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

`t=(370.69-360)/((24.36)/(√(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

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=(\bar X-\mu_o)/((s)/(√(n)))` (1)

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

`t=(370.69-360)/((24.36)/(√(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