# Question 14 – Chi-Square Goodness of Fit Test Problem( Can a dice be considered…)

Question 14

Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?

 Thrown Value 1 2 3 4 5 6 Frequency 182 154 162 175 151 176

Solution Steps

In this case, the questions says to consider whether the dice can be considered regular(consistent). So we use Chi-Square goodness of fit test which is used to determine whether a smple data is consistent with a distribution.

The formular for chi-square is

where
O = Observed values
E = Expected value

Step 1: State the Null and althernate hypothesis

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

Step 2:  Create the Table
I have tranfered the data to excel and created the table as shown in Table 1.

Table 1

Step 3: Calcualte the Expected Value (E)
The expected value is the same as the mean of the sample which I have calculated as shown in the table.

E = 166.67

Step 4: Calculated the Differences
In this step, you need to calculate the Observed Value – Expected Value (O-E) for each of the obseravtion. This I have done and added additional column to the table.

Table 2

Step 5: Calculate the Squared Differences
This is simple the square of the differences calculated in Step 4. I have added another column to our table to hold the squared differences.

Table 3

Step 6: Calculate the Component
This is calculated by dividing the squared difference calculated in step 5 by the expected value for each of the value.
Our table has been updated to include this column as well as the sum of this column

Table 4

Step 7: Calculate the Chi-Square Statistic
The chi-square statistic is calculated as the sum of the last column in our table which is the same as the result of using the formula

Chi-Square Statistic = 4.92

Step 8: Find the P-Value from Tables
You can find the p-value using the calculate chi-square statistic and the degrees of freedom

Degrees of freedom is given by N-1
df = 6 – 1 =5
where N is the number of observation. In this case

See statistical tables here

From the chi-square table we for α = 0.05, df = 5
get a p-value of  11.07