Question

5. (16) A which the sample simulated exercise gave n 23 observations on escape time (sec) for oil workers, from mean and sample standard deviation are 372.99 and 22.51, respectively ippose the investigators had believed a priori that true average escape time would be at most 6 m Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significa nce level of 0.0S. a) State the appropriate hypotheses. b) Calculate the test statistic and determine the P-value. (Round your test statistic to two places and your P-value to three decimal places.) decimal

Answer 1

5)

Solution :

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   $\mu$ $\leq$ 6

H_a : $\mu$ > 6

$\bar x$ = 372.99 / 60 = 6.2165

$\mu$ = 6

s = 0.375167

n = 23

df = n - 1 = 23 - 1 = 22

Test statistic = t = ( $\bar x$ - $\mu$ ) / s / $\sqrt$ n

= (6.2165 - 6) /0.375167 / $\sqrt$ 23 = 2.77

Test statistic = 2.77

This is the right tailed test .

P-value = 0.0056

$\alpha$ = 0.05

P-value < $\alpha$

Reject the null hypothesis .

Cliam is true .