Answer to Question #278214 in Statistics and Probability for nick

Question

Bill is a zoologist who studies Anna’s hummingbirds. Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were:

3.7 2.9 3.8 4.2 4.8 3.1

Let x be a random variable representing weight of Anna’s hummingbirds in this part of Grand Canyon. Assume that it has a normal distribution and 𝜎 = 0.70 grams. It is known that for the population of all Anna’s hummingbirds, the mean is 𝜇 = 4.55. Does this sample data indicate that the mean weight of these birds in this part of Grand Canyon is less than 4.55 grams? Use 𝛼 = 0.01. 

Answer 1

$$mean=\bar{x}=\dfrac{3.7+ 2.9+ 3.8+ 4.2+ 4.8+ 3.1}{6}=3.75$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq4.55$

$H_1:\mu<4.55$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a left-tailed test is $z_c = -2.3263.$

The rejection region for this left-tailed test is $R = \{z: z < -2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{3.75-4.55}{0.70/\sqrt{6}}\approx-2.7994$$

Since it is observed that$z = -2.7994 < -2.3263=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-2.7994)=0.00256,$ and since $p=0.00256<0.01=\alpha,$  it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the mean weight of these birds in this part of Grand Canyon is less than $4.55,$ at the $\alpha = 0.01$ significance level.