Question

1. Are cats finicky eaters? A random sample of 100 house cats found 18 refused to eat dry food.

a. Estimate the true proportion of cats that refuse to eat dry food at the 90% confidence level.

b. Explain the meaning of “90% confident.”

c. What is the margin of error associated with your estimate? How do we interpret this value?

d. The crazy cat lady who lives across the hall from your apartment tells you, “You know? More than 25% of cats refuse to eat dry food.” What would you say to the woman?

Answer 1

a)

sample proportion, = 18/100 = 0.18

b)

sample size, n = 100
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.18 * (1 - 0.18)/100) = 0.0384

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.18 - 1.64 * 0.0384 , 0.18 + 1.64 * 0.0384)
CI = (0.117 , 0.243)

c)

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0384
ME = 0.063

The margin of error of cats that refuse to eat dry food at the 90% confidence level. is 0.063

d)

reject the null jypothesis becaus econfiden ceinterval does not contain 25%