# Testing of mean and variance

Testing of mean and variance. See attached file for full problem description.

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#### Solution Preview

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 ...

#### Solution Summary

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

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