A wild life biologist believes that the median length of the fish in a lake is 36 cm. A random sample of 14 fish yields the following lengths: $$\begin{array} { l l l l l l l l l l l l l l }
21 & 21 & 23 & 25 & 25 & 27 & 28 & 30 & 31 & 33 & 36 & 38 & 42 & 44
\end{array}$$ Test the biologist's hypothesis at $\alpha = 0.05$
A) Reject the claim because the test value 4 is more than the critical value 3.
B) Reject the claim because the test value 4 is more than the critical value 2.
C) Do not reject the claim because the test value 3 is equal to the critical value 3.
D) Do not reject the claim because the test value 3 is more than the critical value 2.

Question

A wild life biologist believes that the median length of the fish in a lake is 36 cm. A random sample of 14 fish yields the following lengths: $$\begin{array} { l l l l l l l l l l l l l l } 
21 & 21 & 23 & 25 & 25 & 27 & 28 & 30 & 31 & 33 & 36 & 38 & 42 & 44
\end{array}$$ Test the biologist's hypothesis at $\alpha = 0.05$
A) Reject the claim because the test value 4 is more than the critical value 3.
B) Reject the claim because the test value 4 is more than the critical value 2.
C) Do not reject the claim because the test value 3 is equal to the critical value 3.
D) Do not reject the claim because the test value 3 is more than the critical value 2.

Quizplus · Accepted Answer

We will use the one-sample t-test since the population standard deviation is unknown.

First, we calculate the sample mean and sample standard deviation:

$$\bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{21+21+23+\cdots+44}{14} = 31.64$$

$$s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} = 8.062$$

Next, we calculate the test statistic:

$$t = \frac{\bar{x}-\mu}{s/\sqrt{n}} = \frac{31.64-36}{8.062/\sqrt{14}} = -2.99$$

The degrees of freedom for this test is 13 (n-1). Using a t-table, we find the critical value to be approximately 2.16 at the 0.05 level of significance.

Since the test statistic (-2.99) is greater than the critical value (-2.16), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the median length of the fish in the lake is different from 36 cm at the 0.05 level of significance.

A Wild Life Biologist Believes That the Median Length of the Fish