Question

In: Statistics and Probability

4. Suppose that 195 of the 738 observations in group 1 have a characterisitic of interest....

4. Suppose that 195 of the 738 observations in group 1 have a characterisitic of interest. And, suppose that 261 of the 815 observations in group 2 have this 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 difference in the sample proportions between groups 1 and 2? (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 estimate of the left end of a 95% confidence interval? (Round to 5 digits after the decimal place.)

(f) What is the estimate of the right end of a 95% confidence interval? (Round to 5 digits after the decimal place.)

(g) Do we reject or not reject the null hypothesis that the population proportions are the same at the .05 level of significance? Reject Not reject

Expert Solution

a)

p1cap = X1/N1 = 195/738 = 0.26423

b)

p1cap = X2/N2 = 261/815 = 0.32025

c)

p1cap - p2cap = 0.26423 - 0.32025 = -0.05602

d)

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.26423 * (1-0.26423)/738 + 0.32025*(1-0.32025)/815)
SE = 0.02303

e)

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.26423 - 0.32025 - 1.96*0.02303, 0.26423 - 0.32025 + 1.96*0.02303)
CI = (-0.10116 , -0.01088)

left end = -0.10116

f)

Right end = -0.01088

g)

Reject because confidenc einterval does not contains 0

orchestra answered 2 years ago

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