Question

Historically, the mean yield of corn in the United States has been 120 bushels per acre with a standard deviation of 12. A survey of 40 farmers this year gives a sample mean yield of 125 bushels per acre . Let p be the mean yield of corn nationally for this year. Supposing that the past standard deviation is still correct, what is the p-value for testing a null hypothesis of mu=120 against an alternative of mu not equal to 20 ? A 0.0041 B 0.0082 C 2.64 D 125

Answer 1

Answer:

$z=\frac{125-120}{\frac{12}{\sqrt{40}}}=2.64$    

P-value

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

$p_v =2*P(z>2.64)=0.0082$  

And the best answer would be

B 0.0082

Step-by-step explanation:

Data given and notation  

$\bar X=125$ represent the mean height for the sample  

$\sigma=12$ represent the population standard deviation

$n=40$ sample size  

$\mu_o =120$ 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 check if the true mean is 120 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 120$  

Alternative hypothesis:$\mu \neq 120$  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-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:  

$z=\frac{125-120}{\frac{12}{\sqrt{40}}}=2.64$    

P-value

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

$p_v =2*P(z>2.64)=0.0082$  

And the best answer would be

B 0.0082