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PolarNik [594]
3 years ago
10

If the animal is in the woods on one observation, then it is four times as likely to be in the woods as the meadows on the next

observation. If the animal is in the meadows on one observation, then it is twice as likely to be in the meadows as the woods on the next observation.
Assume that state 1 is being in the meadows and that state 2 is being in the woods.

If the animal is in the woods on the first observation, what is the probability that it is in the woods on fourth observation
Mathematics
1 answer:
kvv77 [185]3 years ago
4 0
My solution to the problem is as follows:

<span>day 1: 1 
day 2: 0.6667 
day 3: 0.5111 
day 4: 0.4385 <-------
</span>
Therefore, <span>the probability that the animal is in the woods on fourth observation would be 0.4385.

I hope my answer has come to your help. God bless and have a nice day ahead!
</span>
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For the following hypothesis test, H0: μ ≥ 150Ha: μ &lt; 150​the test statistic a. must be positive. b. must be negative. c. can
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Answer:

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".  

And since we are conducting a lower unilateral lest  then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

Step-by-step explanation:

Data given and notation      

\bar X represent the sample mean    

s represent the standard deviation for the sample      

n sample size      

\mu_o =150 represent the value that we want to test    

\alpha 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 determine if the mean is lower than 150, the system of hypothesis would be:      

Null hypothesis:\mu \geq 150      

Alternative hypothesis:\mu < 150      

We don't know the population deviation, so for this case is better apply a t 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".  

And since we are conducting a lower unilateral lest  then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

7 0
3 years ago
Need help seriously
Dima020 [189]
It’s not hard bro you gotta lean probability it’s really easy
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Mr.george owns 425 acres of land.if he divides the land into half-acre plots,how many plots will he have?
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All you do is times 425 by 2
425×2
850
hope this helps
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3 years ago
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