Week 7 examples_lecture_2011
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The document provides examples for statistical hypothesis testing concepts covered in Week 7. It includes examples of identifying the null and alternative hypotheses based on a given claim. It also works through examples of conducting one-tailed and two-tailed hypothesis tests using z-scores and the NORMSDIST function in Excel to determine p-values and compare them to the significance level to determine whether to reject or fail to reject the null hypothesis.
Week 7 examples_lecture_2011
- 1. Week 7 ExamplesFor Week 7 Module
- 3. Week 7 ExamplesRead the question carefully. They say the claim is that their weekly losses are no more than $2500This means the claim would be that the weekly losses are less than or equal to $2500So the claim would be the null hypothesis in this case.The alternative hypothesis would be that their losses are greater than $2500
- 5. Week 7 ExamplesRemember that the null hypothesis always contains equality in that it will either have an = sign, a ≤ sign or a ≥ sign.The alternative hypothesis always has inequality in that it will either have a ≠ sign, a < sign or a > sign.
- 7. Week 7 Examples1 – 0.05 is 0.95
- 10. Week 7 ExamplesWhy is the z = -2.05?Let me show you.
- 11. Week 7 Examples
- 12. Week 7 Examplesz = (575-500)/ (65/sqroot(45)) = 75/9.69 = 7.74
- 13. Week 7 Examples7.74 is not in the range of the critical values +/- 2.33, thus we reject the H0
- 14. Week 7 ExamplesUse “NORMSDIST” function in Excel “=NORMSDIST(-0.8767)” to get 0.190325 or 0.19 rounded to two decimal places. Since it is “two-tailed” (because the Ho has an Equal sign, I must multiply this number by two to get 0.38
- 15. Week 7 ExamplesWith an alpha of 0.05 and looking at the probability of 0.38, we conclude there is notenough evidence to reject the claim.
- 16. Week 7 Examples
- 17. Week 7 Examples1 - 0.01 = 0.99MY ANSWER
- 18. Week 7 ExamplesAlso remember this was “right-tailed”because Ho was p is less than or equalto 0.50
- 19. Week 7 ExamplesP hat or a “p” with a little rooftop on it in this case is 7 over 100since 7 out of 100 had to pay out. p is 5% or 0.05, and n is mynumber of samples or 100.
- 20. Week 7 ExamplesThis would be (.07 -.05)/ sqroot of ((.05(1-.05))/100) or.02/sqroot of (.000475) which is.02/.02179 which is 0.92 rounded to two decimal places.
- 21. Week 7 ExamplesNote the “may” when you can not rejectHo, you still can’t be definitive.
- 22. Week 7 Examples
- 23. Week 7 ExamplesUse “NORMSDIST” function in Excel “=NORMSDIST(1.9596)” to get 0.975 rounded. NOWWE NEED TO SAY 1- THIS NUMBER or 1- 0.975 to get .025 and compare it to our alpha of.05. Since it is less than alpha it supports H1 and thus strongly supports the “claim”.