Testing of mean and variance
Not what you're looking for?
Testing of mean and variance. See attached file for full problem description.
Purchase this Solution
Solution Summary
Testing of mean and variance through students t test and F test are described in the solution.
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 ...
Purchase this Solution
Free BrainMass Quizzes
Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
Know Your Statistical Concepts
Each question is a choice-summary multiple choice question that presents you with a statistical concept and then 4 numbered statements. You must decide which (if any) of the numbered statements is/are true as they relate to the statistical concept.
Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions.
Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.