E 400 trials in which the assistant maintained a neutral expression, the driver stopped in 229 out of the 400 trials. Find a 95% confidence interval for the proportion of drivers who would stop when a neutral expression is maintained. In the 400 trials in which the assistant smiled at the driver, the driver stopped in 277 out of the 400 trials. Find a 95% confidence interval for the proportion of drivers who would stop when the assistant is smiling. What do your results in parts (a) and (b) suggest about the effect of a smile on a driver stopping at a pedestrian crosswalk? Explain briefly. (In Chapter 23, we will consider formal methods for comparing two proportions.)

Step-by-step explanation:

n = 400

Proportion p = 229/400= 0.5725

For 95% confidence interval we use Z value as 1.96

Std error = 0.025

margin of error = 1.96*0.025

Confidence interval 95% = 0.5725±Margin of error

= (0.524, 0.621)

b) When smiled x becomes 277

p = 0.6925

Std error = 0.023

Margin of error= 1.96*0.023

Confidence interval = (0.647, 0.738)

Smiling increases the chances of stopping since mean and conidence interval bounds are showing increasing trend.

the answer to the expression is 83

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)