Question

Given p⎯⎯1p¯1 = 0.87, n1n1 = 494, p⎯⎯2p¯2 = 0.96, n2n2 = 393. (You may find it useful to reference the appropriate table: z table or t table)


a. Construct the 95% confidence interval for the difference between the population proportions. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

b. Is there a difference between the population proportions at the 5% significance level?

• Yes, since the confidence interval includes the value 0.

• No, since the confidence interval includes the value 0.

• Yes, since the confidence interval does not include the value 0.

• No, since the confidence interval does not include the value 0.

Answer 1

Solution:

Part a

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

We are given

P1 = 0.87

P2 = 0.96

N1 = 494

N2 = 393

Confidence level = 95%

Critical Z value = 1.96

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (0.87 – 0.96) ± 1.96*sqrt[(0.87*(1 – 0.87)/494) + (0.96*(1 – 0.96)/393)]

Confidence interval = -0.09 ± 1.96* 0.0181

Confidence interval = -0.09 ± 0.0354

Lower limit = -0.09 - 0.0354 = -0.1254

Upper limit = -0.09 + 0.0354 = -0.0546

Confidence interval = (-0.13, -0.05)

Part b

Is there a difference between the population proportions at the 5% significance level?

Yes, since the confidence interval does not include the value 0.