Question

A simulated exercise gave n = 22 observations on escape time (sec) for oil workers, from which the sample mean and sample standard deviation are 370.46 and 24.89, 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 0.05.

State the appropriate hypotheses.

H₀: μ = 360
H_a: μ < 360H₀: μ = 360
H_a: μ ≤ 360    H₀: μ = 360
H_a: μ ≠ 360H₀: μ = 360
H_a: μ > 360

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t=P-value=

What can you conclude?

Reject the null hypothesis. There is sufficient evidence that true average escape time exceeds 6 min.Reject the null hypothesis. There is not sufficient evidence that true average escape time exceeds 6 min.    Do not reject the null hypothesis. There is sufficient evidence that true average escape time exceeds 6 min.Do not reject the null hypothesis. There is not sufficient evidence that true average escape time exceeds 6 min.

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

HYPOTHESIS TEST-

Suppose, random variable X denotes escape time (in seconds) for oil workers.

We have sample values. But we do not know population standard deviation (or variance). So, we have to perform one sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

[Using R-code '1-pt(1.971143,21)']

Level of significance

We reject our null hypothesis if

Here, we observe that

So, we reject our null hypothesis.

ANSWERS-

Hypotheses are are

Test statistic

We conclude as follows.

Reject the null hypothesis. There is sufficient evidence that true average escape time exceeds 6 minutes.