Testing of mean and variance

Testing of meanand variance. See attached file for full problem description.

Please see the attached file.

Answer 4
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 =

One-Sample 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

Conclusion
Since that Z is grater than 1.96 we reject the null hypothesis the daily yield has not changed.

Answer 5

Descriptive …

Testing of mean and variance through students t test and F test are described in the solution.