During the winter of 2008-2009, the average utility bill for residents of a cert
ID: 3065849 • Letter: D
Question
During the winter of 2008-2009, the average utility bill for residents of a certain state was
$193
per month. A random sample of
60
customers was selected during the winter of 2009-2010, and the average bill was found to be
$185.12
with a sample standard deviation of
$18.4318.43.
Complete parts a and b below.
a) Using
alpha=0.050
does this sample provide enough evidence to conclude that the average utility bill in this state was lower in the winter of 2009-2010 than it was in the winder of 2008-2009?
Determine the hypotheses.
Determine the test statistic
Determine the P Value
Determine the Z Score
Reject or do not reject null hypothesis?
Explanation / Answer
Given :
sample mean is X¯=185.12 and the known population standard deviation =18.43,
and the sample size is n = 60
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: =193
Ha: <193
This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.
(2) Rejection Region
Based on the information provided, the significance level is =0.050, and the critical value for a left-tailed test is zc=1.64.
The rejection region for this left-tailed test is R={z:z<1.64}
(3) Test Statistics
The z-statistic is computed as follows:
z = [ (X¯0) / ( /n ) ] = [ ( 185.12-193) / (18.43/60) ] = -3.312
(4) Decision about the null hypothesis
Since it is observed that z=3.312<zc=1.64, it is then concluded that the null hypothesis is rejected.
P-value : The p-value is p = 0.0005, and since p = 0.0005 < 0.050, it is concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is rejected. There is enough evidence to conclude that the average utility bill in this state was lower in the winter of 2009-2010 than it was in the winder of 2008-2009.
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