Respuesta :
Answer:
Correct option: As the sample size decreases, the margin of error increases.
Step-by-step explanation:
The (1 - α) % confidence interval for population proportion is:
[tex]CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
The margin of error in this confidence interval is:
[tex]\\ MOE=z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
The sample size n is inversely related to the margin of error.
An inverse relationship implies that when one increases the other decreases and vice versa.
In case of MOE also, when n is increased the MOE decreases and when n is decreased the MOE increases.
Compute the new margin of error for n = 125 as follows:
[tex]\\ MOE=z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }=1.96\times \sqrt{\frac{0.40(1-0.40)}{125} }=0.086[/tex]
*Use z-table for the critical value.
For n = 125 the MOE is 0.086.
And for n = 275 the MOE was 0.058.
Thus, as the sample size decreases, the margin of error increases.
