Question

The weights of gallon milk jugs off a filling machine have been normally distributed with a mean \mu _{0}= 380.05 and a standard deviation \sigma = 1.25 grams, respectively. Before deciding if the filling machine needs to be adjusted, a random sample of n= 16 jugs is selected and the sample mean is \overline{x} =381.30 grams. Conduct a hypothesis test at a= 0.05 to test if this mean has changed.

The null and alternative hypotheses should be

a. H_{0}: \mu =380.05
H_{a}: \mu \neq 380.05
b. H_{0}: \mu \leq 380.05
H_{a}: \mu > 380.05
c. H_{0}: \mu \geq 380.05
H_{a}: \mu < 380.05
d. H_{0}: \overline{x} = 380.05
H_{a}: \overline{x} \neq 380.05
e. none of the above

Answer 1

Answer:

a. H_{0}: \mu =380.05

H_{a}: \mu \neq 380.05

$z=\frac{381.30-380.05}{\frac{1.25}{\sqrt{16}}}=4$    

Since is a two sided test the p value would be:  

$p_v =2*P(z>4)=0.0000633$  

the p value is very low and lower than the significance level given so we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 380.05gr and the machine needs a reparation

Step-by-step explanation:

Data given and notation  

$\bar X=381.30$ represent the sample mean

$\sigma=1.25$ represent the population standard deviation

$n=16$ sample size  

$\mu_o =380.05$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

Z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if te true mean is different from 380.05 (needs a reparation), the system of hypothesis would be:  

Null hypothesis:$\mu = 380.05$  

Alternative hypothesis:$\mu \neq 380.05$  

And the best altrnative is:

a. H_{0}: \mu =380.05

H_{a}: \mu \neq 380.05

The statistic for this case would be:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{381.30-380.05}{\frac{1.25}{\sqrt{16}}}=4$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z>4)=0.0000633$  

the p value is very low and lower than the significance level given so we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 380.05gr and the machine needs a reparation