Answer:
Following are the responses to the given question:
Step-by-step explanation:
Please find the table in the attached file.
mean and standard deviation difference: [tex]\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246[/tex]
For point a:
hypotheses are:
[tex]H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\[/tex]
degree of freedom:
[tex]df=n-1=8-1=7[/tex]
From t table, at[tex]\alpha = 0.05[/tex], reject null hypothesis if [tex]t <-1.895[/tex].
test statistic:
[tex]t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55[/tex]
because the [tex]t=-2.553 <-1.895[/tex], removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.
For point b:
From t table, at [tex]\alpha =0.01[/tex], removing the null hypothesis if [tex]t<-2.998[/tex].
because [tex]t=-2.553 >-2.908[/tex], fail to removing the null hypothesis.
The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.
For point c:
Hypotheses are:
[tex]H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5[/tex]
Degree of freedom:
[tex]df=n-1=8-1=7[/tex]
From t table, at [tex]\alpha =0.05[/tex], removing the null hypothesis if [tex]t <-1.895[/tex].
test statistic: [tex]t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986[/tex]
Since [tex]t-1.986 >-1.895[/tex], The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.
From t table, at[tex]\alpha= 0.01[/tex], reject null hypothesis if[tex]t<-2.998[/tex].
Since [tex]t=1.986>-2.998[/tex] , fail to reject null hypothesis.
Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.
