Question

5 Two different fish attractors were compared during 16 time periods spanning 4 years. (Wilbur, R. L. (1978). Two types of fish attractors compared in Lake Tohopekaliga, Florida. Transactions of the American Fisheries Society, 107(5), 689-695.) I present the data as a set of ordered pairs where in each case the first entry is the “pipe attractor” and the second entry is the “brush attractor”

{(6.64,9.73), (7.89,8.21),(1.83,2.17),(.42,.75),(.85,1.61),(.29,.75),(.57,.83),(.63,.56),(.32,.76),(.37,.32),(.00,.48),(.11.,.52),(4.86,5.38),(1.80,2.33),(.23,.91,),(.58,.79)}

(a)(4 pts) Perform the appropriate paired parameteric t-based test to compare the means.

(b)(6 pts) What happens if the two independent samples t-test is used? Make sure to perform all appropriate tests.

Is it possible to answer this questions with the code in Rstudio codes? If not the regular answer will be appreciated.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: \mu_DμD​ = 0

Ha: \mu_DμD​ ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df = 15df=15.

Hence, it is found that the critical value for this two-tailed test is t_c = 2.131tc​=2.131, for \alpha = 0.05α=0.05 and df = 15df=15.

The rejection region for this two-tailed test is R = \{t: |t| > 2.131\}R={t:∣t∣>2.131}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=3.05>tc​=2.131, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0081, and p=0.0081<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.

SOLUTION 2:(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 two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=30. 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 two-tailed test is tc​=2.042, for α=0.05 and df=30.

The rejection region for this two-tailed test is R={t:∣t∣>2.042}.

(3) Test Statistics

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=0.569≤tc​=2.042, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.5733, and since p=0.5733≥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 (pipe attractor) is different than μ2​ (brush attractor), at the 0.05 significance level.