Question

An opinion poll asks an SRS of 1500 adults, "Do you happen to jog?" Suppose that the population proportion who jog (a parameter) is p = 0.15. To estimate p, we use the proportion p-hat in the sample who answer "Yes." Justify the use of a normal approximation and find the following probabilities

Answer 1

a) P(p > 0.16)
b) P(0.14 < p < 0.16)

Answer 2

@Splash_dance can you help with this one also?

Answer 3

First, why do you need to estimate the proportion of those who jog if you are given it in the beginning of your question? (i.e. you're told that p = 15%).

Answer 4

For starters, we can use the Normal distribution (to approximate the Binomial distribution) because the sample size (n = 1500) is so large.

Answer 5

would a be 0.1379

Answer 6

So are you asking:
Given that the population proportion, p, is equal to 0.15, what is the probability that p-hat (the result of your sample) is greater than 0.16?

Answer 7

yes

Answer 8

Well, I might be way off but here it goes:

First, the variance of a Bernoulli random variable is given by:
\[\sigma ^{2} = p(1-p)\]
In addition, when calculating distribution of the sample mean we use the formula:
\[\frac{ sample mean - \mu }{ standard error of sample mean }\]

So, to calculate the P(p>0.16), first we find the z-score associated with p=0.16:
\[= \frac{ 0.16 - 0.15 }{ \sqrt{\frac{ 0.15\times0.85 }{ 1500 }} }\]
which is approximately equal to 1.08465.
Looking up 1.08 in a Z-score table we get:
\[P(Z \le 1.08) = 0.8599\]
Thus, the probability that Z is greater than 1.08 is:
1- 0.8599 = 14.01%

Does this sound right to you?