Question

A Gallup poll was conducted in 2010 in which they surveyed people all over the world and found that 44% of adults in the world believe that global warming is a serious threat to themselves and their families. Assume that this is the population proportion of adults worldwide who believe global warming is a threat (p= 0.44), and suppose we repeatedly select random samples of size 250, and ask them the same question.

(a) Find the mean, and standard error, of the sampling distribution of sample proportions for the given n and p.

(b) Use the Empirical Rule to find the interval centered at p that contains approximately 99.7% of all sample proportions.

(c) Suppose that in a random sample of 250 people, 100 say that global warming is a threat. Find the sample proportion � and the z-score for this sample proportion.  

(d) What is the probability that in a random sample of 250 people, 100 or less will say they think global warming is a threat?

Answer 1

Solution:
Given in the question
Population proportion (p) = 0.44
Number of sample (n) = 250
Solution(a)
We need to calculate mean and standard error of the sampling distribution of sample proportions for given n and p can be calculated as
mean of the sampling distribution of sample proportions = 0.44
Standard error of sampling distribution of sample proportions = sqrt(p*(1-p)/n) = sqrt(0.44*(1-0.44)/250) = 0.0314
Solution(b)
According to empirical rules approximately 99.7% of all sample proportion are between +/-3 standard deviations so Interval can be calculated as
(mean of the sampling distribution - 3*Standard error of sampling distribution) to mean of the sampling distribution + 3*Standard error of sampling distribution)
(0.44 - 3*0.0314) to (0.44 + 3*0.0314)
0.3458 to 0.5342
So interval 0.3458 to 0.5342 centered at p that contains approximately 99.7% of all sample propotion.
Solution(c)
If 100 say that global warning is a threat
So Sample propotion p^ = X/n = 100/250 = 0.4
Z-score for sample proportion can be calculated as
Z-score = (p^ -p)/sqrt(p*(1-p)/n) = (0.4 - 0.44)/sqrt(0.44*(1-0.44)/250) = -0.04/0.0314 = -1.274
Solution(d)
We need to calculate probability that in a random sample of 250 people, 100 or less will say they think global warning is a threat
P(p^<0.4) = ?
From Z table we found p-value
P(p^<0.4) = 0.1020
So there is 10.20% probability that in a random sample of 250 people, 100 or less will say they think global warning is a threat.