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Marina CMI [18]
3 years ago
14

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
Mathematics
1 answer:
STatiana [176]3 years ago
4 0

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

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