.0134 > .01 . do not reject null hypothesis
at a=.05, reject and state that the Midwest is lower
.0134 < .05
reject null hypothesis
2.
CCN and ActMedia provided a television channel targeted to individuals
waiting in supermarket checkout lines. The channel showed news, short
features, and advertisements. The length of the program was based on the
assumption that the population mean time a shopper stands in a
supermarket checkout line is 8 minutes. A sample of actual waiting times will
be used to test this assumption and determine whether actual mean waiting
time differs from this standard.
a.
Formulate the null and alternative hypotheses.
Since we are looking at
difference
Ho:
μ
=
8
Ha: μ ≠
8
b.
A sample of 120 shoppers showed a sample mean waiting time of 8.4
minutes.
If σ = 3, what is the test statistic?
T= 1.46
c.
What is the critical value(s) for a level of significance of .05 and what is
the rejection rule?

Critical Z at .05 = 1.96 so reject if the t value is
greater or less than 1.96
1.46 < 1.96
REJECT
d.
Are waiting times different from the standard of 8 minutes?
e.
No, there is not sufficient evidence that waiting times are
different that 8 minutes.
f.
Compute a 95% confidence interval for the population mean. Does it
support your conclusion?
Using x-bar=8.4 and sx=3 and n=120
7.86 to 8.94 (95% confident that waiting times are between 7.86
and 8.94 minutes.)
Yes, this supports the conclusion that waiting
times are 8 minutes because 8 is in the confidence interval.

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- Fall '14
- Statistics, Normal Distribution, Null hypothesis, Statistical hypothesis testing, Midwest