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?

(a) standard error of the mean increases, width of confidence interval becomes wider.
(b) standard error of the mean increases, width of confidence intervale becomes narrower.
(c) standard error of the mean decreases, width of confidence interval becomes wider.
(d) standard error of the mean decreases, width of confidence interval becomes narrower.
(e) cannot be determined.

(D) Standard error of the mean decreases, width of confidence interval becomes narrower.

Step-by-step explanation:

Given that 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.

When all others remain constant n increases from 25 to 30

Because of this std error becomes

instead of

As a result standard error decreases and in turn margin of error also decreases.

Hence correct option would be

(D) Standard error of the mean decreases, width of confidence interval becomes narrower.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}