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"Orange trees ~ Orange trees are historically known to have mean circumference of 120 mm. A researcher randomly selected 35 orange trees and found that the sample mean circumference for the trees is 115.72 mm and the sample standard deviation is 57.49 mm. Note: Numbers are randomized in each instance of this question. Pay attention to the numbers in this question. The researcher wonders if the actual mean circumference of orange trees is less than the historic value. What is the p-value for this hypothesis test? Give your answer to 4 decimal places."

Respuesta :

Answer:

[tex]t=\frac{115.72-120}{\frac{57.49}{\sqrt{35}}}=-0.440[/tex]    

[tex]df=n-1=35-1=34[/tex]  

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

[tex]p_v =P(t_{(34)}<-0.440)=0.3314[/tex]  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=115.72[/tex] represent the sample mean

[tex]s=57.49[/tex] represent the sample standard deviation

[tex]n=35[/tex] sample size  

[tex]\mu_o =120[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] 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 less than the historical value, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 120[/tex]  

Alternative hypothesis:[tex]\mu < 120[/tex]  

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:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (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:  

[tex]t=\frac{115.72-120}{\frac{57.49}{\sqrt{35}}}=-0.440[/tex]    

P-value

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

[tex]df=n-1=35-1=34[/tex]  

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

[tex]p_v =P(t_{(34)}<-0.440)=0.3314[/tex]  

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