Answer:
We conclude that the mean waiting time is more than or equal to 5 minutes at 5% level of significance.
Step-by-step explanation:
We are given that waiting time is defined as the time the customer enters the line until he or she reaches the teller window.
A random sample of 15 customers is selected. Results found that the sample mean waiting time was 4.287 minutes with a sample standard deviation of 1.638 minutes.
Let
= <u><em>mean waiting time.</em></u>
SO, Null Hypothesis,
:
5 minutes {means that the mean waiting time is more than or equal to 5 minutes}
Alternate Hypothesis,
:
< 5 minutes {means that the mean waiting time is less than 5 minutes}
The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;
T.S. =
~ 
where,
= sample mean waiting time = 4.287 minutes
s = sample standard deviation = 1.638 minutes
n = sample of customers = 15
So, <u><em>the test statistics</em></u> =
~ 
= -1.686
The value of t test statistics is -1.686.
Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.</u>
Since our test statistic is more than the critical value of t as -1.686 > -1.761, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that the mean waiting time is more than or equal to 5 minutes.