Answer :
Here is what we know:
Sample size: 80 people
Mean consumption by a person from the sample: 3.2 pounds
Sample standard deviation: 1.22 pounds
Tuna consumption according to the nutritionist claims: 3.5 pounds
Alpha: 0.07
First, we compute the standard deviation of the mean:
[tex]\sigma_{\mu}=\frac{\sigma}{\sqrt{n}}=\frac{1.22}{\sqrt{80}}\approx0.136\text{ pounds}[/tex]Then, the z score for the nutritionist claims is given by:
[tex]Z=\frac{x-\mu}{\sigma_{\mu}}=\frac{3.5-3.2}{0.136}\approx2.2[/tex]Then, according to a normal table, we have:
[tex]P(Z\ge2.2)+P(Z\leq2.2)=2\cdot(0.5-0.4861)=0.0278[/tex]We can check that 0.0278 < 0.07, therefore the null hypothesis, is rejected.
The null hypothesis is given by:
[tex]\begin{gathered} P(z\leq Z_{null})+P(z\ge Z_{null})\ge1-0.07=0.93 \\ According\text{ to the normal table: }Z_{null}=\pm1.81 \\ =\frac{x_{null}-\mu}{\sigma_{\mu}} \\ \pm1.81=\frac{x_{null}-3.2}{0.136} \\ x_{null}=3.2\pm1.81\cdot0.136\approx3.2\pm0.25 \\ null\text{ hypothesis: 2.95}\leq x\text{ }\leq\text{ 3.45} \\ Alternative\text{ hypothesis: }x\leq2.95\text{ or }x\ge3.45 \end{gathered}[/tex]Null hypothesis: 2.95 <= x <= 3.45
Alternative hypothesis: x <= 2.95 or x >= 3.45
