Respuesta :
Answer:
Yes, it contradict this prior belief as there is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes.
Test statistic t=2.238>tc=1.708.
The null hypothesis is rejected.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the true average escape time is significantly higher than 6 minutes (360 seconds).
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=360\\\\H_a:\mu> 360[/tex]
The significance level is 0.05.
The sample has a size n=26.
The sample mean is M=370.69.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=24.36.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{24.36}{\sqrt{26}}=4.777[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{370.69-360}{4.777}=\dfrac{10.69}{4.777}=2.238[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=26-1=25[/tex]
The critical value for a right-tailed test with a significance level of 0.05 and 25 degrees of freedom is tc=1.708. If the test statistic is bigger than 1.708, it falls in the rejection region and the null hypothesis is rejected.
As the test statistic t=2.238 is bigger than the critical value t=1.708, the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes (360 seconds).
