Answer:
No. He would need $1,080 to perform a test with this error.
Step-by-step explanation:
To answer this question, we have to calculate the sample size. This is the sample size that allow to estimate the average amount of sugar per can with 95% confidence (95% CI).
The difference between the upper and lower limit of the CI have to be equal or less to e=2 mg. The z value for a 95% CI is z=1.96.
[tex]e\leq z\sigma/\sqrt{n}[/tex]
[tex]e=z\sigma/\sqrt{n}\\\\\sqrt{n}=z\sigma/e=\\\\n=(z\sigma/e)^2=(1.96*15/2)^2=14.7^2=216\\\\n=216[/tex]
The minimum sample size needed for this error is 216. At a cost of $5/test, this sample size would cost [tex]n*p=216*5=\$ 1,080[/tex].
This is over the budget for this experiment ($1000).
