-xy - (-x + y) if x = -3 and y= -4

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

4 years ago

Lelia has 5 1/4 hours of homework to do over the weekend. She does 2/3 of her homework on Saturday so sh can get it out of the w

dedylja [7]

Well 5 1/4 hours is 5 hours and 15 minutes and 2/3 of 5 hours and 15 minutes or 315(5*60+15) is 210
hope it helped

5 0

3 years ago

What is the General form of the parabola of (y-3)^2=6(x+8)

MatroZZZ [7]

Answer:

x = 1.5 $t^{2}$ , y = 3 + 2at.

Step-by-step explanation:

For the parabola $Y^{2}$ = 4aX ,

General form will be

(X = a $t^{2}$ , Y = 2at) ,

Thus , for the parabola $(y-3)^{2}$ = 6(x +8)

Here , a = 1.5 and Y from the above equation should be substituted by y - 3 and X must be substituted by x + 8. After substitution of the same we can use the general equation formula for this parabola also.

Thus , general equation comes out to be :-

x + 8 = 1.5 $t^{2}$ , y - 3 = 2at

x = 1.5 $t^{2}$ , y = 3 + 2at.

4 0

4 years ago

Simplify. Consider all cases. |14−(x−6)|

yanalaym [24]

Answer:

20-x; -20+x; -20-x; 20+x

Step-by-step explanation:

|14-x+6|

|20-x|

20-x; -20+x; -20-x; 20+x

4 0

3 years ago

Read 2 more answers

Evaluate m – n if m = –16 and n = 28

natima [27]

Answer:-44

Step-by-step explanation:

-16-(28)=-44

4 0

3 years ago

Read 2 more answers