GIVEN
The study claims that 85% of students are procrastinators. From a random sample of 169 students, 133 identify as procrastinators. The level of significance is 0.05.
SOLUTION
The test statistic (z-score) can be calculated using the formula:
[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}[/tex]
where
[tex]\begin{gathered} \hat{p}=sample\text{ }proportion \\ p=null\text{ }hypothesis \\ q=1-p \\ n=sample\text{ }size \end{gathered}[/tex]
From the question, the following parameters can be seen:
[tex]\begin{gathered} \hat{p}=\frac{133}{169} \\ p=0.85 \\ q=1-0.85=0.15 \\ n=169 \end{gathered}[/tex]
Substitute known values into the formula:
[tex]\begin{gathered} z=\frac{\frac{133}{169}-0.85}{\sqrt{\frac{0.85\cdot0.15}{169}}} \\ \therefore \\ z=-2.29 \end{gathered}[/tex]
Therefore, the test statistic is -2.29.
