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Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna). (Reference: Hummingbirds, K. Long, W. Alther.) 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  = 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 = 0.64 gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is μ = 4.30 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.30 grams? Use α = 0.10.

(a)

What is the level of significance? (Enter a number.)


State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ = 4.3 g; H₁: μ > 4.3 g; right-tailed

H₀: μ < 4.3 g; H₁: μ = 4.3 g; left-tailed    

H₀: μ = 4.3 g; H₁: μ ≠ 4.3 g; two-tailed

H₀: μ = 4.3 g; H₁: μ < 4.3 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 we assume that x has a normal distribution with known σ.

The Student's t, since n is large with unknown σ.

Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.)

(c)

Find (or estimate) the P-value. (Enter a number. Round your answer to four decimal places.)

Answer 1

Solution : -

Giveh that

= 4.3

= 3.75

= 0.64

n = 6

a)

What is the level of significance? (0.10)

The null and alternative hypothesis is ,

H0 :   = 4.3

Ha :    < 4.3

This is the left tailed test .

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

Test statistic = z

= ( - ) / / n

= (3.75 - 4.3 ) / 0.64 / 6

= -2.11

The test statistic = -2.11

P - value = P ( Z < -2.11 ) = 0.0174

( c ) P-value = 0.0174