Question

Computer value of the test statistic round your answer to two decimal places

Answer 1

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:

`z=\frac{\hat{p}-p}{\sqrt{(pq)/(n)}}`

where

`\begin{gathered} \hat{p}=sample\text{ }proportion \\ p=null\text{ }hypothesis \\ q=1-p \\ n=sample\text{ }size \end{gathered}`

From the question, the following parameters can be seen:

`\begin{gathered} \hat{p}=(133)/(169) \\ p=0.85 \\ q=1-0.85=0.15 \\ n=169 \end{gathered}`

Substitute known values into the formula:

`\begin{gathered} z=\frac{(133)/(169)-0.85}{\sqrt{(0.85\cdot0.15)/(169)}} \\ \therefore \\ z=-2.29 \end{gathered}`

Therefore, the test statistic is -2.29.