Respuesta :
Interphase: (450 – 467)^2 = 289 /467 .618
Mitosis: (159 – 170)^2 = 121/170 = .711
Chi square = 1.329
The null hypothesis would be: if treated with the protein, the onion root remains the same. Based on the chi-square value of 1.329, this value is less than the critical value of 3.84. This means we can accept the null hypothesis. In other words, the untreated cells are no different from the treated cells regardless of which phase of mitosis you are viewing.
Answer:
Sample Response:
Chi-square = StartFraction (o minus e) Superscript 2 Over e EndFraction
Interphase: (450 – 467)2 = 289 /467 .618
Mitosis: (159 – 170)2 = 121/170 = .711
Chi square = 1.329
The null hypothesis would be: if treated with the protein, the onion root remains the same. Based on the chi-square value of 1.329, this value is less than the critical value of 3.84. This means we can accept the null hypothesis. In other words, the untreated cells are no different from the treated cells regardless of which phase of mitosis you are viewing.
Explanation:
its the sample response