Answer:
[tex]t_{stat} = -2.03[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 16 days = 384 hours
Sample mean, [tex]\bar{x}[/tex] = 15 days 10 hours = 370 hours
Sample size, n = 19
Sample standard deviation, s = 30 hours
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 384\text{ hours}\\H_A: \mu < 384\text{ hours}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{370 - 384}{\frac{30}{\sqrt{19}} } = -2.03415 \approx -2.03[/tex]
The value of t-statistic is -2.03