Question

Cloud seeding has been studied for many decades as a weather modification procedure. The rainfall in acre-feet from 20 clouds that were selected at random and seeded with silver nitrate are as follows, 18.0, 30.7, 19.8, 27.1, 22.3, 18.8, 31.8, 23.4, 21.2, 27.9, 31.9, 27.1, 25.0, 24.7, 26.9, 21.8, 29.2, 34.8, 26.7, and 31.6.
Assume this sample data follows a normal distribution.
A) Assume that the true standard deviation of the rainfall is σ = 4. Can you support a claim that mean rainfall from seeded clouds exceeds 25 acre-feet? Use ???? = 0.01.
B) Check that rainfall is normally distributed.
C) Compute the power of the test if the true mean rainfall is 27 acre-feet.
D) What sample size would be required to detect a true mean rainfall of 27.5 acre-feet if we wanted the power of the test to be at least 0.9?
E) Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean diameter.

Answer 1

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

X : 18.0, 30.7, 19.8, 27.1, 22.3, 18.8, 31.8, 23.4, 21.2, 27.9, 31.9, 27.1, 25.0, 24.7, 26.9, 21.8, 29.2, 34.8, 26.7,31.6

True standard deviation (σ) = 4

α = 0.01

Mean (m) of the data:

Σ(X) /n = 520.7/20

m = 26.035 = 26.04

Uaing calculator :

Sample standard deviation (s) = 4.785

Null hypothesis : μ = 25

Alternative hypothesis : μ > 25

Test statistic (t) : (m - μ) / (s/√n)

t = (26.04 - 25) / (4.785/√20)

t = 0.972

We can obtain the p value using the pvalue from t score calculator :

df = n - 1 = 20 - 1 = 19

Decision:

If p < 0.01 ; reject null

p(0.972, 19) = 0.1716

Since p > 0.1716 ; we fail to reject the Null