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3 years ago

Either enter an exact answer in terms of \piπpi or use 3.14, point, 14 for \piπpi and enter your answer as a decimal.

Mathematics

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

49π sq. units

Step-by-step explanation:

Find the diagram attached

We can find the area of the given circle

Given

diameter of the circle d = 14

Area of the circle = πd²/4

Substitute the given diameter

Area of the circle = π(14)²/4

Area of the circle = 196π/4

Area of the circle = 49π sq. units

Hence the exact answer in terms of π is 49π sq. units

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We conclude that the true average percentage of organic matter in such soil is different from 3%.

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We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

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Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

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