MATH 203 Sample Problem Unit 3
Bottles of liquid soap are normally distributed in weight with a mean of 64 ounces and a
standard deviation 0.5 ounces.
(a) Let various random samples of size 21 be collected. Compute P(0.47 ≤S≤0.55).
With
n
= 21 and = 0.5:
P(0.47 ≤S≤0.55) =P20 ×0.47
2
0.52≤(n−1)S2
2≤20 ×0.552
0.52
=P17.672 ≤2(20) ≤24.2
( )
≈0.37524 .
Use
2cdf(17.672, 24.2, 20)
(b) Suppose is unknown, but a random sampling of 26 bottles yields
S
= 0.60. Find a
95% confidence interval for the true standard deviation.
Use the 95% chi-square scores from the
2(n−1)
=
2(25)
curve
which are
L
= 13.12 and
R
= 40.65.
(n−1) ×S2
R≤ ≤ (n−1) ×S2
L
25 ×0.602
40.65 ≤ ≤ 25 ×0.602
13.12
0.47 ≤ ≤ 0.828
(c) Using
S
= 0.60 from a sample of 26 bottles, is there evidence, at the 10% level of
significance, to reject the claim that = 0.5? State null and alternative hypotheses, give
the test-statistic and
P
-value, and use the
P
-value to explain your conclusion in detail.
With
S
= 0.6 and
n
= 26 : We test H0: = 0.50 vs. Ha
>0.50
.
The test stat is (n−1)S
2
2=25 ×0.62
0.52=36
. The
P
-value is
2cdf(36, 1E99, 25) ≈0.0716
.
(Use the right tail for Ha
>0.50
.)
If = 0.50 were true, then there is a 7.16% chance
of obtaining an
S
of 0.60 or larger with a sample of size 26.
There is enough evidence to reject H0 with a 10% level of significance
