The Census Bureau reports that households spend an average o

The Census Bureau reports that households spend an average of 31% of their total spending on housing. A homebuilders association in Cleveland wonders if the national finding applies in their area. They interview a sample of 40 households in the Cleveland metropolitan area to learn what percent of their spending goes toward housing. Take m to be the mean percent of spending devoted to housing among all Cleveland households. We want to test the hypotheses

                                                            H₀:       m = 31%

H_a:       m ¹ 31%

The population standard deviation is s = 9.6%.

(a)The survey finds = 28.6% for the 40 households in the sample. What is the value of the test statistic z? Sketch a standard Normal curve and mark z on the axis. Shade the area under the curve that represents the P-value.

(b)Calculate the P-value. Are you convinced that Cleveland differs from the national average?

(c)Find a 90% confidence interval estimate of the average housing expenditure.

Solution

Z-Test For Proportion
a)
Set Up Hypothesis
Null, Cleveland are same from the national average H0:P=0.31
Alternate, Cleveland differs from the national average H1: P!=0.31
Test Statistic
Number of objects in a sample provided(n)=40
No. Of Success Rate ( P )= x/n = 0.29
Success Probability ( Po )=0.31
Failure Probability ( Qo) = 0.69
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.286-0.31/(Sqrt(0.2139)/40)
Zo =-0.33
| Zo | =0.33
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.328 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
b)
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.3282 ) = 0.74276
Hence Value of P0.05 < 0.7428,Here We Do not Reject Ho
Conclude that Cleveland are same from the national average

c)
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)=11.44
Sample Size(n)=40
Sample proportion = x/n =0.286
Confidence Interval = [ 0.286 ±Z a/2 ( Sqrt ( 0.286*0.714) /40)]
= [ 0.286 - 1.64* Sqrt(0.0051) , 0.286 + 1.64* Sqrt(0.0051) ]
= [ 0.1688,0.4032]