Question

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 slader

Answer 1

Answer:

$t=\frac{2.8-3}{\frac{1.2}{\sqrt{64}}}=-1.33$      

The degrees of freedom are given by:

$df=n-1=64-1=63$  

The p value for this case would be given by:

$p_v =P(t_{63}$  

If we use a significance level lower than 9% we have enough evidence to FAIL to reject the null hypothesis that the true mean is greater or equal than 3 but if we use a significance level higher than 9% the conclusion is oppossite we reject the null hypothesis

Step-by-step explanation:

Information given  

$\bar X=2.8$ represent the sample mean

$s=1.2$ represent the standard deviation

$n=64$ sample size      

$\mu_o =3$ represent the value to verify

$\alpha$ represent the significance level

t would represent the statistic (variable of interest)      

$p_v$ represent the p value

Hypothesis to verify

We want to check if the true mean for this case is less than 3 minutes, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 3$      

Alternative hypothesis:$\mu < 3$      

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)      

Replacing we got:

$t=\frac{2.8-3}{\frac{1.2}{\sqrt{64}}}=-1.33$      

The degrees of freedom are given by:

$df=n-1=64-1=63$  

The p value for this case would be given by:

$p_v =P(t_{63}$  

If we use a significance level lower than 9% we have enough evidence to FAIL to reject the null hypothesis that the true mean is greater or equal than 3 but if we use a significance level higher than 9% the conclusion is oppossite we reject the null hypothesis