On November 3 5 2010 the Gallup Organization surveyed 1028

On November 3 – 5, 2010, the Gallup Organization surveyed 1028 adult Americans and found that 463 said they supported a ban on smoking in public places. More recently, on October 15 – 17, 2014, the Gallup Organization surveyed 997 adult Americans and found that 550 supported a ban on smoking in public places.   

Construct a 93% confidence interval for the above data. Show your work using the formulas and verify your work using StatCrunch.

Continued below

Construct a 98% confidence interval for the above data. Show your work using the formulas and verify your work using StatCrunch.

Interpret the confidence interval calculated in part (b) as we learned in class.

At the a = 0.05, is there evidence to conclude that a difference in opinion exists over time? Conduct a full hypothesis test by following the steps below.

State the null and alternative hypotheses.

State the significance level for this problem.

Check the conditions that allow you to use the test statistic, and, if appropriate, calculate the test statistic.

Calculate the p-value and include the probability notation statement.

State whether you reject or do not reject the null hypothesis.

State your conclusion in context of the problem (i.e. interpret your results).

Use StatCrunch to verify your test statistic and p-value.

Solution

a. COMPUTE 93% confidence interval

CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=463
No.Of Observed (n1)=1028
P1= X1/n1=0.45
Proportion 2
No. of chances(X2)=550
No.Of Observed (n2)=997
P2= X2/n2=0.552
C.I = (0.45-0.552) ±Z a/2 * Sqrt( (0.45*0.55/1028) + (0.552*0.448/997) )
=(0.45-0.552) ± 1.81* Sqrt(0)
=-0.101-0.04,-0.101+0.04
=[-0.141,-0.061]

b.COMPUTE 98% confidence interval

C.I = (0.45-0.552) ±Z a/2 * Sqrt( (0.45*0.55/1028) + (0.552*0.448/997) )
=(0.45-0.552) ± 2.33* Sqrt(0)
=-0.101-0.052,-0.101+0.052
=[-0.153,-0.05]

c.
Null , No difference in opinion exists over time Ho: p1 = p2
Alternate , difference in opinion exists over time H1: p1 != p2
Test Statistic
Sample 1 : X1 =463, n1 =1028, P1= X1/n1=0.45
Sample 2 : X2 =550, n2 =997, P2= X2/n2=0.552
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.5
Q^ Value For Proportion= 1-P^=0.5
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.45-0.552)/Sqrt((0.5*0.5(1/1028+1/997))
Zo =-4.556
| Zo | =4.556
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =4.556 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != -4.5564 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

We conclude that difference in opinion exists over time