Question

Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and find the proportion of presidents who were taller than their opponents. Use that result to construct a? 95% confidence interval estimate of the population percentage. Based on the? result, does it appear that greater height is an advantage for presidential? candidates? Why or why? not?

F. Roosevelt 188 182

Harding 183 178

Polk 173 185

Clinton 188 188

J. Adams 170 189

Truman 175 173

J. Q. Adams 171 191

Eisenhower 179 178

Harrison 168 180

G. H. W. Bush 188 173

Carter 177 183

T. Roosevelt 178 175

Hayes 173 178

Buchanan 183 175

Taylor 173 174

Taft 182 178

Harrison 173 168

Hoover 182 180

Coolidge 178 180

Jackson 185 171

1.) ___% < p < ___%

Answer 1

The proportion of presidents who were taller than their opponents is 0.6, and the 95% confidence interval estimate of the population percentage is (0.428, 0.772). Greater height appears to be an advantage for presidential candidates.

Step-by-step explanation:

The proportion of presidents who were taller than their opponents can be calculated by counting the number of presidents who were taller than their opponents and dividing it by the total number of presidents in the sample. In this case, there are 12 presidents who were taller than their opponents out of a total of 20 presidents, so the proportion is 12/20 = 0.6.

To construct a 95% confidence interval estimate of the population percentage, we can use the formula:

CI = p ± Z × √((p × (1-p))/n)

where CI is the confidence interval, p is the proportion, Z is the Z-score for the desired confidence level (1.96 for 95% confidence), and n is the sample size. Plugging in the values, we get:

CI = 0.6 ± 1.96 × √((0.6 × (1-0.6))/20)

Calculating the values, the 95% confidence interval estimate is 0.6 ± 0.172, which can be written as (0.428, 0.772). This means that we can be 95% confident that the population percentage of presidents who were taller than their opponents is between 42.8% and 77.2%.

Based on this result, it appears that greater height is an advantage for presidential candidates, as the proportion of presidents who were taller than their opponents is significantly higher than 50%.

Answer 2

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:

`p=X/n=11/20=0.55`

We can calculate now the standard error of the proportion as:

`\sigma_p=\sqrt{(p(1-p))/(n)}=\sqrt{(0.55*0.45)/(20)}=√(0.012375)=0.11`

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:

`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`

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.