Question

In a sample of 167 children selected randomly from one town, it is found that 37 of them suffer from asthma. At the 0.05 significance level, test the claim that the proportion of all children in the town who suffer from asthma is 11%.

Answer 1

Q1 : Ho : p = 0.11 vs Ha : p $\neq$ 0.11

Level of Significance(l.o.s.) :  $\alpha$ = 0.05

Decision Criteria : Reject Ho at 5% l.o.s. if | Z cal | > Z tab,
where Z tab = Z ($\alpha$/2) = Z (0.025) = 1.96

Calculation :  $\hat{p}$ = 37/167 = 0.2216, n = 167

Z cal =  
P-P p(1-p) n
=  
0.2216 -0.11 10.11*[1-0.11) 167
= 4.6093

Conclusion : Since Z cal > Z tab, we reject Ho at 5% l.o.s. and thus, conclude that the proportion of all children in the town who suffer from asthama is significantly different than 0.11.

Q2 : Ho : p = 0.34 vs Ha : p > 0.34

Level of Significance(l.o.s.) :  $\alpha$ = 0.05

Decision Criteria : Reject Ho at 5% l.o.s. if | Z cal | > Z tab,
where Z tab = Z ($\alpha$) = Z (0.05) = 1.6449

Calculation :  $\hat{p}$ = 96/234 = 0.4103, n = 234

Z cal =  
P-P p(1-p) n
=  $\frac{0.4103-0.34}{\sqrt{\frac{{0.34}*(1-{0.34})}{234}}}$ = 2.2701

Conclusion : Since Z cal > Z tab, we reject Ho at 5% l.o.s. and thus, conclude that the figure is higher for fathers in the town of Littleton.

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