The escape time (sec) for oil workers in a simulated exercise, gave the sample mean 370.69, sample standard deviation 24.36, and number of observations as n =26. Suppose the investigators had believed a priori that true average escape time would be at most 6 minutes. Does the data contradict this prior belief? Assuming normality, test the appropriate hypothesis using a significance level of .05.

And for this case the p value is higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is at most 6 minutes

Step-by-step explanation:

Information given

[tex]\bar X=370.69/60 =6.178[/tex] represent the sample mean

[tex]s=24.36/36=0.68[/tex] represent the standard deviation for the sample

[tex]n=26[/tex] sample size

[tex]\mu_o =6[/tex] represent the value to verify

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value

System of hypothesis

We want to test if the true mean is at least 6 minutes, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 6[/tex]

Alternative hypothesis:[tex]\mu > 6[/tex]

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

Replacing the info we got:

[tex]t=\frac{6.178-6}{\frac{0.68}{\sqrt{26}}}=1.335[/tex]

The degrees of freedom are:

[tex]df=n-1=26-1=25[/tex]

The p value would be given by:

[tex]p_v =P(t_{25}>1.335)=0.097[/tex]

And for this case the p value is higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is at most 6 minutes.