91影视

We are given that $c=99\%=0.99$

$\hat{p}=0.55$

$n=1480$

For confidence interval $1-伪=0.99$, from probability table, we get

role="math" localid="1654447187223" $z_{伪/2}=2.575$

The conditions are:

• Random: As US adults are randomly selected, this condition is satisfied.
• Independence: Sample for $1480$US adults is less than $10\%$of all population.
• Normal: There are $1480(0.55)=814$successes an$1480(1-0.55)=666$failures which is greater than ten.
• All conditions are fulfilled.

Margin of Error:

$E=z_{伪/2}脳\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=2.575脳\sqrt{\frac{0.55(1-0.55)}{1480}}=0.0333$

Confidence Interval:

role="math" localid="1654447528277" $\hat{p}-E=0.55-0.0333=0.5167$

$\hat{p}+E=0.55+0.0333=0.5833$

Hence, confidence interval is$\left(0.5167,0.5833\right)$

The confidence interval $(0.5167,0.5833)$do not have $0.5$. All values are above $0.5$which shows that more than $50\%$of adults are of the view that organic produce is better.