 Globokim8

2021-03-08

The homebuilder’s association reports that $75\mathrm{%}$ of home buyers would like a fireplace in their new home. A local builder finds that 20 out of 30 of his customers wanted a fireplace. Using the p value decision rule determine at the 5% level of significance is there is enough evidence from the sample to conclude that the homebuilders report is too high?
If P is too high in ${H}_{0}$ then what should it be in ${H}_{1}$?
Also construct a $\left(1-\alpha \right)\mathrm{%}$ interval estimate for the true proportion P.
Make sure to state the meaning of P first…and completely interpret the CI for P. joshyoung05M

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 $20\text{/}30=66.67\mathrm{%}$
Therefore, the hypothesis are given as:
${H}_{0}:P=0.75$
${H}_{1}:P<0.75$
To test the above claim, standard normal distribution (z-test) will be used. The formula for the same is given by:
$z=\frac{p-P}{\sqrt{\frac{PQ}{n}}}\approx N\left(0,1\right)$
$z=\frac{0.6667-0.75}{\sqrt{\frac{\left(0.75\right)\left(0.25\right)}{30}}}$
$=\frac{-0.0833}{0.0791}$
$=-1.054$
The critical z - value at $\alpha =0.05$ (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 $\alpha$ level of significance and is accepted otherwise.
$0.1469>0.05-\text{Accept}-{H}_{0}$
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 $0.75\left(75\mathrm{%}\right)$.
Step 2
The formula for $\left(1-\alpha \right)\mathrm{%}$ confidence interval for population proportion (P) is given by:
$C.I=p±{z}_{\alpha \text{/}2}.SE\left(p\right)$
$p=$ sample proportion
level of significance
$SE\left(p\right)=s\mathrm{tan}$ dard error for sample proportion
$p=0.6667$
${z}_{\alpha \text{/}2}=±1.96$
$SE\left(p\right)=\sqrt{\frac{p\left(1-p\right)}{n}}$
$=\sqrt{\frac{\left(0.6667\right)\left(1-0.6667\right)}{30}}$
$=0.086$
$C.I=\left[0.6667±\left(1.96\right)\left(0.086\right)\right]$
$=\left[0.4981,0.8353\right]$
This implies that we are $95\mathrm{%}$ confident that the true population proportion is contained by this interval.

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