Question

g A machine is designed to fill containers with 16 ounces of coffee. A consumer suspects that the machine is not filling the containers completely and the amount of coffee is less than 16 ounces when filled. A sample of 8 containers has a mean of 15.6 ounces of coffee and a standard deviation of 0.3 ounces of coffee. Assume that the distribution of data is normal. At 10% significance level, is there enough evidence to support the consumer’s suspicion that the machine is not filling the containers completely and the amount of coffee in a container is less than 16 ounces when filled?

Answer 1

Answer:

$t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.771$    

The degrees of freedom are given by:

$df=n-1=8-1=7$  

And the p value is given by this probability:

$p_v =P(t_{(7)}$  

Since the p value is lower than the significance level provided of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 16 ounces

Step-by-step explanation:

Information provided

$\bar X=15.6$ represent the sample mean

$s=0.3$ represent the sample standard deviation

$n=8$ sample size  

$\mu_o =16$ represent the value to verify

$\alpha=0.1$ represent the significance level

t would represent the statistic  

$p_v$ represent the p value

Hypothesis to test

We want to verify if amount of coffee in a container is less than 16 ounces when filled, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 16$  

Alternative hypothesis:$\mu < 16$  

The statistic for this case would be given by:

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

Replacing the info given we got:

$t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.771$    

The degrees of freedom are given by:

$df=n-1=8-1=7$  

And the p value is given by this probability:

$p_v =P(t_{(7)}$  

Since the p value is lower than the significance level provided of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 16 ounces