DSCI 3710 LECTURE NOTES Dr. Nick Evangelopoulos
Hypothesis Testing Problem Case 4 (independent two-sample t, Unequal Var.)
Situation: There are two unknown population means
1 and
2. We have information
on two samples coming from these two populations, and we know the sample means
1
x
and
2
x
, the sample standard deviations s1 and s2, and the sample sizes n1 and n2. The
sample sizes are small (at least one under 30). Then, we do a t test for two means.
Example: Refer to the Checker Cab example, text, p.371. The sample information is:
Beltex:
1
x
=3.33, s1 = 0.68, and n1 = 15.
Roadmaster:
2
x
=3.98, s2 = 0.38, and n2 = 15.
Can we conclude that the average blowout times are not the same? Test at a = 0.10.
Solution:
Step1 H0:
1 =
2OR, equivalently: H0:
1 –
2 = 0
HA:
1 ≠
2HA:
1 –
2 ≠ 0
Step 2 Assuming that H0 is true, the following quantity will have a t distribution:
2
2
2
1
2
1
2121
n
s
n
s
xx
t
Step 3 Since the HA here is of the “not-equal-to” type, the test is two-tailed. The
tail probability is a = 0.10. The degrees of freedom are computed by a long formula.
Here, df = 21.9, rounded to 22. Using a t table we find the critical value to be 1.717.
The Decision Rule is “Reject H0 if the observed t value is more extreme than the critical t
value”.
Step 4 The calculated value is:
15
)38(.
15
)68(.
098.333.3
*
22
t
= (3.33 – 3.98) / SQRT(((.68)^2)/15 + ((.38)^2)/15) = –3.23.
Step 5
Conclusion: Reject the null hypothesis H0 since the calculated value (–3.23) is more
extreme than the critical value (1.717). Therefore, we have significant evidence that the
two tire brands have different durabilities.
–tcr tcr
a/2
–t
cr
a/2
t
cr
0.05
–1.717
0.05
1.717
t* = –3.23
