This is all of the information for the question. But I need to find ONLY the Beta(3.5) and find the sample size required using the alpha and Beta. Like seen in the solved example I have attached. A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data (from “Engineering Properties of Soil,” Soil Science, 1998: 93–102). 1.10 5.09 0.97 1.59 4.60 0.32 0.55 1.45 0.14 4.47 1.20 3.50 5.02 4.67 5.22 2.69 3.98 3.17 3.03 2.21 0.69 4.47 3.31 1.17 0.76 1.17 1.57 2.62 1.66 2.05 The values of the sample mean, samp
This is all of the information for the question. But I need to find ONLY the Beta(3.5) and find the
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data (from “Engineering Properties of Soil,” Soil Science, 1998: 93–102).
1.10 5.09 0.97 1.59 4.60 0.32 0.55 1.45
0.14 4.47 1.20 3.50 5.02 4.67 5.22 2.69
3.98 3.17 3.03 2.21 0.69 4.47 3.31 1.17
0.76 1.17 1.57 2.62 1.66 2.05
The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.481, 1.616, and .295, respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level .10 by first determining the P-value. Would your conclusion be different if α = . 05had been used? [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]
![Alternative Hypothesis
Ημ. > με
H₂ μ< Ho
Type II Error Probability B(') for a
Level & Test
a/√₁
0 (2 + 40 = H²)
- 0(-20 + H0 / //-)
1-01-20
но-ме
0 (20/12 + 460/7/2) - (-20/2 + 460/7/2)
H₂ μ‡ Ho
where (2) = the standard normal cdf.
The sample size n for which a level a test also has B(μ') = B at the
alternative value μ' is
[σ(²₂ + ²p)]²
но-ме
[0(²/2 + ²₂)
но-не
[2]
for a one-tailed
(upper or lower) test
for a two-tailed test
(an approximate solution)
EXAMPLE 8.7 Let u denote the true average tread life of a certain type of tire. Consider testing
Hμ = 30,000 versus H:μ>30,000 based on a sample of size n = 16 from
a normal population distribution with = 1500. A test with a = .01 requires
=Z=2.33. The probability of making a type II error when μ = 31,000 is
B(31,000) = 2.33 +
30,000-31,000)
1500/V/16
= (-34) = .3669
Since z, = 1.28, the requirement that the level .01 test also have B(31,000) = .1
necessitates
[1500(2.33 + 1.28)
= (-5.42)² = 29.32
30,000 - 31,000
The sample size must be an integer, so n = 30 tires should be used.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98182d17-9434-4dd7-9261-6c7c628c58da%2F2e8ebece-e8f7-4a1d-b908-fb8ead480612%2Fh3bsqt_processed.png&w=3840&q=75)
Given that the amount of organic matter (%) in the soil of 30 specimen .
1.10 5.09 0.97 1.59 4.60 0.32 0.55 1.45
0.14 4.47 1.20 3.50 5.02 4.67 5.22 2.69
3.98 3.17 3.03 2.21 0.69 4.47 3.31 1.17
0.76 1.17 1.57 2.62 1.66 2.05
The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.481, 1.616, and .295, respectively
Check whether the true average percentage of organic matter in such soil is something other than 3%.
i.e.,
The null hypothesis is,
Ho: The true average percentage of organic matter in such soil is equal to 3%.
Symbolically,
Ho:
The alternative hypothesis is,
H1: The true average percentage of organic matter in such soil is something other than 3%.
Symbolically,
H1:
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