Question 1207008: A new variety of pearl millet is expected to provide an increased yield over the variety presently in use which is about 90 bushels per acre. The new variety of millet produced an average yield of
x = 98 bushels
per acre with a standard deviation of
s = 12.4 bushels
based on 40 one-acre yields.
Find the value of the test statistic for testing the hypotheses that the new variety will increase yield. (Round your answer to two decimal places.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! old = 90 bushels per acre.
new = 98 bushels per acre.
standard deviation of 12.4 bushels an acre based on a study of 40 one acre yields.
sample size appears to be 40.
since the standard deviation is taken from a sample, the use of the t-score is indicated.
degrees of freedom = 39 (sample size minus 1).
standard error = 12.4 / sqrt(40) = 1.9606 rounded to 4 decimal places.
t-score = (98-90)/1.9606 = 4.0803 rounded to decimal places.
area to the right of that t-score with 39 degrees of freedom = 1.07619 * 10^-4.
that's equivalent to .000107619 in standard form.
that's a very high t-score and a very low p-score, indicating the results are significant, indicating that there is a very high probability that the number of bushels in the population has definitely increased due to the introduction of the new variety of pearl millet.
at 99% confidence interval, the two-tailed critical t-score with 39 degrees of freedom is equal to plus or minus 2.7 and the critical p-value is equal to .005.
the test statistics exceed those critical values, with the conclusion being that the results are significant.
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