Question

A simulated exercise gave n = 20 observations on escape time (sec) for oil workers, from which the sample mean and sample standard deviation are 370.42 and 25.74, respectively. Suppose the investigators had believed a priori that true average escape time would be at most 6 min. Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significance level of .05. (Give t to 2 decimal places and the p-value to 3 decimal places.)t =P-value =ConclusionReject the null hypothesis, there is significant evidence that true average escape time exceeds 6 min. Reject the null hypothesis, there is not significant evidence that true average escape time exceeds 6 min. Fail to reject the null hypothesis, there is not significant evidence that true average escape time exceeds 6 min. Fail to reject the null hypothesis, there is significant evidence that true average escape time exceeds 6 min.

Answer 1

Reject the null hypothesis, there is significant evidence that true average escape time exceeds 6 min.

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:

`n=20\\\bar x=370.42\\s=25.74\\\alpha =0.05`

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:

`t=(\bar x-\mu)/(\s/√(n))=(370.42-360)/(25.74/√(20))=1.81`

Thus, the test statistic value is 1.81.

Compute the p-value of the test as follows:

`\text{p-value}=P(t_(n-1)<t)`

`=P(t_(20-1)<1.81)\\\\=P(t_(19)<1.81)\\\\=0.044`

*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.