Mathematics, 15.04.2020 20:17 adrian1742

A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.05 years, with sample standard deviation s = 0.78 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01.

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Answer from: eguzmandpandoracom

$t=\frac{2.05-1.75}{\frac{0.78}{\sqrt{41}}}=2.463$

$p_v =P(t_{(40)}2.463)=0.00909$

If we compare the p value and the significance level given $\alpha=0.01$ 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 the true mean is higher than 1.75 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=2.05$ represent the sample mean

s=0.78 represent the sample standard deviation

n=41 sample size

$\mu_o =1.75$ represent the value that we want to test

$\alpha=0.01$ 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 true mean is higher than 1.75, the system of hypothesis would be:

Null hypothesis: $\mu \leq 1.75$

Alternative hypothesis: $\mu 1.75$

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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:

$t=\frac{2.05-1.75}{\frac{0.78}{\sqrt{41}}}=2.463$

P-value

The first step is calculate the degrees of freedom, on this case:

df=n-1=41-1=40

Since is a one sided test the p value would be:

$p_v =P(t_{(40)}2.463)=0.00909$

Conclusion

Answer from: Quest

i found two solutions, n = 2 • ± √3 = ± 3.4641

step-by-step explanation:

Answer from: Quest

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

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