Question

A machine in the student lounge dispenses coffee. The average cup of coffee is supposed to contain 7.0 ounces. A random sample of seven cups of coffee from this machine show the average content to be 7.4 ounces with a standard deviation of 0.70 ounce. Do you think that the machine has slipped out of adjustment and that the average amount of coffee per cup is different from 7 ounces

Answer 1

Answer:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$    

The degrees of freedom are

$df=n-1=7-1=6$  

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$  

The p value is a higher value and using a significance levels of 5% or 10% we see that the p value is higher than the significance level and then we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is different from 7 ounces.

Step-by-step explanation:

Information provided

$\bar X=7.4$ represent the sample mean

$s=0.7$ represent the sample standard deviation

$n=7$ sample size  

$\mu_o 7$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the average amount of coffee per cup is different from 7 ounces, the system of hypothesis would be:  

Null hypothesis:$\mu = 7$  

Alternative hypothesis:$\mu \neq 7$  

The statistic is given by:

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

Replacing the info given we got:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$    

The degrees of freedom are

$df=n-1=7-1=6$  

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$  

The p value is a higher value and using a significance levels of 5% or 10% we see that the p value is higher than the significance level and then we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is different from 7 ounces.