Question

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.

Answer 1

Null hypothesis:
`\mu \leq 6`

Alternative hypothesis:
`\mu > 6`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(6.178-6)/((0.68)/(√(26)))=1.335`

`p_v =P(t_(25)>1.335)=0.097`

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

Explanation:

Information given

`\bar X=370.69/60 =6.178` represent the sample mean

`s=24.36/36=0.68` represent the standard deviation for the sample

`n=26` sample size

`\mu_o =6` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` 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:
`\mu \leq 6`

Alternative hypothesis:
`\mu > 6`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(6.178-6)/((0.68)/(√(26)))=1.335`

The degrees of freedom are:

`df=n-1=26-1=25`

The p value would be given by:

`p_v =P(t_(25)>1.335)=0.097`

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.