Since 2007, a psychological association has supported an annual nationwide survey to examine stress across the United States. One year, a total of 760 millennials (18- to 33-year-olds) were asked to indicate their average stress level (on a 10-point scale) during a month. The mean score was 5.5. Assume that the population standard deviation is 2.8. Give the margin of error for a 95% confidence interval. (Enter your answer to three decimal places.)

The margin of error for a 95% confidence interval is 0.199.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

Assume that the population standard deviation is 2.8.

This means that

760 millennials (18- to 33-year-olds)

This means that

Give the margin of error for a 95% confidence interval.

The margin of error for a 95% confidence interval is 0.199.

2

step-by-step explanation:

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}