are observed values and what values are expected values

Step 1: State the null and alternate hypothesis
From the question, we are required to verify the null hypothesis whether the product information corresponds to a given specified distribution. Therefore we can state the null hypothesis as:

H0: the data is consistent with a specified distribution
Ha: the data is not consistent with a specified distributions


Step 2: Identify the observed and expected values
In this case, the observed values are the values given in the table that is

A = 184
B = 145
C = 100
D = 68
E = 63

As for the expected values, the question gives us the percentages (of the total) of the expected values
Total = 184 + 145 + 100 + 68 + 63 = 560

So we can calculate the expected values based on the percentages

Expected value for A = 560 * 0.35 = 196
Expected value for B = 560 * 0.25 = 140
Expected value for C = 560 * 0.20 = 112
Expected value for D = 560 * 0.10 = 56
Expected value for E = 560 * 0.10 = 56

Now we have the two important sets of values, the set of observed values and the set of expected values.


Step 3: Calculate the Chi-Square Test Statistic
Now that we have calculated the expected values, the rest of the problem is quite easy. We can get the chi-square statistic using the formular.


If we use the actual values, thene we have the following calculation:

Step 4: Look up the critical value from tables
We need to look at the table of chi-square distribution.

From the question, alpha = 0.01
Degrees of freedom, df = 5 – 1 = 4

The critical value for alpha = 0.01 and 4 degrees of freedom is written as

K0.1,4 = 7.729


Step 5: State your conclustion
Since the calculated value of the test statistic is less than the critical value, we therefore accept the null hypothesis that the content of the checked sac corresponds to the advertised distribution.