# Imagine that you are conducting a poll to determine the percentage of adults who gamble at least once

Imagine that you are conducting a poll to determine the percentage of adults who gamble at least once a month. as your sample size increases (let us say from 100 to 400 cases), which of the following becomes true?

a. confidence interval becomes wider
b. margin of error becomes smaller
c. amount of sampling error increases
d. margin of error increases

## This Post Has 3 Comments

1. anonymous9723 says:

Option B) Margin of error becomes smaller

Step-by-step explanation:

We are given the following information in the question:

A poll to determine the percentage of adults who gamble at least once a month.

We have to find what happens to the confidence interval as the sample size increases.

Margin of error =

$\text{Critical Value}\times \displaystyle\frac{\sigma}{\sqrt{n}}$

Confidence interval:

$\mu \pm \text{Margin of error}$

As the sample size increases, the margin of error decreases.

As the margin of error decreases the width of the confidence interval decreases.

Option B) Margin of error becomes smaller

2. rosy2019 says:

B) The margin of error becomes smaller

Step-by-step explanation:

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 width of the confidence interval is given by:

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

So as n increases, the width, or margin of error, becomes smaller.

As your sample size increases (let us say from 100 to 400 cases), which of the following becomes true?

B) The margin of error becomes smaller

3. jahmanilittlejotz1pe says:

Option C

Step-by-step explanation:

The given function is: $h(t)=-16t^2+80t+108$.

$h(5)=-16(5)^2+80(5)+108=108$.

$h(3)=-16(3)^2+80(3)+108=204$.

The average rate of change of this function on the interval t=3 to t=5 is given by:

$\frac{h(5)-h(3)}{5-3}$

$\frac{204-108}{2}=-48$

Since the average rate of change is negative on the interval 3-5 seconds, it means the ball is moving downwards at an average rate of 48 feet per second.

The correct choice is C.