Question

Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna).† 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 as follows. 3.7 2.9 3.8 4.2 4.8 3.1 The sample mean is x = 3.75 grams. Let x be a random variable representing weights of hummingbirds in this part of the Grand Canyon. We assume that x has a normal distribution and σ = 1.00 gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is μ = 4.75 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.75 grams? Use α = 0.01. (a) What is the level of significance? 0.01 State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H0: μ = 4.75 g; H1: μ ≠ 4.75 g; two-tailed H0: μ = 4.75 g; H1: μ > 4.75 g; right-tailed H0: μ < 4.75 g; H1: μ = 4.75 g; left-tailed H0: μ = 4.75 g; H1: μ < 4.75 g; left-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The standard normal, since we assume that x has a normal distribution with unknown σ. The standard normal, since we assume that x has a normal distribution with known σ. The Student's t, since n is large with unknown σ. The Student's t, since we assume that x has a normal distribution with known σ. Compute the z value of the sample test statistic. (Round your answer to two decimal places.) (c) Find (or estimate) the P-value. (Round your answer to four decimal places.)

Answer 1

(a) Level of significance= 0.01

Ho: = 4.75

H1: < 4.75

Null hypothesis states the the mean weight of these birds in this part of the Grand Canyon is 4.75

Alternative hypothesis states the the mean weight of these birds in this part of the Grand Canyon is less than 4.75

This is a left tailed test : H0: μ = 4.75 g; H1: μ < 4.75 g; left-tailed

(b) The standard normal, since we assume that x has a normal distribution with known σ.

n = 6

sample mean = 3.75

population sd = 1

z = - 2.45

(c) p value

P ( Z < − 2.4495 ) = 1 − P ( Z < 2.4495 ) = 1−0.9929 = 0.0071

p value = 0.0071

The z-critical value for a left-tailed test, for a significance level of α=0.01 zc = − 2.33

As the z stat falls in the rejection area, we reject the Null hypothesis.

As the p value (0.0071) is less than level of significance (0.01), we reject the Null hypothesis.