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 34 bees a choice between 2 flowers: one with, and one without a crab spider. In 22 of the 34 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.

Answer 1

According to the data given to us, 22 of the 34 trials, the bees picked the flower that had the spider and rest the other trials, the bees chose the spiderless flower . Therefore in order to test whether the bees correctly distinguished between the flower with crab spider and the flower without crab spider, we define the test hypothesis as;

[H_{0}:p=0.50]   

(the bees correctly identify the flower with spider)

Against the alternative hypothesis as:

[H_{1}:p\neq 0.50]

(the bees not correctly identify the flower with spider).

As [x=22] of the [n=34] trials, the bees picked the flower that had the spider , then sample proportion is determined as:

[\hat{p}=\frac{x}{n}=\frac{22}{34}=0.6471]

To test this hypothesis , the test statistics [Z] is defined as:

[Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}]

[=\frac{0.6471-0.50}{\sqrt{\frac{0.50(1-0.50)}{34}}}]

   [=\frac{0.1471}{0.08575}]

   [=1.7155]

The decision rule for null hypothesis is

[Z_{calculated}=1.7155

and according to the p value approaches

P value of two tail test is calculated as:

[P-value=P(\left |Z \right |\geq 1.7155)]

   [=2[1-P(Z<1.7155)]]

   [=2[1-0.9568 ]]

   [=2\times 0.04312]

   [=0.0862]

As [P-Value=0.0862] is greater than the level of significance [\alpha =0.05] , we failed to reject the null hypothesis and therefore we have sufficient evidence to conclude that the bees  correctly distinguished between the flower with crab spider and the flower without crab spider.

[P(Z<1.7155)] is calculated in Excell as: