A manufacturer of coffee vending machines has designed a new, less expensive machine. The current machine is known to dispense (into cups) an average of 7 fl. oz., with a standard deviation of .2 fl. oz. When the new machine is tested using 15 cups, the mean and the standard deviation of the fills are found to be 7 fl. oz. and .219 fl. oz. Test H0: σ = .2 versus Ha: σ ≠ .2 at levels of significance .05 and .01. Assume normality. (Round your answer to 4 decimal places.)

Respuesta :

Answer:

[tex] t=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

Step-by-step explanation:

Data given

[tex]\mu=7[/tex] population mean (variable of interest)

[tex]\sigma=0.2[/tex] represent the population standard deviation

[tex]s=0.219[/tex] represent the sample deviation

n=15 represent the sample size

[tex]\alpha=0.05,0.01[/tex] represent the values for the significance level  

Hypothesis test

On this case we want to check if the population standard deviation is equal or not to 0.2, so the system of hypothesis are:

H0: [tex]\sigma = 0.2[/tex]

H1: [tex]\sigma \neq 0.2[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

P value

We know the degrees of freedom of the distribution 14 on this case and we can find the p value like this:

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

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