In a sample of 28 houses in a city 7 contained lead paint a

In a sample of 28 houses in a city, 7 contained lead paint.

a. Create a 90% confidence interval estimate for the proportion of houses in that city containing lead paint. Use formulas & chart

b. With .05 significance, test the claim that 30% of houses in that city contain lead paint. Use the P-value method with formulas.

c. Explain why we cannot use the techniques from chapter 8 to test the claim that fewer than 15% of the houses in that city contain lead paint.

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)=7
Sample Size(n)=28
Sample proportion = x/n =0.25
Confidence Interval = [ 0.25 ±Z a/2 ( Sqrt ( 0.25*0.75) /28)]
= [ 0.25 - 1.64* Sqrt(0.007) , 0.25 + 1.64* Sqrt(0.007) ]
= [ 0.116,0.384]

b)
Set Up Hypothesis
Null, 30% of houses in that city contain lead paint H0:P=0.3
Alternate, H1: P!=0.3
Test Statistic
No. Of Success chances Observed (x)=7
Number of objects in a sample provided(n)=28
No. Of Success Rate ( P )= x/n = 0.25
Success Probability ( Po )=0.3
Failure Probability ( Qo) = 0.7
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.25-0.3/(Sqrt(0.21)/28)
Zo =-0.5774
| Zo | =0.5774
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.577 & | 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 != -0.57735 ) = 0.5637
Hence Value of P0.05 < 0.5637,Here We Do not Reject Ho