Answer:
(a) <em>H₀</em>: <em>μ</em> = 10 vs. <em>Hₐ</em>: <em>μ</em> < 10.
(b) The level of significance is 0.05.
Step-by-step explanation:
A new system is used to reduce the time customers spend waiting for teller service during peak hours at a bank.
A single mean test can be used to determine whether the waiting time has reduced.
(a)
The hypothesis to test whether the new system is effective or not is:
<em>H₀</em>: The mean waiting time is 10 minutes, i.e. <em>μ</em> = 10.
<em>Hₐ</em>: The mean waiting time is less than 10 minutes, i.e. <em>μ</em> < 10.
(b)
The information provided is:
![\bar x=9.5\\s=2.2\\n=70](https://tex.z-dn.net/?f=%5Cbar%20x%3D9.5%5C%5Cs%3D2.2%5C%5Cn%3D70)
Compute the test statistic value as follows:
![t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{9.5-10}{2.2/\sqrt{70}}=-1.902](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7Bs%2F%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B9.5-10%7D%7B2.2%2F%5Csqrt%7B70%7D%7D%3D-1.902)
The test statistic value is <em>t</em> = -1.902.
Compute the <em>p</em>-value of the test as follows:
![p-value=P(t_{n-1}](https://tex.z-dn.net/?f=p-value%3DP%28t_%7Bn-1%7D%3Ct%29)
![=P(t_{69}1.902)\\=0.031](https://tex.z-dn.net/?f=%3DP%28t_%7B69%7D%3C-1.902%29%5C%5C%3DP%28t_%7B69%7D%3E1.902%29%5C%5C%3D0.031)
The null hypothesis will be rejected if the <em>p</em>-value of the test is less than the significance level (<em>α</em>).
The <em>p</em>-value obtained is 0.031.
To reject the null hypothesis the value of <em>α</em> should be more than 0.031.
The most commonly used values of <em>α</em> are: 0.01, 0.05 and 0.10.
So, the least value of <em>α</em> at which we can conclude that the wait times have decreased is, <em>α</em> = 0.05.
Thus, the level of significance is 0.05.