Different colors of floor tiles are going to be installed in a random pattern with the frequency of each color being same. A sample of 200 tiles laid exactly as they were packaged was observed and the following was found: Color and number of tiles orange-32, white-36, blue-48, green-50, violet-34.Test the hypothesis at the 5 percent level of significance for the uniform color distribution.
We need to check whether the color distribution so uniform so hypotheses are:
H0: The color distribution is uniform.
Ha: The color distribution is not uniform.
If all the five colors are equally likely then expected frequency for each color is 200 /5 = 40.
Following table shows the calculations for chi square test statistics:
O | E | (O-E)^2/E |
32 | 40 | 1.6 |
36 | 40 | 0.4 |
48 | 40 | 1.6 |
50 | 40 | 2.5 |
34 | 40 | 0.9 |
Total | 7 |
Following is the test statistics:
Degree of freedom: df =5-1=4
The p-value is: 0.1359
Since p-value is greater than 0.05 so we fail to reject the null hypothesis. That is we can conclude that the color distribution is uniform.
Excel function used for p-value: "=CHIDIST(7,4)"
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