40 is what percent of 49
2 answers:
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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:
The hypothesis for the test can be defined as follows:
<em>H</em>₀: The true average escape time for oil workers is more than 360 seconds, i.e. <em>μ</em> > 360.
<em>Hₐ</em>: The true average escape time for oil workers is at most 360 seconds, i.e. <em>μ</em> ≤ 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:
Thus, the test statistic value is 1.81.
Compute the p-value of the test as follows:
*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.
Answer:
<u>Given equation:</u>
<u>Solving for p:</u>
p - any real number
We can state the initial expression is correct for any value of p.