.
1-617-275-8164

Sample SAS Assignment

Need Help on a Physics Assignment, Email us or call us at 1-617-275-8164

Conclusion:

Here the calculated t value is 3.40(equal variances).

The tabulated t value at 18 df at 5% level of significance is 2.10.

So our calculated t value is greater than the tabulated t value so we reject the null hypothesis. So
we can conclude that there is a significant difference between type1 and type2. .

SAS program for this problem:

data problem1;
input type nox @@;
cards;

1 1.65 1 1.78 1 1.12 1 1.26 1 1.21 1 1.89 1 1.13 1 1.35 1 1.96
1 1.70 1 1.87 1 1.14
2 1.11 2 1.06 2 1.08 2 1.09 2 1.05 2 1.15 2 1.12 2 1.14;


proc print;
title 'emission control device comparison';
run;
proc ttest;
class type;
run;           

    (b)   From the results labeled “garage paired t comparison,” based on the data in problem 3,
            indicate your conclusion from testing H0: μD = 0 against Ha: μD =/ 0, with α = .01. Give an 
            interpretation of the attained level of significance shown on the printout


Answer:  Null hypothesis (H0): μD = 0
              Alternate hypothesis (H1): μD |= 0 
              Number of observations =n=10


garage paired t comparison

Obscarnbrgarage1garage2diff
118.69.3-0.7
2211.211.10.1
3310.510.40.1
442.33.5-1.2
554.04.7-0.7
667.37.8-0.5
776.36.9-0.6
887.27.3-0.1
993.24.0-0.8
10105.86.2-0.4

garage paired t comparison
The UNIVARIATE Procedure
          Variable: diff;
 

Moments


N 10Sum Weithts10
Mean -0.48Sum Observations-4.8
Std Deviation 0.4157991Variance0.17288889
Skewness 0.03199464Kurtosis-0.4595706
Uncorrected SS 3.86Corrected SS1.556
Coeff Variation -86.624812Std Error Mean0.13148722





Free Trial

Submit your homework or assignment.