Question

For a recent report on sleep deprivation, the Centers for Disease Control and Prevention interviewed 11508 residents of California and 4860 residents of Oregon. In California, 932 respondents reported getting insufficient rest or sleep during each of the preceding 30 days, while 452 of the respondents from Oregon reported the same.

Calculate a 95 % confidence interval for the difference between the proportions of Californians and Oregonians, pCalifornia−pOregonpCalifornia−pOregon, who report getting insufficient rest. Round your answer to 4 decimal places.

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Answer 1

95% confidence interval for the difference between the proportions of Californians and Oregonians is (-0.0070, -0.0130).

Explanation:

Confidence interval for a proportion is given as p +/- margin of error (E)

Californians

p = 932/11508 = 0.08

n = 11508

C = 95% = 0.95

Significance level = 1-C = 1-0.95 = 0.05 = 5%

critical value corresponding to infinity degrees of freedom and 5% significance level is 1.96

E = critical value × sqrt[p(1-p) ÷ n] = 1.96 × sqrt[0.08(1-0.08) ÷ 11508] = 1.96 × 0.00253 = 0.0050

Lower limit = p - E = 0.08 - 0.0050 = 0.0750

Upper limit = p + E = 0.08 + 0.0050 = 0.0850

Oregonians

p = 452/4860 = 0.09

n = 4860

critical value = 1.96

E = 1.96 × sqrt[0.09(1-0.09) ÷ 4860] = 1.96 × 0.0041 = 0.0080

Lower limit = p - E = 0.09 - 0.0080 = 0.0820

Upper limit = p + E = 0.09 + 0.0080 = 0.0980

Difference in lower limit of proportion = 0.0750 - 0.0820 = -0.0070

Difference in upper limit of proportion = 0.0850 - 0.0980 = -0.0130

95% confidence interval for the difference in proportion is between a lower limit of -0.0070 and an upper limit of -0.0130.