Question

What is the mean number of hours that high school students spend on homework per week? It is known that the standard deviation of the population is 2.2 hours. In a random sample of 40 students, it is found that they average 13 hours per week on homework.

a) What is the standard deviation for this sample?
b) What is the critical value for a 90% confidence interval?
c) What is the critical value for a 95% confidence interval?

Answer 1

@jim_thompson5910

Answer 2

a) use the formula

s = sigma/sqrt(n)

in this case, sigma = 2.2 and n = 40

Answer 3

) 2.2/sqrt(40)
= 0.3478 <--- for a

Answer 4

good

Answer 5

that's the std dev for the sample of 40

Answer 6

one sec while i get a calc ready

Answer 7

ok so thats a

Answer 8

yes

Answer 9

ok

Answer 10

ok go here and tell me what you see after the |z| in the last row

http://www.wolframalpha.com/input/?i=normal+cdf&a=*MC.normal+cdf-_*Formula.dflt-&f2=0.90&x=12&y=13&f=NormalProbabilities.pr_0.90&a=*FVarOpt.1-_***NormalProbabilities.pr--.***NormalProbabilities.z--.**NormalProbabilities.l-.*NormalProbabilities.r---.*--&a=*FVarOpt.2-_**-.***NormalProbabilities.mu--.**NormalProbabilities.sigma---.**NormalProbabilities.pr---

Answer 11

i seee .9?

Answer 12

before that, to the left a bit

Answer 13

P(|z| < ....)

Answer 14

what goes in the .... place

Answer 15

it's on the line where you see "confidence level"

Answer 16

1.645?

Answer 17

good, that's your critical value for the 90% CI

CI = confidence interval

Answer 18

now change that 0.90 to 0.95 and recompute

Answer 19

ok so b is 1.645?

Answer 20

yep

Answer 21

tell me what you get for c

Answer 22

umm

Answer 23

1.96

Answer 24

?

Answer 25

you got it

Answer 26

ook cool! d?

Answer 27

i don't see part d

Answer 28

here:

Answer 29

d) What is the 90% confidence interval? Be sure to include the formula with your numbers plugged in. Use the model: x plus or minus z times sigma divided by the square root of n
e) What is the 95% confidence interval?

Answer 30

ok for part d, we'll break it into two parts

Answer 31

ok

Answer 32

first part

lower limit

L = xbar - C*s

L = 13 - 1.645*0.3478 ... note we found the value for s back in part a)

L = 12.427869

Answer 33

second part

upper limit

U = xbar + C*s

U = 13 + 1.645*0.3478

U = 13.572131

Answer 34

So the CI (L, U) turns into (12.427869, 13.572131)

ie (L, U) = (12.427869, 13.572131)

Answer 35

therefore, the 90% CI is (12.427869, 13.572131)

and you can round this however you need to or leave it like it is

Answer 36

for part e), you're doing the exact same thing as part d...but...you're using 1.96 for C instead of 1.645 for C

basically, C = 1.96 in part e) while everything else stays the same

Answer 37

so D is (12.427869, 13.572131)

Answer 38

yeah pretty much

Answer 39

ok so E?

Answer 40

@jim_thompson5910

Answer 41

do the same thing in part d), but use C = 1.96 this time

Answer 42

whats C?

Answer 43

can u write it ouut? @jim_thompson5910

Answer 44

just go back to the work for D and instead of using C = 1.645, use C = 1.96

Answer 45

L = 13 minus 1.96 * 0.3478
= 12.427869

for the upper limit: U = 13 plus 1.96 * 0.3478
= 13.57213

Answer 46

well not the equal part, but the formula?

Answer 47

good, you just need to fix the results and you're done

Answer 48

13- .681688= 12.318312

Answer 49

13.681688

Answer 50

good on both

Answer 51

so your 95% CI is (12.318312, 13.681688)

Answer 52

is it
ANSWER = (12.318312, 13.681688)

Answer 53

yes

Answer 54

@jim_thompson5910 can u help with a fe wmore