Question

A sample of 100 janitors in a certain school district found that the average time each janitor had worked for the school district was 8 years. Suppose we know that the standard deviation of the population is s=5 years. Construct a 95% confidence interval for the mean time of the population of janitors have spent with the school district.

8±1.98

8±0.82

8±0.71

8±1.53

8±0.98

Answer 1

@whpalmer4

Answer 2

@poopsiedoodle

Answer 3

@nikato

Answer 4

@beccaboo333

Answer 5

Sorry, I don't know :/

Answer 6

?

Answer 7

I don't really know, but you can see if this helps

http://m.wikihow.com/Calculate-Confidence-Interval

Answer 8

@agent0smith

Answer 9

A.?

Answer 10

help guys!

Answer 11

This might help:
http://www.wolframalpha.com/input/?i=confidence+interval+mean&a=FSelect_**TInterval-.dflt-&f2=0.95&f=TInterval.c%5Cu005f0.95&f3=100&f=TInterval.n_100&f4=5&f=TInterval.s_5&f5=8&f=TInterval.xbar_8

Answer 12

Thank you @ybarrap

Answer 13

you're welcome

Answer 14

D.?

Answer 15

okay last one

Answer 16

The same sample of 25 seniors from the urban school district with the mean and standard deviation N(450, 100). A 95% confidence interval for µ for the population of seniors with a margin of error of ± 25 is used. What is the smallest sample size we can take to achieve this same margin of error?

Answer 17

@agent0smith your turn! :)

Answer 18

@ranga

Answer 19

@phi

Answer 20

I think this might be more accurate since the samples are so high:

http://www.wolframalpha.com/input/?i=confidence+interval+mean&a=FSelect_**ZInterval-.dflt-&f2=0.95&f=ZInterval.c%5Cu005f0.95&f3=100&f=ZInterval.n%5Cu005f100&f4=5&f=ZInterval.sigma_5&f5=8&f=ZInterval.xbar%5Cu005f8

This uses the z-score rather than the student's T distribution with are based on small samples.

Answer 21

This means
$$
\bar{x}=8\pm0.98
$$

Answer 22

well here are my choices and i say it would be 62?

26

45

50

62

69

Answer 23

(I was still on the first, lol)

Answer 24

OHh Sorry! yes I knew about that! isaw the link

Answer 25

thats why i said thanks before lol

Answer 26

I know from the first question that my confidence interval would be 8 plinus 0.98

Answer 27

but i need help on this other quesiton

The same sample of 25 seniors from the urban school district with the mean and standard deviation N(450, 100). A 95% confidence interval for µ for the population of seniors with a margin of error of ± 25 is used. What is the smallest sample size we can take to achieve this same margin of error?

26

45

50

62

69

Answer 28

Margin of error is inversely related to the square root of the sample size.

Answer 29

@ranga i already no that, you're no help

Answer 30

*know

Answer 31

(Please be polite when someone is trying to guide you through these problems)

Confidence interval is,
$$
\large
\mu\pm\cfrac{z_{(1-c)\sigma}}{\sqrt n}
$$


We want margin of error to be \(\pm25\), so
$$
\large{
25=\cfrac{z_{(1-c)/2}\sigma}{\sqrt n}\\
\sqrt n=\cfrac{z_{(1-c)/2}\sigma}{25}\\
n=\left (\cfrac{z_{(1-c)/2}\sigma}{25}\right )\\
}
$$
Where \(c=0.95\) and \(\sigma=100\).

This is the smallest sample size that will satisfy the requirement.

We now just need \(\large z_{(1-c)/2}=z_{0.025}\)

Do you know how to find this?

Answer 32

@ybarrap ?

Answer 33

hmm yes hold on

Answer 34

50

Answer 35

@ybarrap

Answer 36

We need \(\large z_{0.025}\) to plug in the equation above, I'd have to look it up, I thought you were doing that.

Answer 37

i did......

Answer 38

what was it?

Answer 39

my calculator does it for me so i dont know, i have a n-spire im just saying overall it is c

Answer 40

because i plugged in those numbers into that equation you gave me

Answer 41

ok, let me look it up

Answer 42

alright

Answer 43

I need to go, I'll check back later if you still need validation

Answer 44

@agent0smith :(

Answer 45

I've been on this question for the past hour please help me

Answer 46

i was saying that it was 50?