Bradford T. answered 12/30/20
Retired Engineer / Upper level math instructor
Part A:
Let like = 0 and don't like = 1
xbar = (0(207) + 1(93))/300 = 0.31
xbar(population) = 0.31(1000) = 310 people are not satisfied with their quality of care
Part B: To find the margin of error for 95% confidence, need to find the sample standard deviation, S.
σx-bar = σ/√samplesize = σ/√300 ≈ S/17.3
S2 = (207(0-.31)2 + 93(1-0.31)2)/299 = 0.28 --> S = 0.53
S/17.3 = 0.53/17.3 = 0.03
2 Margins of error = ±0.06