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MAXImum [283]
3 years ago
5

A study was conducted to compare the overall length of two species of lake trout, labeled A and B. Ten fish from each species we

re randomly selected from each of three lakes and their respective lengths were recorded.(a) What is the experimental unit?(b) What is the population?(c) What is the sample?(d) What is the response variable?(e) Is this an experiment or an observational study?
Mathematics
1 answer:
Fofino [41]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that Ten fish from each species were randomly selected from each of three lakes and their respective lengths were recorded

a) Experimental unit is the length of two species of fish of lake trout.

b) Population is the set of all species in the lake

c) Sample is the twenty taken for testing

d) Response variable is the length of the fish

e) This is an observational study as already existing is recorded without any external control to change.

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kow [346]

Answer:

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

\bar X=23500 represent the sample mean  

s=3900 represent the sample standard deviation      

n=100 sample size      

\mu_o =2000 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 true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:\mu \leq 2000      

Alternative hypothesis:\mu > 2000      

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".  

Calculate the statistic      

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

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

df=n-1=100-1=99  

Since is a one-side upper test the p value would be:      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

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