Q1a) z-Test of a Single Population Mean
SOLUTION
Given
Sample mean: 
Hypothesized population mean: 
Population standard deviation: 
Sample size: 
Hypotheses
Null Hypothesis
![\[H_0: \mu = 48 \]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-72e98845811acfef51bd8647603c84d1_l3.png "Rendered by QuickLaTeX.com")
(The average customer watches 48 hours or less of a sport channel per month.)
Alternative Hypothesis
![\[H_1: \mu > 48 \]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-3eb6d5eab33c1efae1e04d219910d56e_l3.png "Rendered by QuickLaTeX.com")
(The average customer watches more than 48 hours of a sport channel per month.)
Test Statistic
We use the formula for the z-test of one population mean:
![\[z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-258eb3c028d839ad73647ca26818d489_l3.png "Rendered by QuickLaTeX.com")
![\[z = \frac{49.6 - 48}{9/\sqrt{48}}\]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-20646f517acf34ee09bb61a2aae88137_l3.png "Rendered by QuickLaTeX.com")
![\[z = \frac{1.6}{1.3} \]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-087359eadf446ccab130d07985a2a1ea_l3.png "Rendered by QuickLaTeX.com")
![\[z = 1.23\]](https://statsdoesntsuck.com/wp-content/ql-cache/quicklatex.com-cfb5ef4dfd65275437b8616e1168a28c_l3.png "Rendered by QuickLaTeX.com")
Rejection Region
Critical Value
For a significance level 
, and since this is a right-tailed test (due to 
), we find the critical z-value from the z-table as 
.
Decision Rule
Reject the 
if 
Conclusion
Since the calculated z-value of 
is not greater than the critical z-value of , we do not reject the
Therefore, at the 1% significance level, we do not have sufficient evidence to conclude that the average customer watches more than 48 hours of a sport channel per month.
This means that the claim by the Advance cable network that the average customer watches 48 hours or less per month stands.