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A simulated exercise gave n = 22 observations on escape time (sec) for oil workers, from which the sample mean and sample standard deviation are 370.23 and 25.74, respectively. Suppose the investigators had believed a priori that true average escape time would be at most 6 min.

1. Does the data contradict this prior belief?
2. Assuming normality, test the appropriate hypotheses using a significance level of 0.05. State the appropriate hypotheses.

User Yatg
by
6.7k points

1 Answer

3 votes

Answer:

Explanation:

Let us create hypotheses as


H_0:  \bar x = 6\\H_a: \bar x \leq 6

(left tailed test at 5% significance level)

Sample size n =22

Sample mean =370.23

Sample std deviation =25.74

Since population std deviation is not known we can use t test only

Std error of mean =
(25.74)/(√(22) ) \\=5.488

Test statistic t = mean differene/std error

=
(370.23-6(60))/(5.488) \\=1.86

p value for t one tailed for df 21= 0.0384

Since p <0.05 we reject null hypothesis

The true average time is contradicting the prior belief.

User Kumar V
by
6.9k points
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