Download 1. You are testing H0 : µ = 10 vs H a : µ < 10 based on a random

1. You are testing H0 : µ = 10 vs Ha : µ < 10 based on a random sample of
20 observations from a normal distribution. The data gave
ȳ = 8, s = 4. The value of the t statistic is
a.-0.5
b. -10
c.-2.24
answer: c.
t=
8−10
√
4/ 20
= −2.24.
2. You are testing H0 : µ = 10 vs Ha : µ < 10 based on a random sample of
20 observations from a normal distribution. The t statistic is -2.25. The
p-value of the test is
(0.0182).
R code: pt(-2.25,19).
3. Data on the blood cholesterol levels of 24 rats (milligrams per deciliter of
blood) give ȳ = 85, s = 12. A 95% CI for the mean blood cholesterol of
rats is
a. 79.9 to 90.1
b. 80.2 to 89.8
c.84.0 to 86.0.
answer : a.
√
85 ± 2.069 ∗ 12/ 24 = (79.9, 90.1).
4. The following is a summary of stress limit for specimens using two types
of wood.
Wood Type
Red Oak
Fir
sample size
15
15
sample mean
8.50
7.70
sample sd
0.80
1.25
Assume the two population variances are different. Use smaller sample
size-1 as the degrees of freedom.
What t critical value will you use in getting a 90% CI for the difference
between the average stress limit for red oak and for fir?
t=1.761.
The 90% CI is from
p
8.5 − 7.7 ± 1.761 0.802 /15 + 1.252 /15 = (0.13, 1.47).
Test whether the true average stress limit for red oak is different from that
for fir.
1
H0 : µ1 − µ2 = 0, Ha : µ1 − µ2 6= 0.
The t statistic is t = √ 28.5−7.7 2
0.8 /15+1.25 /15
= 2.09.
The p-value is 0.055.
R code: 2(1-pt(2.09,14)).
The next two parts assume the two population variances are equal.
The pooled sample standard deviation sp is
s2p =
14∗0.82 +14∗1.252
30−2
= 1.05.
Test again whether the two population means are equal. What is the t
statistic now?
t = 2.09.
What degrees of freedom will you use for this test? (d.f.=28.). the p-value
is (0.046).
2