Question

One hundred volunteers were divided into two​ equal-sized groups. Each volunteer took a math test that involved transforming strings of eight digits into a new string that fit a set of given​ rules, as well as a​ third, hidden rule. Prior to taking the​ test, one group received 8 hours of​ sleep, while the other group stayed awake all night. The scientists monitored the volunteers to determine whether and when they figured out the rule. Of the volunteers who​ slept, 37 discovered the​ rule; of the volunteers who stayed​ awake 15 discovered the rule. What can you infer about the proportions of volunteers in the two groups who discover the​rule? Support your answer with a 90​% confidence interval.

Let p1 wbe the proportion of volunteers who figured out the third rule in the group that slept and let p2 be the proportion of volunteers who figured out the third rule in the group that stayed awake all night.

The 90% confidence interval for (p1 - p2) is (___, ___).

Is there sufficient evidence?

Answer 1

Here, , n1 = 50 , n2 = 50
p1cap = 0.74 , p2cap = 0.3

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.74 * (1-0.74)/50 + 0.3*(1-0.3)/50)
SE = 0.0897

For 0.9 CI, z-value = 1.64
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.74 - 0.3 - 1.64*0.0897, 0.74 - 0.3 + 1.64*0.0897)
CI = (0.2929 , 0.5871)

There is sufficient evidence that the volunteers received sleep before the test performed better