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## Statistical Analysis

by: Mr. Alex Berge

14

0

2

# Statistical Analysis STAT 401

Marketplace > University of Idaho > Statistics > STAT 401 > Statistical Analysis
Mr. Alex Berge
UI
GPA 3.55

Christopher Williams

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COURSE
PROF.
Christopher Williams
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Statistics

This 2 page Class Notes was uploaded by Mr. Alex Berge on Friday October 23, 2015. The Class Notes belongs to STAT 401 at University of Idaho taught by Christopher Williams in Fall. Since its upload, it has received 14 views. For similar materials see /class/227938/stat-401-university-of-idaho in Statistics at University of Idaho.

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Date Created: 10/23/15
1 Solutions to review problenls 1 Here are solutions to the rst set of review problems 477 The sampling distribution of is normal with mean pg 60 and standard deviation 07 TAE 5E 125 Since the distribution of is normal approximately 95 of the values of should fall within pg i 196 07 which is 60 i 196125 giving the interval 5755 6245 541 a A 99 con dence interval for p is given by it t 00514 s 3147 i2977504 3147 i387 which gives the interval 276 3534 b The question is a bit unclear on exactly what a null hypothesis might be but if we choose H0 p 35 then at the a 01 signi cance level we will fail to reject H0 since the 99 con dence interval includes 5 1165 a Yes the plotted points seem to follow a line b From the printout 1251 3583 mi 1166 a 33 5 2112397 272 MSE 1069 b From the printout 5561 696 c For this research hypothesis H0 l 0 and Ha l gt 0 since they are interested in detecting a positive relationship The p value in the printout for H0 l 0 is p 0004 but the printout is for the twosided alternative hypothesis Ha l 7E 0 Thus to get the p value for our oneesided alternative hypothesis we divide the printed 17 value by 2 yielding p 00042 0002 1130 a The plot looks good there could be one or more in uential points b The estimated regression equation is 9978 5192 mi and the residual standard deviation is 35 V MSE 148999 1221 A A c A 95 con dence interval for l is given by l i t 02528 56 1 yielding 5192 i 2048 586 or 519 i 12 giving an interval of 507 531 1131 a and b From the printout t 8853 with a p value ofp lt 0001 1132 a F 783726 and p lt 0001 b They are equal because both tests are testing the same null hypothesis when we do simple linear regression The test statistics are related by t2 F 1 Solutions to review problems 1 Here are solutions to the rst set of review problems 4187 The sampling distribution of is normal with mean My 60 and standard deviation 0 a 5m 1125 Since the distribution of is normal approximately 95 of the values of should fall within Myi1l96 03 which is 60 i 11961125 giving the interval 5755 6245 5155 a A 99 con dence interval for M is given by i t00514 s 31147 i21977 504E 3147 i387 which gives the interval 276 3534 b The question is a bit unclear on exactly what a null hypothesis might be but if we choose H0 M 35 then at the a 01 signi cance level we will fail to reject H0 since the 99 con dence interval includes 5 11135 a Yes the plotted points seem to follow a line b From the printout 12151 358311 1 1136 a 33 T 3y 7 3 MSE 1069 b From the printout 316731 6196 c For this research hypothesis H0 Bl 0 and Ha 61 gt 0 since they are interested in detecting a positive relationship The p value in the printout for H0 Bl 0 is p 0004 but the printout is for the twosided alternative hypothesis Ha 61 f 0 Thus to get the p value for our onesided alternative hypothesis we divide the printed p value by 2 yielding p 00042 0002 11139 a See the plot using SAS or SYSTATl b The estimated regression equation is 99178 51191 and the residual standard deviation is as xMSE 148999 121211 A A c A 95 con dence interval for 61 is given by 61 i t02523 816151 yielding 5 19 i2048 0586 or 519 i112 giving an interval of 507 5131 11140 a and b From the printout t 88153 with a p value of p lt 0001 11141 a F 7837126 and p lt 0001 b They are equal because both tests are testing the same null hypothesis when we do simple linear regression The test statistics are related by t2 Fl

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