Question

a) Bromeliads are tropical flowering plants. Many are epiphytes that attach to trees and obtain moisture and nutrients from air and rain. Their leaf bases form cups that collect water and are home to the larvae of many insects. In an experiment in Costa Rica, Jacqueline Ngai and Diane Srivastava studied whether added nitrogen increases the productivity of bromeliad plants. Bromeliads were randomly assigned to nitrogen or control groups. Here are data on the number of new leaves produced over a seven-month period:

Control 11 13 16 15 15 11 12 Nitrogen 15 14 15 16 17 18 17 13

Answer 1

1 represents the control group and 2 represents Nitrogen.

The sample means are shown below:

Also, the sample standard deviations are:

s1​=2.058663, s2​=1.685018

and the sample sizes are n1​=7 and n2​=8.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.176 and FU​=5.119, and since F = 1.493, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

The significance level is α=0.05, and the degrees of freedom are df = 13. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is tc​=−1.771, for α=0.05 and df = 13.

The rejection region for this left-tailed test is R = { t : t < −1.771}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

​​

(4) Decision about the null hypothesis

Since it is observed that t = −1.297 ≥ tc ​= −1.771, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1086, and since p=0.1086≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the 0.05 significance level.

Therefore, there is not enough evidence to claim that the added nitrogen increases the productivity of bromeliad plants