Daniellelovee
  • Daniellelovee
http://k12.kitaboo.com/k12/ebookpdf/maths05/17501_HS_PS_chapter06.pdf
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Daniellelovee
  • Daniellelovee
pg 235 question 14 @Directrix @ganeshie8
Daniellelovee
  • Daniellelovee
please help me
Daniellelovee
  • Daniellelovee
@ganeshie8

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More answers

anonymous
  • anonymous
Please post your actual questions here, one at a time.
ganeshie8
  • ganeshie8
@CShrix
Daniellelovee
  • Daniellelovee
is that there is a graph and like I said it is in page 235
Daniellelovee
  • Daniellelovee
a table not a graph sorry @Cardinal_Carlo
anonymous
  • anonymous
So it's Question 14 pp. 235? Construct a 99% confidence interval for the difference between the percent of all tennis players under the age of 18 and all tennis players in the 18–34 age group who play tennis regularly. Use a complete sentence to describe the confidence interval.
Daniellelovee
  • Daniellelovee
yes
Daniellelovee
  • Daniellelovee
@Luigi0210
Daniellelovee
  • Daniellelovee
@ganeshie8 would it be the formula inpage 232?
ganeshie8
  • ganeshie8
what is the zcritical for 99% confidence level ?
Daniellelovee
  • Daniellelovee
2.58
ganeshie8
  • ganeshie8
yes, lets use the formula in page 232
Daniellelovee
  • Daniellelovee
ok
Daniellelovee
  • Daniellelovee
thank you so much for helping me
ganeshie8
  • ganeshie8
|dw:1450254597763:dw|
ganeshie8
  • ganeshie8
find the percentage of tennis players in both groups : under 18 18-34
Daniellelovee
  • Daniellelovee
83+102=180
ganeshie8
  • ganeshie8
No
ganeshie8
  • ganeshie8
how many ppl in under18 group play tennis ?
Daniellelovee
  • Daniellelovee
28?
Daniellelovee
  • Daniellelovee
would it be 102/28=x/100?
ganeshie8
  • ganeshie8
No
ganeshie8
  • ganeshie8
did you take time to read the question ?
ganeshie8
  • ganeshie8
Look at the table. The first row represents Under18 group
Daniellelovee
  • Daniellelovee
oh sorry I got ahead of myself is 19
ganeshie8
  • ganeshie8
|dw:1450254846019:dw|
ganeshie8
  • ganeshie8
Under18 group : 19 play tennis out of 83 surveyed
Daniellelovee
  • Daniellelovee
therefore 0.2289
Daniellelovee
  • Daniellelovee
19/83=x/100
ganeshie8
  • ganeshie8
Yes
ganeshie8
  • ganeshie8
so \(p_1 = 0.2289\)
ganeshie8
  • ganeshie8
similarly find \(p_2\) from 18-34 group
Daniellelovee
  • Daniellelovee
is actualy 0.27 sorry
Daniellelovee
  • Daniellelovee
0.086
Daniellelovee
  • Daniellelovee
9/102=x/100
ganeshie8
  • ganeshie8
Ok. We need perccentages actually
Daniellelovee
  • Daniellelovee
27% and 8%
ganeshie8
  • ganeshie8
\(p_1 = 19/83*100 = 22.9\) \(p_2 = 9/102*100 =8.8\)
Daniellelovee
  • Daniellelovee
ohh ok
ganeshie8
  • ganeshie8
plug them in the margin of error formula in step2
Daniellelovee
  • Daniellelovee
is n1=83? and n2=102?
ganeshie8
  • ganeshie8
|dw:1450255197123:dw|
ganeshie8
  • ganeshie8
Yes
Daniellelovee
  • Daniellelovee
13.9
Daniellelovee
  • Daniellelovee
and now substract the p?
ganeshie8
  • ganeshie8
\[2.58\cdot\sqrt{\dfrac{22.9(100-22.9)}{83}+\dfrac{8.8(100-8.8)}{102}}\]
ganeshie8
  • ganeshie8
that is the margin of error right ?
Daniellelovee
  • Daniellelovee
13.9
ganeshie8
  • ganeshie8
simplify
Daniellelovee
  • Daniellelovee
yes
Daniellelovee
  • Daniellelovee
alright I got 13.9
ganeshie8
  • ganeshie8
Oh youhave done the simplification already! nice
ganeshie8
  • ganeshie8
margin of error = 13.9 difference between sample proportions = 22.9-8.8 = 14.1
ganeshie8
  • ganeshie8
construct the confidence interval
ganeshie8
  • ganeshie8
do we get (14.1 - 13.9, 14.1 + 13.9) ?
Daniellelovee
  • Daniellelovee
alright thank you very much for your help
ganeshie8
  • ganeshie8
np :)
anonymous
  • anonymous
Hehe, thanks for the tag @ganeshie8 X)

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