Brian L. answered • 11/25/19

PharmD from St. John's University; Pharmacist in Westchester County

Let n_{1} = the number (sample size) of Europeans surveyed = 12083

Let n_{2} = the number (sample size) of Americans surveyed = 862

Let x_{1 }= the number of Europeans who thought genetic engineering was risky = 7628

Let x_{2} = the number of Americans who thought genetic engineering was risky = 448

Let p_{1} = the proportion of Europeans who thought genetic engineering was risky

p_{1} = (x_{1})÷(n_{1}) = 7628 ÷ 12083 = 0.631

Let p_{2} = the proportion of Americans who thought genetic engineering was risky

p_{2} = (x_{2})÷(n_{2}) = 448 ÷ 862 = 0.520

Even though this is considered a binomial random variable, the normal distribution may be used as a sampling distribution because n_{1}p_{1} and n_{2}p_{2} are both greater than 5.

The standard error (SE) is going to be [(p_{1}q_{1 }÷ n_{1})+((p_{2}q_{2 }÷ n_{2})]^{1/2} = √(p_{1}q_{1 }÷ n_{1})+((p_{2}q_{2 }÷ n_{2})

Note that q_{1} = (1-p_{1}) and q_{2} = (1-p_{2}). Stated another way, q_{1 }is the proportion of Europeans who did NOT find genetic engineering risky and q_{2} is the proportion of Americans who did NOT find genetic engineering risky.

The estimator for the difference in proportions of persons who found genetic engineering risky is going to be (p_{1} - p_{2}).

For a 97% confidence interval, the critical Z value is 2.17.

The 97% CI for difference in proportions is

(p_{1} - p_{2}) ± 2.17√(p_{1}q_{1 }÷ n_{1})+((p_{2}q_{2 }÷ n_{2})=(0.631-0.520)±√[ (0.631 x 0.369 ÷ 12083) + (0.520 x 0.480 ÷ 862)]

=0.111 ± 0.038

97% CI for difference in proportions (Europe - United States) is 0.073 to 0.149

97% CI for difference in PERCENTAGE is **7.3% to 14.9%**

Brian L.

11/26/19

Jacqueline A.

Thank you so much!11/26/19