In: Math
1a. Suppose x has a distribution with μ = 21 and
σ = 16.
(a) If a random sample of size n = 39 is drawn, find μx, σx and P(21 ≤ x ≤ 23). (Round σx to two decimal places and the probability to four decimal places.)
μx = |
σx = |
P(21 ≤ x bar ≤ 23) = |
(b) If a random sample of size n = 55 is drawn, find
μx, σx
and P(21 ≤ x ≤ 23). (Round
σx to two decimal places and the
probability to four decimal places.)
μx = | ||||||||||||||||||||
σx = | ||||||||||||||||||||
P(21 ≤ x bar ≤ 23) 1b.Find P(69 ≤ x ≤ 74). (Round your answer to four decimal places.) 1c. 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.
(a) What is the level of significance? (b) What is the value of the sample test statistic? (Round your
answer to two decimal places.) 1D. 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†.
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 ratio of a certain stock index is μ = 19. Let x be a random variable representing the P/E ratio of all large U.S. bank stocks. We assume that x has a normal distribution and σ = 3.6. Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 19? Use α = 0.01. (a) What is the level of significance? (b)What is the value of the sample test statistic? (Round your
answer to two decimal places.) |
1)a) = 21
= = 16/ = 2.56
P(21 < < 23)
= P((21 - )/() < ( - )/() < (23 - )/())
= P((21 - 21)/2.56 < Z < (23 - 21)/2.56)
= P(0 < Z < 0.78)
= P(Z < 0.78) - P(Z < 0)
= 0.7823 - 0.5000
= 0.2823
b) = 21
= = 16/ = 2.16
P(21 < < 23)
= P((21 - )/() < ( - )/() < (23 - )/())
= P((21 - 21)/2.16 < Z < (23 - 21)/2.16)
= P(0 < Z < 0.93)
= P(Z < 0.93) - P(Z < 0)
= 0.8238 - 0.5000
= 0.3238
1c)a) Level of significance = 0.05
The test statistic z = ()/()
= (3.75 - 4.55)/(0.92/)
= -2.13
c) P-value = P(Z < -2.13)
= 0.0166
1D)a) Level of significance = 0.01
b) The test statistic z = ()/()
= (17.1 - 19)/(3.6/)
= -1.97
c) P-value = P(Z < -1.97)
= 0.0244