Question

The homebuilder’s association reports that 75% of home buyers would like a fireplace in their new home. A local builder finds that 20 out of 30 of his

Confidence intervals
ANSWERED
asked 2021-03-08
The homebuilder’s association reports that \(75\%\) 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 \(\displaystyle{H}_{{0}}\) then what should it be in \(\displaystyle{H}_{{1}}\)?
Also construct a \(\displaystyle{\left({1}-\alpha\right)}\%\) interval estimate for the true proportion P.
Make sure to state the meaning of P first…and completely interpret the CI for P.

Answers (1)

2021-03-09

Step 1
According to the homebuilder's report, the population proportion is \(75\%\), but on the basis of the sample selected, the sample proportion (an unbiased estimate of population proportion) is \(\displaystyle{20}\text{/}{30}={66.67}\%\)
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:
\(\displaystyle{z}=\frac{{{p}-{P}}}{\sqrt{{\frac{{{P}{Q}}}{{n}}}}}\approx{N}{\left({0},{1}\right)}\)
\(\displaystyle{z}=\frac{{{0.6667}-{0.75}}}{\sqrt{{\frac{{{\left({0.75}\right)}{\left({0.25}\right)}}}{{30}}}}}\)
\(\displaystyle=\frac{{-{0.0833}}}{{0.0791}}\)
\(\displaystyle=-{1.054}\)
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 \(0.75 (75\%)\).
Step 2
The formula for \(\displaystyle{\left({1}-\alpha\right)}\%\) confidence interval for population proportion (P) is given by:
\(\displaystyle{C}.{I}={p}\pm{z}_{{\alpha\text{/}{2}}}.{S}{E}{\left({p}\right)}\)
\(\displaystyle{p}=\) 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
\(\displaystyle{S}{E}{\left({p}\right)}={s} \tan{}\) dard error for sample proportion
\(\displaystyle{p}={0.6667}\)
\(\displaystyle{z}_{{\alpha\text{/}{2}}}=\pm{1.96}\)
\(\displaystyle{S}{E}{\left({p}\right)}=\sqrt{{\frac{{{p}{\left({1}-{p}\right)}}}{{n}}}}\)
\(\displaystyle=\sqrt{{\frac{{{\left({0.6667}\right)}{\left({1}-{0.6667}\right)}}}{{30}}}}\)
\(\displaystyle={0.086}\)
\(\displaystyle{C}.{I}={\left[{0.6667}\pm{\left({1.96}\right)}{\left({0.086}\right)}\right]}\)
\(\displaystyle={\left[{0.4981},{0.8353}\right]}\)
This implies that we are \(95\%\) confident that the true population proportion is contained by this interval.

0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-05-14
When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
asked 2021-08-04
A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a \(\displaystyle{99}\%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with \(\displaystyle{99}\%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
asked 2021-02-21
Marty asked some of his classmates to rate their level ofstress out of 10, with 10 being very high. He also asked them tomeasure th enumber of minutes it took them to get from home toschool. A random selection of his results is listed below.

A.) Explain what a positive value for thecoefficient of correlation indicates.
asked 2021-08-01
Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a \(\displaystyle{95}\%\) two-sided confidence interval on the death rate from lung cancer.
a) Construct a \(\displaystyle{95}\%\) two-sides confidence interval on the death rate from lung cancer. Round your answer 3 decimal places.
\(\displaystyle?\leq{p}\leq?\)
b) Using the point estimate of p obtained from the preliminary sample what sample size is needed to be \(\displaystyle{95}\%\) confident that the error in estimatimating the true value of p is less than 0.00?
c) How large must the sample if we wish to be at least \(\displaystyle{95}\%\) confident that the error in estimating p is less than 0.03 regardless of the value of p?
...