Question

A recent study was done to see if the dissatisfaction rates at two local stores are different. Store A had a random sample of 650 people revealing that 6% were dissatisfied; Store B had a random sample of 475 people reveal that 5.2632% were dissatisfied. Does this show, at 99% confidence, that the dissatisfaction rates were different? Be sure to state the interval.

		I am 99% confident that the difference in dissatisfaction rates at the two stores is between -.0283 and .0430. We failed to show a difference.
		I am 99% confident that the difference in dissatisfaction rates at the two stores is between -.0283 and .0430. The evidence suggests the rates are different.
		I am 99% confident that the difference in dissatisfaction rates at the two stores is between -.0683 and .0115. The evidence suggests the rates are different.
		I am 99% confident that the difference in dissatisfaction rates at the two stores is between -.0683 and .0115. We failed to show a difference.

Answer 1

Here, , n1 = 650 , n2 = 475
p1cap = 0.06 , p2cap = 0.0526

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.06 * (1-0.06)/650 + 0.0526*(1-0.0526)/475)
SE = 0.0138

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.06 - 0.0526 - 2.58*0.0138, 0.06 - 0.0526 + 2.58*0.0138)
CI = (-0.0283 , 0.043)

  
I am 99% confident that the difference in dissatisfaction rates at the two stores is between -.0283 and .0430. We failed to show a difference.

Option A