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

Answer:

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

Step-by-step 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:

 $t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{370.42-360}{25.74/\sqrt{20}}=1.81$

Thus, the test statistic value is 1.81.

Compute the p-value of the test as follows:

 $\text{p-value}=P(t_{n-1}$

             $=P(t_{20-1}$

*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.