To assess Mendel’s law of segregation using tomatoes, a true-breeding tall variety (SS) is crossed with a true-breeding short variety (ss). The heterozygous F1 tall plants (Ss) were crossed to produce two sets of F2 data, as follows. Set I Set II 30 tall 300 tall 5 short 50 short Part A What is the chi square value for Set I? Enter your answer to two decimal places (for example: 1.23). nothing Request Answer Part B Based on the chi square value for Set I, you would Based on the chi square value for Set I, you would fail to reject the null hypothesis. reject the null hypothesis. Request Answer Part C What is the chi square value for Set II? Enter your answer to two decimal places (for example: 1.23). nothing Request Answer Part D Based on the chi square value for Set II, you would Based on the chi square value for Set II, you would reject the null hypothesis. fail to reject the null hypothesis. Request Answer Part E Based on the results of these test, which of the following is true? Based on the results of these test, which of the following is true? a. Dataset size affects the conclusion reached; b. the larger the dataset the less confidence in the result. c. Dataset size affects the conclusion reached; d. the larger the dataset the more confidence in the result. e. Dataset size has no effect on the conclusion reached. Request Answer

> Both Ss and SS are tall, while the ss are short. This results into that that the 3/4 of tomatoes is expected to be tall and 1/4 is expected to be short.

The chi-square test needs the expected and observed values. Observed values were given. To calculate the expected, we need to know the total of each set.

[ chi-square value = (observed-expected)^2/expected ]

Parts A and B:

• Set 1: 30 tall + 5 short = 35 total observed

Tall(expected) = 35 * 3/4 = 26

Short(expected) = 35 * 1/4 = 9

Tall: [ chi-square value = (30-26)^2/26 = 0.62 ]

Short: [ chi-square value = (5-9)^2/9 = 1.78 ]

chi-square value = 0.62 + 1.78 = 2.40

→ We have n=2 categories for which we are calculating. Degree of freedom would be then n-1, which is 1.

→ The Null Hypothesis states that there is no significant difference between observed and expected frequencies.

→ A value is considered significant, when p < 0.05

Now, we need to look up to the table values: The chi-square value we obtained, 2.40, lies between 0.25 and 0.10 for degree of freedom 1. Therefore, the value is not significant, meaning that we accept null hypothesis for Set I.

Similarly, we do Parts C and D:

• Set 2: 300 tall + 50 short = 350 total observed

Tall(expected) = 350 * 3/4 = 263

Short(expected) = 350 * 1/4 = 87

Tall: [ chi-square value = (300-263)^2/263 = 5.21 ]

Short: [ chi-square value = (50-87)^2/87 = 15.74 ]

chi-square value = 5.21 + 15.74 = 20.95

→ Now, we need to look up to the table values: The chi-square value we obtained, 20.95, belongs to p<0.01(not shown on table) for degree of freedom 1. Therefore, the value is significant, meaning that we reject null hypothesis for Set II.

Part E:

The correct is D. the larger the dataset the more confidence in the result, which means that if you have bigger dataset, you will obtain more precise and accurate confidence intervals. And that gives accurate results.