I am interested in the population proportion intending to pu

I am interested in the population proportion intending to purchase an iWatch. I claim that the proportion, p, of the population that is intending to purchase it is less than 10%. From a sample of 1000 people, 90 intend to purchase it

a. What are your null and alternative hypotheses? Calculate the p-value and state your conclusion

b. Find the 90% confidence interval for p. Does the interval contain 10%? What does “90% confidence” mean?

Solution

a.
Set Up Hypothesis
Null, H0:P>0.1
Alternate, H1: P<0.1
Test Statistic
No. Of Success chances Observed (x)=90
Number of objects in a sample provided(n)=1000
No. Of Success Rate ( P )= x/n = 0.09
Success Probability ( Po )=0.1
Failure Probability ( Qo) = 0.9
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.09-0.1/(Sqrt(0.09)/1000)
Zo =-1.0541
| Zo | =1.0541
P-Value: Left Tail -Ha : ( P < -1.05409 ) = 0.14592
Hence Value of P0.05 < 0.14592,Here We Do not Reject Ho
the population that is intending to purchase is n\'t less than 10%

b)
Confidence Interval For Proportion
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)=90
Sample Size(n)=1000
Sample proportion = x/n =0.09
Confidence Interval = [ 0.09 ±Z a/2 ( Sqrt ( 0.09*0.91) /1000)]
= [ 0.09 - 1.645* Sqrt(0) , 0.09 + 1.65* Sqrt(0) ]
= [ 0.075,0.105]