A sample of size =n72 is drawn from a population whose standard deviation is =σ25. Part 1 of 2 (a) Find the margin of error for a 95% confidence interval for μ. Round the answer to at least three decimal places. The margin of error for a 95% confidence interval for μ is . Part 2 of 2 (b) If the sample size were =n89, would the margin of error be larger or smaller?

a) margin of error ME = 5.77

b) Margin of error becomes smaller

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

x+/-ME

Where margin of error ME = zr/√n

a)

Given that;

Mean = x

Standard deviation r = 25

Number of samples n = 72

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

ME = 1.96(25/√72)

ME = 1.96(2.946278254943)

ME = 5.774705379690

ME = 5.77

b)

For n = 89

ME = 1.96(25/√89)

ME = 1.96(2.649994700015)

ME = 5.193989612031

ME = 5.19

5.19 is smaller than 5.77 in a) above. So,

Margin of error becomes smaller

This is the answer

0.02

0.1

0.12

0.21