To test whether or not female quails were actually selecting males based on the
ID: 3176890 • Letter: T
Question
To test whether or not female quails were actually selecting males based on the characteristics of their plume, a mate-choice study is conducted. Male California quails have their plumes snipped off, and then a fake plume is glued on in its place. Half the males get a replacement the same size as the mean of the original plumes, and half get a plume 50% larger. Trials are then conducted in which a female quail is placed halfway between two male quails, each in an enclosure with a glass side. One male has a regular plume, the other a big plume. The trial is scored for one type of male when the female moves within a certain distance of that male. Each quail is used in only one trial.
Test arena. If the female enters Zone 1, the trial is scored for Male 1; if Zone 2, then Male 2.
A total of 50 trials are conducted for this experiment. The results are shown below.
Answer the following questions based on this information.
1. What is the null hypothesis being tested by the goodness-of-fit test? Choose the best answer.
2.What is the reference, or expected, relative frequency for the big-plumed males? Note that your answer should be a fraction.
3.What is the *expected* count for the big-plumed males?
4. While the absolute best approach to this goodness-of-fit question would be the binomial test, the sample size is big enough that we can use the chi-squared test.
Calculate the chi-squared value and provide it below.
5. Assuming this is a case with just one constraint on the data (that's the typical case), what is the degrees of freedom?
6.What is the P-value for the chi-squared test, to three decimal places? You will probably need to use Excel or a statistical program to get this.
7.
What is your interpretation of your results? Assume the standard alpha of 0.05.
Male type Times chosen Normal-plumed 16 Big-plumed 34Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: That the number of choices for each male is the same.
Alternative hypothesis: The number of choices for each male is different.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 2 - 1 = 1
(Ei) = n * pi
(E1) = 50 * 0.50 = 25
(E2) = 50 * 0.50 = 25
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (16 - 25)2 / 25 ] + [ (34 - 25)2 / 25 ]
2 = 3.24 + 3.24
2 = 6.48
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and 2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 19.58.
We use the Chi-Square Distribution Calculator to find P(2 > 6.48) = 0.0109
Interpret results. Since the P-value (0.0109) is less than the significance level (0.05), we cannot accept the null hypothesis.
From this we conclude that we have sufficient evidence in the favor of the claim that the number of choices for each male is different.
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