Chapter 11.1, Problem 7E

To determine

To find: chi square goodness of fit statistic and test the significance using the predicted and actual count of smooth and wrinkled peas plants using random and independent trials conducted by Mendel.

Answer to Problem 7E

Since the chi square statistic is highly insignificant, we fail to reject null hypothesis and hence conclude that there is no significant difference between observed and expected count of peas. In other words, we may say Mendel prediction is right after carrying chi square goodness of fit test.

Explanation of Solution

Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. To check his understanding, he conducted an experiment which met Random and Independent conditions. He could find that 423 smooth and 133 wrinkled peas were there.

The expected count of smooth peas among 423+133=556 plants is 3/4×556=417.

Similarly expected count of wrinkled peas among 423+133=556 plants is 1/4×556=139.

The observed counts are 423 and 133 for smooth and green peas. The chi square statistic is calculated using formula is x²=∑(O-E)²/E where O is observed count and E is expected count of each peas type.

Calculation table is given below

  

Hence total chi square is 0.0863+0.2590=0.3453 . The p value of this test statistic is 0.4432 which very much higher than level of significance of 5% or 1%, we fail to reject null hypotheis.

Conclusion: Since the chi square statistic is highly insignificant, we fail to reject null hypothesis and hence conclude that there is no significant difference between observed and expected count of peas. In other words, we may say Mendel prediction is right.