**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.

Watch the video explanation here

Get the excel sheet for this question

The formular for chi-square is

where

O = Observed values

E = Expected value

**Step 1: State the Null and althernate hypothesis**

**H _{0}:** the data is consistent with a specified distribution

**H**: the data is not consistent with a specified distributions

_{a}**Step 2: Create the Table**

I have tranfered the data to excel and created the table as shown in 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.

**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.

**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

**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

From the chi-square table we for α = 0.05, df = 5

get a p-value of **11.07**

**Step 9: State your Conclusion**

In this case the calculated test statistic from the sample which is 4.92 is less than the critical value from the table which is 11.07.

Therefore we accept the null hypothesis