2. A study by the National Athletic Trainers Association surveyed 1679 high scho

Question

2. A study by the National Athletic Trainers Association surveyed 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used to improve athletic performance. Give and interpret a 95% confidence interval for the proportion of all high school freshmen in Illinois who have used steroids.| a) b) Is there a significant difference between the proportions of freshmen and seniors who have used steroids using the 4 C's (context, conditions, calculations, and conclusion)? Context: Conditions: Calculation (use MiniTab): Conclusion:

Explanation / Answer

Q2.
PART A
TRADITIONAL METHOD
given that,
sample one, x1 =34, n1 =1679, p1= x1/n1=0.02
sample two, x2 =24, n2 =1366, p2= x2/n2=0.018
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.02*0.018/1679) +(0.018 * 0.982/1366))
=0.005
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.005
=0.01
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.02-0.018) ±0.01]
= [ -0.007 , 0.012]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =34, n1 =1679, p1= x1/n1=0.02
sample two, x2 =24, n2 =1366, p2= x2/n2=0.018
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.02-0.018) ± 1.96 * 0.005]
= [ -0.007 , 0.012 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.007 , 0.012] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2
PART B.
Given that,
sample one, x1 =34, n1 =1679, p1= x1/n1=0.02
sample two, x2 =24, n2 =1366, p2= x2/n2=0.018
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.02-0.018)/sqrt((0.019*0.981(1/1679+1/1366))
zo =0.538
| zo | =0.538
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =0.538 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.5382 ) = 0.5904
hence value of p0.05 < 0.5904,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.538
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.5904

we don't have diffrence in proportions of freshman and seniors

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