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

`t=(2.8-3)/((1.2)/(√(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)<-1.33)=0.0942`

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

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=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing we got:

`t=(2.8-3)/((1.2)/(√(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)<-1.33)=0.0942`

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