Question

After extensive measurements of the time necessary to complete the first homework assignment, a teacher determines that there is a population mean of 100 and a stardard deviation of 20. If she samples a class of 60 students and calculates a mean of 96 minutes, what is the z statistic (round to the nearest 2 decimal places, don't forget a negative sign if necessary)?

Answer 1

Answer: -1.55

Explanation:

Given : Mean :
`\mu=100`

Standard deviation :
`\sigma=20`

Sample size :
`n=60`

Sample mean :
`\overline{x}=96`

The test statistic for the population mean is given by :-

`z=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}`

`\Rightarrow\ z=(96-100)/((20)/(√(60)))=-1.54919333848\approx-1.55`

Hence, the value of z statistic = -1.55