Please help ASAP thank youu!!

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

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 $\mu$ = true average percentage of organic matter

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

Alternate Hypothesis, $H_A$ : $\mu \neq$ 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 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 percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -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 sufficient evidence to 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 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 insufficient evidence to reject our null hypothesis 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.

8 0

3 years ago

Use the digits 2, 3, and 5 to create a fraction and a whole number with a product greater than 2

kupik [55]

The fractions which cannbe formed are: 2/3, 3/2, 2/5, 5/2, 3/5 and 5/3.
Now, we find the products with the third number:
2/3*5= 10/3 = 3.33 > 2
3/2*5 = 15 >2= 7.5 > 2
2/5*3= 6/5 (rejected)
5/2*3 = 15/2 > 2
3/5*2= 6/5 (rejected)
5/3*2 = 15/2 = 7.5> 2

5 0

3 years ago

Read 2 more answers

Solve The Problems

spayn [35]

X - 7 > 10
x > 10 + 7
x > 17

3x < = 21
x < = 21/3
x < = 7

5a - 2 < 18
5a < 18 + 2
5a < 20
a < 20/5
a < 4

2t + 8 > = -4 (t + 1)
2t + 8 > = -4t - 4
2t + 4t > = -4 - 8
6t > = - 12
t > = -12/6
t > = -2

4 0

3 years ago

Read 2 more answers

How many sets of three consecutive integers are there in which the sum of the three integers equals their product?

skad [1K]

Answer:

3

Step-by-step explanation:

since the 3 integers are consecutive, we are dealing with x, x+1, x+2.

and their sum is the same as their product :

x + (x + 1) + (x + 2) = x(x + 1)(x + 2)

3x + 3 = x(x² + 3x + 2) = x³ + 3x² + 2x

x³ + 3x² - x - 3 = 0

this is a polynomial of third degree.

and as such it has 3 solutions.

of course, it could be that some of them are the same or are even in the realm of complex numbers (i = sqrt(-1)), but usually these 3 solutions are different real numbers.

I tried x=1 just to see, and, hey, it is a solution for this equation.

x = 1 means that the other 2 consecutive integers are 2 and 3.

and indeed, 1+2+3 = 1×2×3 = 6.

now it is easier to find the other 2 solutions, as a zero solution can be expressed as a factor of the whole expression.

for x = 1 the factor term is (x - 1), as this term is then turning 0, when x = 1.

I can divide the main expression by this factor and then analyze the quotient about the other 2 solutions.

x³ + 3x² - x - 3 : x - 1 = x² + 4x + 3

- x³ - x²

----------------

0 4x² - x

- 4x² - 4x

-----------------------

0 + 3x - 3

- 3x - 3

---------------------------

0 0

so, the original expression can be written as

(x² + 4x + 3)(x - 1).

now we need to find the 2 zero solutions for x²+4x+3

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 4

c = 3

so,

x = (-4 ± sqrt(4² - 4×1×3))/(2×1) =

= (-4 ± sqrt(16 - 12))/2 = (-4 ± sqrt(4))/2 =

= (-4 ± 2)/2 = -2 ± 1

x1 = -2 + 1 = -1

x2 = -2 - 1 = -3

so, we have the additional solutions :

-1 0 1

-3 -2 -1

-1 + 0 + 1 = -1×0×1 = 0

-3 + -2 + -1 = -3×-2×-1 = -6

and there we have it fully proven :

there are 3 different sets of 3 consecutive integers with the same sum as product.

4 0

2 years ago

suppose that unit price p in dollars for a quantity demanded X of a certain product of a company is given by p=-0.08+620 (0 <

77julia77 [94]

I am sorry, I would love to help but I am not sure. Good luck anyways. :(

7 0

3 years ago