Answer:
Option C is correct.
The value of the z statistic for testing equality of the proportion of smokers in 1995 and 2010 = 2.05
Explanation:
Complete Question
In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study of national smoking trends, two random samples of U.S. adults were selected in different years: The first sample, taken in 1995, involved 4276 adults, of which 1642 were smokers. The second sample, taken in 2010, involved 3908 adults, of which 1415 were smokers. The samples are to be compared to determine whether the proportion of U.S. adults that smoke declined during the 15-year period between the samples.
Let p₁ be the proportion of all U.S. adults that smoked in 1995. Let p₂ denote the proportion of all U.S. adults that smoked in 2010.
The value of the z statistic for testing equality of the proportion of smokers in 1995 and 2010 is
A. z = 1.39.
B. z = 1.66.
C. z = 2.05.
D. z = 4.23.
Solution
The test statistic for two samples test is given as
z = (p₁ - p₂)/σ
σ = √{[p₁(1-p₁)/n₁] + [p₂(1-p₂)/n₂]}
p₁ = Proportion of smokers in 1995 = (1642/4276) = 0.384
n₁ = 4276
p₂ = Proportion of smokers in 2010 = (1415/3908) = 0.362
n₂ = 3908
σ = √{[0.384(1-0.384)/4276] + [0.362(1-0.362)/3908]}
σ = √[(0.384×0.616/4276) + (0.362×0.638/3908)] = √(0.0000553189 + 0.0000590983) = √0.0001144172 = 0.0106965976 = 0.010697
z = (p₁ - p₂)/σ
z = (0.384 - 0.362)/0.010697
z = 2.05
Hope this Helps!!!