Answers - Millersville University
Answers - Millersville University
Answers - Millersville University
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“There is not sufficient sample data to support the claim that the mean<br />
length of commercials is less than 30 seconds.”<br />
5. A pharmaceutical company makes a pill intended for children susceptible to seizures.<br />
The pill is supposed to contain 20.0 mg of phenobarbital. A random sample of 20 pills<br />
yielded the amounts (in mg) listed below. Do the pills contain the proper amount of<br />
phenobarbital at the α = 0.01 significance level? (Hint: The sample standard deviation<br />
of the weights is s = 3.5.)<br />
27.5 26.0 22.9 23.5 23.0<br />
23.9 26.3 20.9 22.5 21.7<br />
28.4 16.1 19.3 17.2 17.9<br />
21.7 23.0 16.5 24.1 22.3<br />
(a) State the original claim in symbolic form.<br />
If µ represents the mean amount of phenobarbital measured in milligrams in the<br />
pills, then the claim above can be expressed in the symbolic form: µ = 20.0.<br />
(b) In non-technical language describe the type II error for this hypothesis test.<br />
A type II error occurs when a we fail to reject a false null hypothesis. In this<br />
case a type II error occurs when we fail to reject claim that the pills contain an<br />
average of 20.0 milligrams of phenobarbital, when in fact they do not contain that<br />
amount.<br />
(c) Find the critical value of t for this hypothesis test.<br />
Since the sample size is only n = 20, which is considered a small sample, we must<br />
use the Student-t Distribution. The degrees of freedom becomes df = 19. The<br />
original claim is the null hypothesis. The alternative hypothesis is the claim that<br />
µ = 20.0 and thus we are conducting a two-tailed test. The critical value of t is<br />
the value of t for a two-tailed test with α = 0.01 and df = 19. From the Student-t<br />
table this value is tα/2 = ±2.86.<br />
(d) Do the pills contain the proper amount of phenobarbital at the α = 0.01 significance<br />
level?<br />
We must calculate the sample mean before we can find the test statistic.<br />
The test statistic is then<br />
t0 =<br />
x =<br />
x<br />
n<br />
= 444.7<br />
20<br />
≈ 22.24<br />
x − µ<br />
s/ √ 22.24 − 20.0<br />
=<br />
n 3.5/ √ ≈ 2.862.<br />
20<br />
The test statistic is (barely) larger than the critical value. Thus we reject the null<br />
hypothesis. We conclude that the sample data warrant rejection of the claim that<br />
the mean amount of phenobarbital in the pills is 20.0 milligrams.
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