OPPapers.com Essay Index >> Miscellaneous >> Stats 2
We have many free term papers and essays on Stats 2. We also have a wide variety of research papers and book reports available to you for free. You can browse our collection of term papers or use our search engine.
stats 2 ADM 2304 E - Statistical Methods Assignment #1 Problem 1 a) Test and CI for Two Proportions: bmi_A, bmi_B Null Hypothesis: No significant difference Alternative
stats Homework Assignment # 4 Problem MBS-2. Rush Hour Traffic in ActivStats (Tests for a Mean Homework) In order to reduce the average number of vehicles that use
Euro disney case Technical Documentation Stats & Matchups Contents 1. Document information 3 1.1 Version management 3 2. Introduction 4 3. Program architecture 5
I Like Playing With Monkeys Kevin Millar - Great '03, 0kay '04, mediocre '05 and same stats in '06 2.1 million this year - not the answer Bruce Chen - w/Bos in '03
Stats Homework CE 103 Statistics Homework Problems Homework #3: Problems 1-2 are due in class on February 27, 2007. 1. For the data on page 45, problem 2.8 in SS,
Submitted by britt85 on January 30, 2007
Category: Miscellaneous
Words: 1127 | Pages: 5
Views: 193
Popularity Rank: 73,520
Average Member Grade: N/A (Add a Comment / Grade this Paper)
ADM 2304 E - Statistical Methods Assignment #1
Problem 1
a)
Test and CI for Two Proportions: bmi_A, bmi_B
Null Hypothesis: No significant difference
Alternative Hypothesis: Significant difference
Event = 1
Variable X N Sample p
bmi_A 192 1342 0.143070
bmi_B 97 720 0.134722
Difference = p (bmi_A) - p (bmi_B)
Estimate for difference = 0.143070 0.134722 => 0.00834782
90% CI for difference = (-0.0178288, 0.0345244)
Test for difference = 0 (vs not = 0) Z = 0.52 P-Value = 0.600
Therefore, based on these results we would not reject the null hypothesis. There is no significant difference in the weights of both cities.
Manual
Pooled P-hat = (192 + 97) / (1342 + 720)
= 289 / 2062
= 0.140155
Z-stat = (0.143070 0.134722) 0 / sqrt[(0.140155) x (1-0.140155) x (1/1342) x(1/720)]
= 0.008348 / 0.016037
= 0.520546
b)
Confidence Interval = (p1 p2) +/- Zá/2 * [sqrt ((p1(1-p1)/n1) + p2(1-p2)/n2)]
= (0.143070 0.134722) +/- 1.645 * [sqrt(0.14307(1-0.14307)/1342) + 0.134722(1-0.134722)/720)
= 0.008348 +/- 0.026179
Therefore this equals the interval of
(-0.0178288, 0.0345244)
The confidence interval calculated above is consistent with the conclusion in question A.
Problem 2
a)
Test and CI for One Proportion: C3
Test of p = 0.5 vs p not = 0.5
Event = 1
Variable X N Sample p 95% CI Exact P-Value
C3 1247 1937 0.643779 (0.621990, 0.665132) 0.000
b & c)
Test and CI for One Proportion: C4, C5, C6, C7, C8, C9, C10, C11, ......
You must Login to view the entire paper.
If you are not a member yet, Sign Up for free!