Question

7. We want to know whether the difference in the population proportions between group 1 and group 2 is less than 0. Specifically, we want to test H0 : p1 − p2 ≥ 0 versus Ha : p1 − p2 < 0. We know that 468 of the 729 observations in group 1 have the characterisitic. And, we know that 463 of the 684 observations in group 2 have the characterisitic.

(a) What is the sample proportion for group 1? (round to 5 digits after the decimal place)

(b) What is the sample proportion for group 2? (round to 5 digits after the decimal place)

(c) What is the pooled estimator for p? (round to 5 digits after the decimal place)

(d) What is the standard error for the difference in the sample proportions? (Use ˜σp1−p2 and round to 5 digits after the decimal place.)

(e) What is the value of the test statistic? (Round to 2 digits after the decimal place.)

(f) What is the p-value of the test? (Round to 3 digits after the decimal place.)

(g) Do we reject or not reject the null hypothesis at the .05 level of significance? reject or not reject

Answer 1

a)

p1cap = X1/N1 = 468/729 = 0.64198

b)

p1cap = X2/N2 = 463/684 = 0.67690

c)

pcap = (X1 + X2)/(N1 + N2) = (468+463)/(729+684) = 0.65888

d)

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.64198 * (1-0.64198)/729 + 0.6769*(1-0.6769)/684)
SE = 0.02520

e)

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.64198-0.6769)/sqrt(0.65888*(1-0.65888)*(1/729 + 1/684))
z = -1.38

f)

P-value Approach
P-value = 0.084

g)

As P-value >= 0.05, fail to reject null hypothesis.