Question

The following confidence interval for the population proportion for how many U.S. adults do not get enough fruits and vegetables in their diet is constructed at what level of confidence, if there were 2050 people in the survey? (0.878,0.903)

Answer 1

Answer:

The interval is constructed at 93% confidence.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between these two bounds, divided by 2.

Confidence interval of proportions concepts:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this problem, we have that:

2050 people, so n = 2050.

Lower bound: 0.878

Upper bound: 0.903

$\pi = \frac{0.878 + 0.903}{2} = 0.8905$

$M = \frac{0.903 - 0.878}{2} = 0.0125$

Confidence level:

We have to find z.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.0125 = z\sqrt{\frac{0.8905*0.1095}{2050}}$

$0.0069z = 0.0125$

$z = \frac{0.0125}{0.0069}$

$z = 1.81$

$z = 1.81$ has a pvalue of 0.965.

That is:

$]1 - \frac{\alpha}{2} = 0.965$

$\frac{\alpha}{2} = 0.035$

$\alpha = 2*0.035$

$\alpha = 0.07$

Finally

$1 - \alpha = 1 - 0.07 = 0.93$

The interval is constructed at 93% confidence.