An analyst takes a random sample of 25 firms in the telecommunications industry and constructs a confidence interval for the mean return for the prior year. Holding all else constant, if he increased the sample size to 30 firms, how are the standard error of the mean and the width of the confidence interval affected

The standard error decreases and the width of the confidence interval also decreases.

Step-by-step explanation:

The standard error of a distribution (E) is given as:

Where n is the sample size, is the z score of he confidence and is the standard deviation.

The sample size is inversely proportional to the standard error. If the sample size is increased and everything else is constant, the standard would decrease since they are inversely proportional to each other. The confidence interval = μ ± E = (μ - E, μ + E). μ is the mean

The width of the confidence interval = μ + E - (μ - E) = μ + E - μ + E = 2E

The width of the confidence interval is 2E, therefore as the sample size increase, the margin of error decrease and since the width of the confidence interval is directly proportional to the margin of error, the width of the confidence interval also decreases.

the system of inequalities is

step-by-step explanation:

step 1

find the equation of the dashed line

we have the points

(0,0) and (3,1)

the line passes through the origin, so, is a direct variation

find the slope

the equation of the dashed line is

the solution of the inequality a s the shaded area above the dashed line

therefore

the equation of the inequality a is

step 2

find the equation of the solid line

we have the points

(4,0) and (0,4)

find the slope

the equation of the solid line is

the solution of the inequality b s the shaded area below the solid line

therefore

the equation of the inequality b is

step 3

the system of inequalities is

no

step-by-step explanation: