Angel bought 8 oranges and 11 apples for a total of $15.22. He knows that each orange costs $0.50 and reasoned that each apple m
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Answer:
We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.
We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.
Step-by-step explanation:
We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each 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.
Let = <u><em>true average percentage of organic matter</em></u>
So, Null Hypothesis, :
= 3% {means that the true average percentage of organic matter in such soil is 3%}
Alternate Hypothesis, :
3% {means that the true average percentage of organic matter in such soil is something other than 3%}
The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean percentage of organic matter = 2.481%
s = sample standard deviation = 1.616%
n = sample of soil specimens = 30
So, <u><em>the test statistics</em></u> = ~
= -1.76
The value of t-test statistics is -1.76.
(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.
Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.
Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.
(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.
Since the value of our test statistics lies within the range of critical values of t, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.
Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.
Given:
μ = 50 g, the mean
Let σ = the standard deviation.
For the random variable x =55 g, the z-score is
z = (55 - 50)/σ
We know that P(x > 55) =0.14 (14%).
From standard tables,
P(x > 55) = 0.14 => z = 0.36 (see the figure below).
Therefore
(55 - 50)/σ = 0.36
5/σ = 0.36
σ = 13.9
Answer: 13.9 grams
μ = 50 g, the mean
Let σ = the standard deviation.
For the random variable x =55 g, the z-score is
z = (55 - 50)/σ
We know that P(x > 55) =0.14 (14%).
From standard tables,
P(x > 55) = 0.14 => z = 0.36 (see the figure below).
Therefore
(55 - 50)/σ = 0.36
5/σ = 0.36
σ = 13.9
Answer: 13.9 grams