Respuesta :
Answer:
Both the approach gave the same conclusion:
There is enough evidence to support the claim that the mean waiting time is shorter than six minutes
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 6 minutes
Sample mean, [tex]\bar{x}[/tex] = 5.46 minutes
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, s = 2.475 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 6\text{ minutes}\\H_A: \mu < 6\text{ minutes}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{5.36 - 6}{\frac{2.475}{\sqrt{100}} } = -2.5858[/tex]
a) Use the critical value approach
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = -1.6603[/tex]Since, the calculated test statistic is less than the critical value of t, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that new system reduces the time customers spend waiting for teller service during peak hours from the current 9 to 10 minutes to less than 6 minutes.
b) Use the p-value approach
The p-value can be calculated as:
P-value = 0.0055
Since the p-value is less than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that new system reduces the time customers spend waiting for teller service during peak hours from the current 9 to 10 minutes to less than 6 minutes.
