Answer: c. increasing sample size
d. decreasing confidence level
Step-by-step explanation:
Formula for margin of error :
[tex]E=z\dfrac{\sigma}{\sqrt{n}}[/tex] ,[tex]\sigma[/tex] = Population standard deviation.
[tex]E=t\dfrac{s}{\sqrt{n}}[/tex] , s= sample standard deviation.
[tex]E=z\sqrt{\dfrac{p(1-p)}{n}}[/tex], p= sample proportion.
Here z and t are critical values.
n= sample size.
Margin of error has critical t-value of z-value in the numerator .
⇒ Margin of error is proportional to the critical value.
⇒ Margin of error is proportional to the confidence level.
(The critical t-value of z-value is proportional the confidence level increases.)
So , Margin of error can be reduced by decreasing confidence level.
Also , Margin of error has square root of the sample size in the denominator.
⇒ Margin of error is inversely proportional to the sample size.
So , Margin of error can be reduced by increasing sample size.
Hence, the correct options area :
c. increasing sample size
d. decreasing confidence level
