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.

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

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if the probability is 1/12, 75 people will receive a $25 gift card

step-by-step explanation: