Respuesta :
Answer:
1.) 33% < p < 66%
As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.
If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.
Step-by-step explanation:
Out of this sample, we have 11 presidents, out of 20, that were taller than their oponent.
Then, the proportion of presindents that were taller than their oponent can be calculated as:
[tex]p=X/n=11/20=0.55[/tex]
We can calculate now the standard error of the proportion as:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.55*0.45}{20}}=\sqrt{0.012375}=0.11[/tex]
For a 95% confidence interval, the z-value is z=1.96 (we can loook up this value in the standarized normal distribution table).
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z\cdot \sigma_p=0.55-1.96*0.11=0.55-0.22=0.33\\\\UL=p+z\cdot \sigma_p=0.55+1.96*0.11=0.55+0.22=0.66[/tex]
As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.
If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.
Answer:
[tex]0.33\leq p \leq0.77[/tex]
Step-by-step explanation:
Given information:
Confidence,
[tex]c[/tex]=95%=0.95
Significance level,
[tex]\alpha=1-c=1-0.95=0.05[/tex]
[tex]\frac{\alpha}{2}=\frac{0.05}{2}=0.025[/tex]
[tex]z_{\frac{\alpha}{2}}=1.96[/tex]
Total number of president [tex]n[/tex]=20
Number of president who were taller than their opponent [tex]X=11[/tex]
[tex]\widehat{p}=\frac{X}{n}=\frac{11}{20}=0.55[/tex]
Then lower and upper bounds are,
[tex]\widehat{p}-z_{\frac{\alpha}{2}\times} \sqrt\frac{\widehat{p}(1-\widehat{p})}{n}\leq p \leq\widehat{p}+z_{\frac{\alpha}{2}}\times \sqrt\frac{\widehat{p}(1-\widehat{p})}{n}[/tex]
On substitution,
[tex]{0.55}-1.96\times} \sqrt\frac{{0.55}(1-{0.55})}{20}\leq p\leq{0.55}+1.96\times} \sqrt\frac{{0.55}(1-{0.55})}{20}[/tex]
[tex]0.55-0.22\leq p \leq0.55+0.22[/tex]
[tex]0.33\leq p \leq0.77[/tex]
If the greater height ia an advantage then taller candidates should have won more than 50%.in this case if greater height does not appear an advantage because the confidence level includes 50%.
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https://brainly.com/question/2782108?referrer=searchResults
