1. A random sample of 64 customers at a drive-through bank window is observed, and it is found that the teller spends an average of 2.8 minutes with each customer, with a standard deviation of 1.2 minutes. Is there sufficient evidence to conclude that the teller spends less than 3 minutes with each customer

The 95% confidence interval is 2.5 < u <3.1.

Step-by-step explanation:

The provided sample mean is X = 2.8 and the sample standard deviation is s = 1.2, and the sample size is n = 64.

1. Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

H0 u = 3

Ha: u < 3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2. Rejection Region Based

on the information provided, the significance level is alpha = 0.05, and the critical value for a left-tailed test is t c = - 1.669.

The rejection region for this left-tailed test is R = t : t < - 1.669

3. Test Statistics

The t-statistic is computed as follows:

t = (X - uo) / [s/n^ (1/2) ] =

replacing

t = (2.8 - 3) / [1.2/64 ^ (1/2) ]

t = -1.333

4. Decision about the null hypothesis

Since it is observed that t = - 1.333 > t c = - 1.669, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0936, and since p = 0.0936 = > 0.05, it is concluded that the null hypothesis is not rejected.

5. Conclusion It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 3, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 2.5 < u <3.1.