The homebuilder’s association reports that 75% of home buyers would like a fireplace in their...

Step 1
According to the homebuilder's report, the population proportion is $75\mathrm{\%}$, but on the basis of the sample selected, the sample proportion (an unbiased estimate of population proportion) is ${\displaystyle 20\text{/}30=66.67\mathrm{\%}}$
Therefore, the hypothesis are given as:
${\displaystyle H_0:P=0.75}$
${\displaystyle H_1:P<0.75}$
To test the above claim, standard normal distribution (z-test) will be used. The formula for the same is given by:
${\displaystyle z=\frac{p-P}{\sqrt{\frac{PQ}{n}}}\approx N(0,1)}$
${\displaystyle z=\frac{0.6667-0.75}{\sqrt{\frac{\left(0.75\right)\left(0.25\right)}{30}}}}$
${\displaystyle =\frac{-0.0833}{0.0791}}$
${\displaystyle =-1.054}$
The critical z - value at ${\displaystyle \alpha =0.05}$ (left - tailed) is -1.645
The p - value which is the probability value for rejecting null hypothesis is 0.1469
According to the p-value approach, null hypothesis is rejected when p-value is less than the ${\displaystyle \alpha }$ level of significance and is accepted otherwise.
${\displaystyle 0.1469>0.05-\text{Accept}-H_0}$
Therefore, there are insufficient evidence to reject null hypothesis, concluding that the homebuilders report is not too high and that the population proportion for home buyers who would like fireplace in their house is $0.75(75\mathrm{\%})$.
Step 2
The formula for ${\displaystyle (1-\alpha )\mathrm{\%}}$ confidence interval for population proportion (P) is given by:
${\displaystyle C.I=p\pm z_{\alpha \text{/}2}.SE\left(p\right)}$
${\displaystyle p=}$ sample proportion
${\displaystyle z_{\alpha \text{/}2}=\text{critical}z-\text{value at}\alpha =0.025\left(0.05\text{/}2\right)}$ level of significance
${\displaystyle SE\left(p\right)=s\tan }$ dard error for sample proportion
${\displaystyle p=0.6667}$
${\displaystyle z_{\alpha \text{/}2}=\pm 1.96}$
${\displaystyle SE\left(p\right)=\sqrt{\frac{p(1-p)}{n}}}$
${\displaystyle =\sqrt{\frac{\left(0.6667\right)(1-0.6667)}{30}}}$
${\displaystyle =0.086}$
${\displaystyle C.I=[0.6667\pm \left(1.96\right)\left(0.086\right)]}$
${\displaystyle =[0.4981,0.8353]}$
This implies that we are $95\mathrm{\%}$ confident that the true population proportion is contained by this interval.