Question

A study of cats and dogs found that 21 of 50 cats and 32 of 55 dogs slept more than 10 hours per day. At the .05 level of significance, does this sample data provide sufficient evidence to conclude that a difference exists between the proportion of cats and the proportion of dogs that sleep more than 10 hours per day? O No, there is not sufficient evidence to reject the hypothesis that the proportions are the samebecause the test value - 1.92 is inside the interval (-1.96, 1.96). No, there is not sufficient evidence to reject the hypothesis that the proportions are the same, because the test value - 1.09 is inside the interval (-1.96, 1.96). O No, there is not sufficient evidence to say that there is a difference in the proportions because the test value - 1.66 is inside the interval (-1.96, 1.96). O No, there is not sufficient information to conclude that the proportions are different because the test value -0.52 is inside the interval (-1.96, 1.96).

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

p1cap = X1/N1 = 21/50 = 0.42
p1cap = X2/N2 = 32/55 = 0.5818
pcap = (X1 + X2)/(N1 + N2) = (21+32)/(50+55) = 0.5048

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.42-0.5818)/sqrt(0.5048*(1-0.5048)*(1/50 + 1/55))
z = -1.66

No, there is not sufficient evidence to say that there is a difference in the proortions because the test value -.66 is inside the interval (-1.96 , 1.96)