umthumaL3e

2022-12-03

A local newspaper in a large city wants to assess support for the construction of a highway bypass around the central business district to reduce downtown traffic. They survey a random sample of 1152 residents and find that 543 of them support the bypass. Construct and interpret a 95% confidence interval to estimate the proportion of residents who support construction of the bypass.

kriteria0b1

Beginner2022-12-04Added 10 answers

given that,

possible chances (x)=543

size(n)=1152

success rate ( p )= x/n = 0.4714

point of estimate = proportion = 0.4714

standard error $=\sqrt{\frac{(0.4714\cdot 0.5286)}{1152}}$

= 0.0147

margin of error $=Z\frac{a}{2}\cdot \text{(standard error)}$

where,

$Z\frac{a}{2}=Z$-table value

level of significance, $\alpha =0.05$

from standard normal table, two tailed $z\frac{\alpha}{2}=1.96$

$\text{margin of error}=1.96\cdot 0.0147$

= 0.0288

$CI=[p\pm \text{margin of error}]$

$\text{confidence interval}=[0.4714\pm 0.0288]$

$=[0.4425,0.5002]$

possible chances (x)=543

size(n)=1152

success rate ( p )= x/n = 0.4714

point of estimate = proportion = 0.4714

standard error $=\sqrt{\frac{(0.4714\cdot 0.5286)}{1152}}$

= 0.0147

margin of error $=Z\frac{a}{2}\cdot \text{(standard error)}$

where,

$Z\frac{a}{2}=Z$-table value

level of significance, $\alpha =0.05$

from standard normal table, two tailed $z\frac{\alpha}{2}=1.96$

$\text{margin of error}=1.96\cdot 0.0147$

= 0.0288

$CI=[p\pm \text{margin of error}]$

$\text{confidence interval}=[0.4714\pm 0.0288]$

$=[0.4425,0.5002]$

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