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tino4ka555 [31]
3 years ago
15

We just saw an example of "Survival of the fittest" with the rabbits that survived. We often use this as an expression to descri

be natural selection. Use another example (any example you can think of and IN YOUR OWN WORDS) to explain how survival of the fittest (aka natural selection) leads to evolution.
Mathematics
1 answer:
GarryVolchara [31]3 years ago
3 0

Answer:

If only the fittest survive in certain enviorments other animals have to evolve to live in different places

Step-by-step explanation:

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What is the volume, in cubic centimeters, of a cylinder with a height of 12 centimeters and a base radius of 6 centimeters, to t
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Which expression is equivalent to 11 - (-3 5/8)
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3 years ago
To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin
Murrr4er [49]

Answer:

a) t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

p_v =2*P(t_{15}>1.733)=0.104  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

b) t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

p_v =2*P(t_{15}>2.6)=0.0201  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation  

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

\bar X=31.3 represent the sample mean  

s=3 represent the sample standard deviation  

n=16 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 30, the system of hypothesis are :  

Null hypothesis:\mu = 30  

Alternative hypothesis:\mu \neq 30  

Since we don't know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>1.733)=0.104  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>2.6)=0.0201  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0
3 years ago
A room is 20 ft by 15 ft. The carpeting selected is $
MrMuchimi
The room is 20 / 3 yd by 15 / 3 = 5 yd, which is (20 / 3) x 5 = 33.33 square yards. So the cost of carpeting would be 33.33 x 37 = $1233.33
3 0
3 years ago
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