Question

Part 1) 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 ? = 0.86 gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is ? = 4.60 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.60 grams? Use ? = 0.10.

(a) What is the level of significance?

(b) 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.)

Part 2) The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios.†

24

16

22

14

12

13

17

22

15

19

23

13

11

18

The sample mean is x?17.1.

Generally speaking, a low P/E ratio indicates a "value" or bargain stock. Suppose a recent copy of a magazine indicated that the P/E ratioof a certain stock index is ?= 18.Let xbe a random variable representing the P/E ratioof all large U.S. bankstocks. We assume that xhas a normal distribution and ?= 3.6.Do these data indicate that the P/E ratioof all U.S. bankstocks is less than 18? Use ?= 0.01.

(a) What is the level of significance?

(b) Compute the zvalue 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

Part 1) Weight of Anna's hummingbird

a. Level of significance, = 0.1

b.

Sample mean, 3.75

Population mean, 4.60

Population standard deviation, 0.86

Number of observation, n = 6

Therefore,

Z = -2.42

c)

p-value can be calculated using excel with the function NORMSDIST. Type =NORMSDIST(-2.42) in any of the cell to get the p-value. The resulting value is 0.00776. Hence, the p-value is 0.0078.

since p-value is less than the level of significance of 0.1 we would reject the null hypothesis and conclude that the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.60 grams.

Part 2) P/E ratio

a. Level of significance, = 0.01

b.

Sample mean, 17.1

Population mean, 18

Population standard deviation, 3.6

Number of observation, n = 14

Therefore,

Z = -0.94

c)

p-value can be calculated using excel with the function NORMSDIST. Type =NORMSDIST(-0.94) in any of the cell to get the p-value. The resulting value is 0.173609. Hence, the p-value is 0.1736.

since p-value is greater than the level of significance of 0.01 we would not reject the null hypothesis and conclude that the data indicate that the P/E ratio of all U.S. bankstocks is eqaul to 18.