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.
.
proc print;
title 'emission control device comparison'; run;
proc ttest; classtype;
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
Obs
carnbr
garage1
garage2
diff
1
1
8.6
9.3
-0.7
2
2
11.2
11.1
0.1
3
3
10.5
10.4
0.1
4
4
2.3
3.5
-1.2
5
5
4.0
4.7
-0.7
6
6
7.3
7.8
-0.5
7
7
6.3
6.9
-0.6
8
8
7.2
7.3
-0.1
9
9
3.2
4.0
-0.8
10
10
5.8
6.2
-0.4
garage paired t comparison
The UNIVARIATE Procedure
Variable: diff;