A recent survey on drug use and health measured the prevalen

A recent survey on drug use and health measured the prevalence of drug use in the United States. This survey is designed to provide estimates on the use of illicit drugs, alcohol, and tobacco among members of United States households aged 12 and older. In addition, numerous demographic variables like gender,age, race, etc. are recorded. Assume that this is a legitimate statistical survey and that valid inferences for the US population may be drawn from its results.

In the 2007 survey, 57 out of 820 respondents aged 12 to 15 years old indicated that they had tried marijuana. Investigators would like to estimate the true proportion for the population of 12 to 15 year olds in the US that have tried marijuana. To do this, they wish to construct a 99% confidence interval.

A) The point estimate for the true proportion of 12 to 15 year olds that have tried marijuana is:
(Round the answer to three decimal places.)




B) Calculate the 99% margin of error:  
(Round the answer to three decimal places.)


C) Find the 99% confidence interval for the proportion of 12 to 15 year olds that have tried marijuana.
(Round the answer to three decimal places.)

______ to _______

In 1995, a similar survey was taken that indicated that 7% of the adolescents in this age group had tried marijuana. Assuming 7% was the true value in 1995, you want to perform a test to determine whether the proportion of kids in this age group that have tried marijuana has changed since 1995. Let = 0.01.

D) The null/alternative setting for this problem is:

a. Ho: 0.07 vs Ha: = 0.07
b. Ho: 0.07 vs Ha: = 0.07     
c. Ho: 7 vs Ha: = 7
d. Ho: = 0.07 vs Ha: 0.07
e. Ho: = 0.07 vs Ha: 0.07
f. Ho: = 7 vs Ha: 7
g. Ho: 7 vs Ha: = 7
h. Ho: 7 vs Ha: = 7

E) The test statistic is:  
(Round the answer to two decimal places.)


F) The p-value is:  
(Round the answer to four decimal places.)

G) The proper conclusion and corresponding interpretation is:

a. Reject Ho conclude that the true proportion of 12 to 15 year olds that have tried marijuana has changed since 1995
b. FTR Ho: we cannot conclude that the true proportion of 12 to 15 year olds that have tried marijuana has changed since 1995     
c. FTR Ha: conclude that the true proportion of 12 to 15 year olds that have tried marijuana has changed since 1995
d. Reject Ha we cannot conclude that the true proportion of 12 to 15 year olds that have tried marijuana has changed since 1995

Solution

[ANSWER]
A.
respondents aged 12 to 15 years(x)=57
Sample Size(n)=820
Sample proportion = x/n =0.07

B.
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=57
Sample Size(n)=820
Sample proportion =0.07
Margin of Error = Z a/2 * ( Sqrt ( (0.07*0.93) /820) )
= 2.576* Sqrt(0)
=0.023

C.
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.07 ±Z a/2 ( Sqrt ( 0.07*0.93) /820)]
= [ 0.07 - 2.576* Sqrt(0) , 0.07 + 2.58* Sqrt(0) ]
= [ 0.047,0.093]

Set Up Hypothesis
Null, H0:P=0.07
Alternate, H1: P!=0.07
Test Statistic
No. Of Success chances Observed (x)=57
Number of objects in a sample provided(n)=820
No. Of Success Rate ( P )= x/n = 0.0695
Success Probability ( Po )=0.07
Failure Probability ( Qo) = 0.93
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.06951-0.07/(Sqrt(0.0651)/820)
Zo =-0.0547
| Zo | =0.0547
Critical Value
The Value of |Z | at LOS 0.01% is 2.58
We got |Zo| =0.055 & | Z | =2.58
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.05475 ) = 0.95634
Hence Value of P0.01 < 0.9563,Here We Do not Reject Ho

D.
d. Ho: = 0.07 vs Ha: 0.07

E.
Zo =-0.0547 ~ -0.05

F.( P != -0.05475 ) = 0.95634

G. b. FTR Ho: we cannot conclude that the true proportion of 12 to 15 year olds that have tried marijuana has changed since 1995