The crab spider, Thomisus spectabilis , sits on flowers and preys upon visiting
ID: 3063274 • Letter: T
Question
The crab spider, Thomisus spectabilis, sits on flowers and preys upon visiting honeybees. Do honeybees distinguish between flowers that have crab spiders and flowers that do not? To test this, Heiling et al. (2003) gave 33 bees a choice between 2 flowers: one with, and one without a crab spider. In 20 of the 33 trials, the bees picked the flower that had the spider. In the other trials, the bees chose the spiderless flower.
With these data, carry out the appropriate hypothesis test (one- or two-tailed), using the normal approximation to the binomial distribution to determine Z. For a one-tailed test, use the formula =(1-NORM.DIST(Z,0,1,TRUE) in Excel calculate P. For a two-tailed test, use the formula =2(1-NORM.DIST(Z,0,1,TRUE).
State your answer for the value of P to three decimal places, and include the leading zero.
Do all of the math in Excel
DO NOT round the value of Z.
Substitute the cell (e.g. B1) for Z in the formula for P.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.55
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.12247
z = (p1 - p2) / SE
z = 1.73
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 1.73 or greater than 1.73
Thus, the P-value = 0.0836
Interpret results. Since the P-value (0.0836) is greater than the significance level (0.05), we have to accept the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is no significance difference choice between 2 flowers.
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