Question

Don't involve regression

1. A scientist collects fish from two lakes. The first lake is known to have large fish and the second is known to have small fish.

The fish caught from the first lake have weights, in grams, of:

c(48.2, 45.9, 34, 52.9, 29.8, 45.4, 35.2, 42.2)

and fish caught from the second lake has weights of:

c(55.3, 39.7, 45.9, 34.7, 37.1, 49.7, 12, 37.7, 45.6, 59.2)

Is there evidence that the first group is actually more heavy than the second?

1. State a sensible null hypothesis
2. State the precise definition of p-value and explain what “more extreme” means in this context
3. Is a one-sided or two-sided test needed? justify
4. Perform a student t-test using R and interpret

Answer 1

As per the provided information,

There are two sample data:

Sample 1: first lake containing large fish.

Sample 2: second Lake containing small fish.

Let x denote the sample 1.

Let y denote the sample 2.

The researcher wants to test that first group is actually more heavy than the second group.

(a).

The statistical hypothesis is,

Null hypothesis (Ho): The mean weight of fishes in the first group is less than or equal to the mean weight of fishes in the second group.

Alternative hypothesis (Ha): The mean weight of fishes in the first group is actually heavier than the mean weight of fishes in the second group.

In mathematical terms,

(b). P-value can be defined as the chance of considering that the null statement is true; that is, the tested claim is incorrect. If the p-value is low, then there is a high chance of rejecting the null; otherwise, the null is accepted. In this context, the term “more extreme” can be interpreted as that the data values lie at a far distance from the mean value, which results in the small p-value. If the p-value is smaller, then there is a higher chance of rejecting Ho.

(c).

It is a right tailed test or one-sided test as the direction of testing is defined in the claim.

(d).

Coding in R:

> x <- c(48.2, 45.9, 34, 52.9, 29.8, 45.4, 35.2, 42.2) > y <- c(55.3, 39.7, 45.9, 34.7, 37.1, 49.7, 12, 37.7, 45.6, 59.2) > t.test(x,y, alternative = "greater", var.equal = TRUE)

R Output:

Two Sample t-test data: x and y t = 0.0018843, df = 16, p-value = 0.4993 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: -9.255261       Inf sample estimates: mean of x mean of y     41.70         41.69

Assume the significance level is 0.05.

Since the p-value (0.4993) is greater than the significance level 0.05, so the researcher fail to reject the null hypothesis. Therefore, it can be concluded that there is insufficient evidence to support that the first group is actually heavier than the second.