Question 1207008
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.