Question

A sample of 26 offshore oil workers took part in a simulated escape exercise, and their escape time (unit: second) were observed. The sample mean and sample standard deviation are 370.69 and 24.36, respectively. 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

Answer 1

Answer:

$t=\frac{370.69-360}{\frac{24.36}{\sqrt{26}}}=2.238$  

The degrees of freedom are given by:

$df = n-1= 26-1=25$

And the p value would be:

$p_v =P(t_{25}>2.238)=0.0172$  

If we use a 5% of significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 360 second or 6 minutes. We need to be careful since if we use a significance level of 1% the result change

Step-by-step explanation:

Information given

$\bar X=370.69$ represent the sample mean

$s=24.36$ represent the sample standard deviation

$n=26$ sample size  

$\mu_o =6*60 =360 s$ represent the value to verify

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to check if the true mean is at most 360 seconds, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 360$  

Alternative hypothesis:$\mu > 360$  

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

We can replace in formula (1) the info given like this:  

$t=\frac{370.69-360}{\frac{24.36}{\sqrt{26}}}=2.238$  

The degrees of freedom are given by:

$df = n-1= 26-1=25$

And the p value would be:

$p_v =P(t_{25}>2.238)=0.0172$  

If we use a 5% of significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 360 second or 6 minutes. We need to be careful since if we use a significance level of 1% the result change