Testing of meanand variance. See attached file for full problem description.
Please see the attached file.
Here we test the null hypothesis H0:
i.e the average daily yield =880.
The test statistic used is . If calculated value of Z is less than , we reject the null hypothesis.
Here Z =
Test of mu = 880 vs not = 880
The assumed standard deviation = 21
N Mean SE Mean 95% CI Z P
50 871.000 2.970 (865.179, 876.821) 3.03 0.002
Since that Z is grater than 1.96 we reject the null hypothesis the daily yield has not changed.
Testing of mean and variance through students t test and F test are described in the solution.