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:
![\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](https://tex.z-dn.net/?f=%5Cbar%7Bd%7D%3D%5Cfrac%7B%5CSigma%20d%7D%7Bn%7D%20%3D%5Cfrac%7B-4-6-.......-4-4%7D%7B8%7D%3D-4.125%20%5C%5C%5C%5CS_d%3D%5Csqrt%7B%5Cfrac%7B%5CSigma%20%28d-%5Cbar%7Bd%7D%29%5E2%20%7D%7Bn-1%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%28-4%20%2B%204.125%29%5E2%20%2B.......%2B%28-4%20%2B4.125%29%5E2%20%7D%7B8-1%7D%7D%3D%201.246)
For point a:
hypotheses are:
degree of freedom:
From t table, at
, reject null hypothesis if
.
test statistic:
because the
, 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
, removing the null hypothesis if
.
because
, 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:
Degree of freedom:
From t table, at
, removing the null hypothesis if
.
test statistic: ![t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%7Bd%7D-%5Cmu_d%7D%7B%5Cfrac%7Bs_d%7D%7B%5Csqrt%7Bn%7D%7D%7D%20%3D%5Cfrac%7B-4.125-%28-5%29%7D%7B%5Cfrac%7B1.246%7D%7B%5Csqrt%7B8%7D%7D%7D%3D1.986)
Since
, 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
, reject null hypothesis if
.
Since
, 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.