Answer:
Decision is reject the [tex]H_0[/tex] as There is sufficient evidence to support the claim that the mean time is less then 5 minutes.
Step-by-step explanation:
Consider the provided information.
The mean waiting time for a bus during rush hour is less than 5 minutes.
Thus the null hypothesis is:
[tex]H_0: \mu\geq 5[/tex]
[tex]H_1: \mu< 5[/tex]
A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At α = 0.01,
Now use the table to find the z value, where α = 0.01,
[tex]t_L = -2.3264[/tex]
The value of n is 20, sample mean is 3.7, population mean is 5 and standard deviation is 2.1.
Now, use the formula: Standardized test statistic for z-scores =[tex]t=\frac{\bar x-\mu_0}{{s}/{\sqrt{n} }}[/tex].
Substitute the respective values in the above formula.
[tex]t=\frac{3.7-5}{{2.1}/{\sqrt{20} }}[/tex]
[tex]t=\frac{-1.3}{0.47}[/tex]
[tex]t=-2.768465115\approx-2.77[/tex]
As, the test statistic is in the reject interval, so reject [tex]H_0[/tex]
There is sufficient evidence to support the claim that the mean time is less then 5 minutes.