Question

Please in RStudio

3. A zoologist in Auckland studies kiwi behaviour and catches kiwi in two locations, A and B. He is interested in the weights of kiwi birds in the two locations; he weighs the kiwi and records their weight before releasing them. The dataset is as follows:

kiwi_A <- c(1.10, 1.50, 1.10, 1.25, 1.25, 1.34, 1.53, 1.82, 1.31)

kiwi_B <- c(1.11, 1.25, 1.02, 1.12, 1.00, 0.94, 1.18, 1.02, 1.66)

The scientist does not know which of the sites is likely to have heavier kiwi.

(a) State a sensible null hypothesis

(b) Is a one-sided test or a two sided test needed? Why?

(c) Perform a t-test on the dataset and interpret.

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Thank you so much!!

Answer 1

We want to test the zoologist claim that whether there is a significant difference in the weighs of the kiwi in two locations.

a) The null hypothesis is:

Null Hypothesis: There is no significant mean difference in the weighs of the kiwi in two locations.

i.e.

Alternative Hypothesis: There is a significant mean difference in the weighs of the kiwi in two locations.

i.e.

b) The test is two tail test since, the scientist does not know which of the sites is likely to have heavier kiwi.

Therefore the alternative hypothesis is non directional.

Hence this is a two tail test.

c) ### By using R-studio:

> kiwi_A <- c(1.10, 1.50, 1.10, 1.25, 1.25, 1.34, 1.53, 1.82, 1.31)
> kiwi_B <- c(1.11, 1.25, 1.02, 1.12, 1.00, 0.94, 1.18, 1.02, 1.66)
> t.test(kiwi_A,kiwi_B)

Welch Two Sample t-test

data: kiwi_A and kiwi_B
t = 2.008, df = 15.941, p-value = 0.0619
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01183297 0.43405519
sample estimates:
mean of x mean of y
1.355556 1.144444

Here p value is 0.0616>0.05, Therefore we are unable to reject the null hypothesis at 5% level of significance.

Therefore we do not have sufficient evidence to claim that there is a significant mean difference in the weighs of the kiwi in two locations.