{"id":137,"date":"2018-03-13T04:23:00","date_gmt":"2018-03-13T03:23:00","guid":{"rendered":"https:\/\/kindsonthegenius.com\/blog\/2018\/03\/13\/question-14-chi-square-goodness-of-fit-test-problem-can-a-dice-be-considered\/"},"modified":"2020-08-22T20:22:48","modified_gmt":"2020-08-22T18:22:48","slug":"question-14-chi-square-goodness-of-fit-test-problem-can-a-dice-be-considered","status":"publish","type":"post","link":"https:\/\/kindsonthegenius.com\/blog\/question-14-chi-square-goodness-of-fit-test-problem-can-a-dice-be-considered\/","title":{"rendered":"Question 14 – Chi-Square Goodness of Fit Test Problem( Can a dice be considered…)"},"content":{"rendered":"
\n
Question 14<\/b><\/span><\/span><\/p>\n
\n

Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?<\/p>\n

<\/a><\/div>\n\n\n\n\n
Thrown Value<\/b><\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n<\/tr>\n
Frequency<\/b><\/td>\n182<\/td>\n154<\/td>\n162<\/td>\n175<\/td>\n151<\/td>\n176<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution Steps <\/b><\/span><\/span><\/p>\n

\n

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.<\/p>\n

Watch the video explanation here<\/a>
Get the excel sheet for this question <\/a><\/p>\n

The formular for chi-square is<\/p>\n

<\/a><\/div>\n

where
O = Observed values
E = Expected value <\/p>\n

Step 1: State the Null and althernate hypothesis<\/b><\/p>\n

H0<\/sub>:<\/b> the data is consistent with a specified distribution
Ha<\/sub><\/b>: the data is not consistent with a specified distributions<\/p>\n

Step 2:  Create the Table<\/b>
I have tranfered the data to excel and created the table as shown in Table 1.<\/p>\n

 <\/a><\/div>\n
Table 1 <\/div>\n

<\/ins>
Step 3: Calcualte the Expected Value (E)<\/b>
The expected value is the same as the mean of the sample which I have calculated as shown in the table.<\/p>\n

E = 166.67<\/p>\n

Step 4: Calculated the Differences<\/b>
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.<\/p>\n

 <\/a><\/div>\n
Table 2 <\/div>\n

Step 5: Calculate the Squared Differences<\/b>
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.<\/p>\n

 <\/a><\/div>\n
Table 3 <\/div>\n

Step 6: Calculate the Component<\/b>
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<\/p>\n

<\/a><\/div>\n
<\/div>\n
Table 4 <\/div>\n

Step 7: Calculate the Chi-Square Statistic<\/b>
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<\/p>\n

Chi-Square Statistic = 4.92<\/p>\n

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

Degrees of freedom is given by N-1
df = 6 – 1 =5
where N is the number of observation. In this case <\/p>\n

See statistical tables here <\/a><\/p>\n

From the chi-square table we for \u03b1 = 0.05, df = 5<\/span>
get a p-value of  11.07<\/b><\/p>\n

Step 9: State your Conclusion<\/b>
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<\/p>\n

Get this excel file<\/a><\/p>\n

<\/div>\n<\/div>\n

<\/ins>
<\/ins><\/p>\n","protected":false},"excerpt":{"rendered":"

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