DSCI 3710 LECTURE NOTES Dr. Nick Evangelopoulos
Hypothesis Testing Problem Case 5 (independent two-sample t, Equal 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
population standard deviations are assumed to be equal (
1 =
2). 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. Now assume that the two tire
populations have the same variance. 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:
21
2121
11
nn
s
xx
t
p
, where the pooled st. dev. is
)2(
)1()1(
21
2
22
2
11
nn
snsn
s
p
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 df = n1 + n2 – 2 = 28. Using a t table we find the critical
value to be 1.701.
The Decision Rule is “Reject H0 if the observed t value is more extreme than the critical t”.
Step 4 The pooled st. deviation is sp = SQRT((14*0.68^2+14*0.38^2)/28) =
0.55. The calculated t value is:
15
1
15
1
55.0
098.333.3
*
t
= (3.33 – 3.98) / (SQRT(1/15 + 1/15) = –3.25.
Step 5
Conclusion: Reject the null hypothesis H0 since the calculated value (–3.25) is more
extreme than the critical value (1.701). Therefore, we have significant evidence that the
two tire brands have different durabilities.
–tcr tcr
a/2
–tcr
a/2
tcr
0.05
–1.701
0.05
1.701
t* = –3.23
