Find a 95 confidence interval for theta and explain how to i

Find a 95% confidence interval for theta and explain how to interpret this interval. Based on the data, carry out a test of H-0 : theta = 0.1 vs H_1 : theta = 0.1. Report an approximate p-value and explain your conclusion.

Solution

a.
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
Mean(x)=80
Sample Size(n)=831
Sample proportion = x/n =0.096
Confidence Interval = [ 0.096 ±Z a/2 ( Sqrt ( 0.096*0.904) /831)]
= [ 0.096 - 1.96* Sqrt(0) , 0.096 + 1.96* Sqrt(0) ]
= [ 0.076,0.116]

b.
Set Up Hypothesis
Null, H0:P=0.1
Alternate, H1: P!=0.1
Test Statistic
No. Of Success chances Observed (x)=80
Number of objects in a sample provided(n)=831
No. Of Success Rate ( P )= x/n = 0.0963
Success Probability ( Po )=0.1
Failure Probability ( Qo) = 0.9
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.09627-0.1/(Sqrt(0.09)/831)
Zo =-0.3585
| Zo | =0.3585
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.35846 ) = 0.72
Hence Value of P0.05 < 0.72,Here We Do not Reject Ho