Food fight A $2016$survey of $1480$randomly selected U.S. adults found that $55\%$of respondents agreed with the following statement: “Organic produce is better for health than conventionally grown produce.”

a. Construct and interpret a $99\%$confidence interval for the proportion of all U.S. adults who think that organic produce is better for health than conventionally grown produce.

b. Does the interval from part (a) provide convincing evidence that a majority of all U.S. adults think that organic produce is better for health? Explain your answer.

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

$\hat{p}=0.55$

$n=1480$

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

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

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:

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

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

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.