AP Biology : Perform chi squared test

Study concepts, example questions & explanations for AP Biology

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Example Questions

Example Question #71 : Inheritance

In a population of flowers, red is dominant to white. A true-breeding white flower is crossed with a heterozygous flower. Determine the expected ratios of this cross. Given observed values: 63 white flowers, 37 red flowers, determine:

1) The chi squared value

2) The degrees of freedom

3) The p-value

Possible Answers:

1) 6.76    

2) 1    

3) p>0.025

1) 6.76    

2) 2    

3) p=0.05

1) 14.44    

2) 2    

3) p>0.005

1) 3.35    

2) 4    

3) p=0.25

Correct answer:

1) 6.76    

2) 1    

3) p>0.025

Explanation:

First, determine the expected ratios using a Punnet Square. Given that red is dominant to white, the genotype of a true-breeding white flower can be denoted rr, and the genotype of a heterozygous flower can be denoted Rr. This cross will produce 50% Rr, 50% rr. Then, converting these percentages to decimals (50%= 0.50), and multiplying by the total population size of the observed population (0.50*100), gives expected values of 50 Rr (red flowers) and 50 rr (white flowers). These expected values and the observed values can then be plugged into the chi square equation . The equation will be . Degrees of freedom is n-1, so 2-1= 1. To determine p-value, use critical values table. The chi square value of 6.76 with 1 degrees of freedom will fall between a critical value corresponding with a p-value of 0.05 and 0.025. Thus, p>0.025

Example Question #1 : Perform Chi Squared Test

In a chi squared test, the greater the difference between the observed and expected frequencies of a trait, the ___ the p-value.

Possible Answers:

lower

more positive

greater

more negative

Correct answer:

lower

Explanation:

 A larger difference between observed and expected values will result in a larger chi square value. Using the critical values table, for a given degrees of freedom, as chi square value increases, p-value decreases.

Example Question #1 : Perform Chi Squared Test

If the p-value determined by a chi-square test is low, which is possible

Possible Answers:

The null hypothesis is accepted

The chi square value is also low

The null hypothesis is rejected

The alternative hypothesis is rejected

Correct answer:

The null hypothesis is rejected

Explanation:

When the p-value is equal to or below the significance level (alpha), the null hypothesis is rejected. For the p-value to be low, the chi square value would need to be large (large difference between observed and expected values). The null hypothesis would not be accepted (under no circumstance is a null hypothesis “accepted”; onlay rejected or failed to reject). The alternative hypothesis would be accepted if the p-value is equal to or below the significance level (alpha).

Example Question #1 : Perform Chi Squared Test

Which of the following chi square values would likely result in rejection of the null hypothesis?

Possible Answers:

13.8

0.35

7.36

1.97

Correct answer:

13.8

Explanation:

A larger difference between observed and expected values will result in a larger chi square value. Using the critical values table, for a given degrees of freedom, as chi square value increases, p-value decreases. A small p-value results in rejection of the null hypothesis.

Example Question #1 : Perform Chi Squared Test

Which is not a possible critical value of a chi square test?

Possible Answers:

-1.8

0

180

1.8

Correct answer:

-1.8

Explanation:

Critical values cannot be negative. The chi squared formula  will not result in a negative value as the numerator is squared.

Example Question #1 : Perform Chi Squared Test

If the chi square value is larger than the critical value at a given level of significance, what can be stated?

Possible Answers:

The null hypothesis is accepted

The alternative hypothesis is rejected

The null hypothesis fails to be rejected

The null hypothesis is rejected

Correct answer:

The null hypothesis is rejected

Explanation:

If the chi square value is large, this indicates a large difference between the observed and expected values. This will subsequently result in a small p-value when using the critical value table. When the p-value is equal to or smaller than the significance level (alpha), the null hypothesis is rejected.

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