Question

In a large city A, 20% of a random sample of 900 school students had
defective eye sight In other large city B, 15% of a random sample
of 1600 children had the same defect Then the standard error of
the difference in proportion is
(a) 0.0156 (b) 0.145
)
(C) 0.126
(d) 0.0129
оа
ob
Ос
Od
Finish
Clear Response​

Answer 1

Answer:

don't know soory have a nice day

Answer 2

Answer:

The standard error of the difference in proportion calculated is $0.0156$

Therefore, option a) $0.0156$ is correct.

Step-by-step explanation:

Given data,

In a large city A;

The proportion of the random sample, p₁ = $20\%$ = $\frac{20}{100}$ = $0.2$

The number of school students who had defective eyesight, n₁ =$900$

In another large city B

The proportion of the random sample, p₂ = $15\%$ = $\frac{15}{100}$ = $0.15$

The number of school students who had defective eyesight, n₂ = $1600$

The standard error of the difference in proportion =?

As we know,

• The proportion of samples having the defective eyesight:
• P = $\frac{n_{1}p_{1} + n_{2}p_{2} }{n_{1} +n_{2} }$
• P = $\frac{900\times0.2+ 1600 \times 0.15}{900+1600}$
• P = $\frac{180 +240}{2500}$
• P = $\frac{420}{2500}$ = $0.168$

Now,

• The proportion of samples that do not have defective eyesight:
• Q = $1-P$
• Q = $1- 0.168$
• Q = $0.832$

Now, the standard error:

• e = $\sqrt{PQ (\frac{1}{n_{1} } + \frac{1}{n_{2} }) }$
• e = $\sqrt{0.168\times 0.832 (\frac{1}{900 } + \frac{1}{1600 })$
• e = $\sqrt{0.1397 (\frac{1600+900}{1440000 })$
• e = $\sqrt{0.1397 (\frac{2500}{1440000 })$
• e = $\sqrt{0.1397 (\frac{25}{14400 })$
• e = $0.01555$ ≈ $0.0156$

Hence, the standard error of the difference in proportion = $0.0156$.