Suppose you own a restaurant and have a cook whose ability and attitude you are suspicious of. One of the dishes on the menu is duck cassoulet, which uses duck legs that have been slow fried over a couple of hours in oil that does not exceed a temperature of 175 degrees. This is a time consuming and monotonous process, but one that results in excellent meat that you sell for a large mark-up. You suspect your cook is lazy and doesn't properly monitor and maintain the oil temperature. You take a random sample of 15 duck legs and take them to a forensics lab where you are able to discover the maximum temperature the meat has reached. Within your sample the mean maximum temperature of the duck legs is 180 degrees with a standard deviation of 4 degrees. Meat cooked precisely to 175 degrees is what your cook is supposed to do.
A) Which of the following is true about a hypothesis test for the claim that your employee is capable (meaning he doesn't over-fry the meat) at the 90% confidence level?
Group of answer choices:
O Reject the null with a test statistic value of 1.83
O Reject the null with a test statistic value of 2.17
O Fail to reject the null with a test statistic value of 1.59
O Fail to the null with a test statistic value of 1.47
O None of the above are true

Respuesta :

Answer:

[tex]t=\frac{180-175}{\frac{4}{\sqrt{15}}}=4.84[/tex]    

[tex]df=n-1=15-1=14[/tex]  

[tex]p_v =P(t_{(14)}>4.84)=0.000131[/tex]  

We got a very low value for the p value so then we have enough evidence to reject the null hypothesis at any significance level commonly used. And the best conclusion based on the possible options are:

None of the above are true

Step-by-step explanation:

Data given

[tex]\bar X=180[/tex] represent the sample mean

[tex]s=4[/tex] represent the sample standard deviation

[tex]n=15[/tex] sample size  

[tex]\mu_o =175[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is above the limit of 175 degrees, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 175[/tex]  

Alternative hypothesis:[tex]\mu > 175[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

And replacing we got:

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

[tex]t=\frac{180-175}{\frac{4}{\sqrt{15}}}=4.84[/tex]    

P-value

The degrees of freedom are given by:

[tex]df=n-1=15-1=14[/tex]  

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

[tex]p_v =P(t_{(14)}>4.84)=0.000131[/tex]  

We got a very low value for the p value so then we have enough evidence to reject the null hypothesis at any significance level commonly used. And the best conclusion based on the possible options are:

None of the above are true