Answer:
Reject the null hypothesis, there is significant evidence that true average escape time exceeds 6 min.
Step-by-step explanation:
In this case we need to test whether the data contradict the prior belief that the true average escape time for oil workers would be at most 6 min or 360 seconds.
The information provided is:
[tex]n=20\\\bar x=370.42\\s=25.74\\\alpha =0.05[/tex]
The hypothesis for the test can be defined as follows:
H₀: The true average escape time for oil workers is more than 360 seconds, i.e. μ > 360.
Hₐ: The true average escape time for oil workers is at most 360 seconds, i.e. μ ≤ 360.
As the population standard deviation is not known we will use a t-test for single mean.
Compute the test statistic value as follows:
[tex]t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{370.42-360}{25.74/\sqrt{20}}=1.81[/tex]
Thus, the test statistic value is 1.81.
Compute the p-value of the test as follows:
[tex]\text{p-value}=P(t_{n-1}<t)[/tex]
[tex]=P(t_{20-1}<1.81)\\\\=P(t_{19}<1.81)\\\\=0.044[/tex]
*Use a t-table.
Thus, the p-value of the test is 0.044.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
p-value = 0.044 < α = 0.05
The null hypothesis will be rejected at 5% level of significance.
Thus, concluding that the true average escape time would be at most 6 min.