Statistics

Week 7 Discussion Example : Week 7 Discussion Board Example – Test of Hypothesis – Corn Yield per Acre In 2013, the national average corn yield in the U.S. was 158.8 bushels per acre. A county agricultural agent wanted to determine if this year’s rate in her county is above the national average. She collected information from 25 different corn growers and found a sample mean of 161.3 bushels per acre with a standard deviation of 6.2 bushels. She asked you to conduct a test of hypothesis to determine if the county average exceeds the 2013 national average. Use what you have learned about hypothesis testing to answer the following questions. What type of test should you perform? W hich of the three equation for the test statistic should you use? Why did you choose these? You may assume production is normally distributed. We were asked to determine if the county average exceeds the national average so we are only testing to see if ̅> 0. This makes it an upper 1 -tail test. Since we are testing an average value, and we are not given the population standard deviation, σ, we should use the equation for testing of the mean, when σ is not known. State your null and alternate hypotheses. Why did you choose these values and mathematical operators? The 158.8 bushels per acre is the 2013 national average. We used the Less Than or Equal sign in the null hypothesis because we assume the county average is less than or equal to the national = ̅ − √ ⁄ 0: ≤ 158.8 : > 158 .8 average, until the test provides evidence to the contrary. We used the Greater Than sign in the alternate hypothesis because this is the claim we are testing. That the county averag e is greater the 2013 national average. What is the value of your test statistic? (Clearly show how you arrived at this value) ̅ = 161.3 s = 6.2 n = 25 tStat = (161.3 – 158.8) / (6.2 / Sqrt(25)) tStat = 2.0161 Interpret the test statistic: Choose an appropriate confidence level, then evaluate the test statistic using either the critical values or the p -value approach. (You are expected to use EITHER the Critical values, OR the P -value method. NOT BOTH). Why did you choose the confidence level that you did? Clearly state the outcome of your test of hypothesis. Reading the problem statement, there does not appear to be anything of a critical nature riding on this analysis, therefore I chose a 90% confidence level . Using the critical values method , I need to find the t -value that separates the upper 10% of the distribution. Since this is a 1 -tail test all of α is n the upper tail. I used the Excel function, T.INV(1 - α, n -1) to find the critical value. T.INV(0.90, 24) = 1.3178 Since the tests statistic is larger than the critical value (in the Reject Region) REJECT the Null Hypothesis. Using the p -value method , I need to find the probability of getting a t -value equal to or greater than the test statistic, if the n ull hypothesis is true. Because this is an upper 1 -tailed test I used 1 minus the Excel function T.DIST(tStat, n -1) to find the probability. = ̅ − √ ⁄ 1 - T.DIST(2.0161, 24) = 0.028 Since the p -value is smaller than α (0.10) REJECT the Null Hypothesis. What does yo ur outcome mean in statistical terms? For the critical method: We can be at least 90% confident that the mean of the distribution that the sample was pulled from is greater than 158.8. For the p -value method: We can be 97.2% confident that the mean of the distribution that the sample was pulled from is greater than 158.8. What does your outcome mean in terms of the problem? For the critical method: We can be at least 90% confident that the average corn yield in the county is greater than the 2013 nationa l average of 158.8 bushels per acre. For the p -value method: We can be 97.2% confident that the average corn yield in the county is greater than the 2013 national average of 158.8 bushels per acre.