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Did the treatment have an effect? The investigators expected the control group to adjust their breeding date the next year, whereas the well-fed supplemented group had no reason to change. The report continues: "But in the following year, food-supplemented females were more out of synchrony with the caterpillar peak than the controls." Here are the data (days behind caterpillar peak):

Carry out an appropriate test and show that it leads to the quoted conclusion.

Short Answer

Expert verified

There is sufficient evidence to support the claim that the food-supplemented females were more out of synchrony with the caterpillar peak than the controls..

Step by step solution

01

Given Information

The mean is the sum of all values divided by the number of values:

x1=4x2=11.3

nis the number of values in the data set?

The standard deviation is the square root of the sum of squared deviations from the mean divided by n-1:

s1=3.1093

s2=3.9256

02

Explanation

Determine the hypothesis:

H0:1=2

Ha:1<2

Determine the test statistic:

localid="1650516568809" t=x1-x2s12n1+s22n2=4-11.33.109326+3.925627-3.739

Determine the degrees of freedom:

localid="1650516589517" df=minn1-1,n2-1=min(7-1,6-1)=5

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the rowdf=5:

0.005<P<0.01

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

P<0.05RejectH0

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