Suppose simple random sample of size 50 is obtained from...

Suppose a simple random sample of size 50 is obtained from a population whose size is N = 25,000 and whose population proportion with a specified characteristic is p = 0.6. (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below: Approximately normal because n < 0.05N and np(1 - p) > 10. Approximately normal because n > 0.05N and np(1 - p) < 10. Not normal because n < 0.05N and np(1 - p) < 10. Not normal because n < 0.05N and np(1 - p) > 10. Determine the mean of the sampling distribution of p: μp = p = 0.6 (Round to one decimal place as needed) Determine the standard deviation of the sampling distribution of p: σp = √(p(1 - p)/n) = √(0.6(1 - 0.6)/50) (Round to six decimal places as needed) (b) What is the probability of obtaining x = 31 or more individuals with the characteristic? That is, what is P(p ≥ 0.62)? P(p ≥ 0.62) = (Round to four decimal places as needed) (c) What is the probability of obtaining x = 26 or fewer individuals with the characteristic? That is, what is P(p ≤ 0.52)? P(p ≤ 0.52) = (Round to four decimal places as needed)

Transcript

00:01 A sample of size 50 is taken from a population.

00:04 So sample size 50, the proportion of a particular characteristic p is 0 .6.

00:10 Okay, so we want to describe the sampling distribution of p hat.

00:15 So that's the proportion of a particular sample that has this characteristic.

00:21 So first of all, let's look at this.

00:24 We have a sample taken from a population.

00:26 I would like to start out here with a binomial distribution.

00:31 For that to be true, these need to be considered independent trials.

00:35 They also need to be considered independent to apply the central limit theorem, which we will be applying in the next step.

00:45 Okay, so can i consider these independent? well, the sample is much smaller than the population.

00:52 It's less than or equal to 5 % of the population.

00:57 So i can say the sample is so much smaller than the population that yes, i can consider these to be independent.

01:04 Next, i want to take a normal approximation to the binomial.

01:10 And for that, i need the binomial to be reasonably symmetric.

01:14 Some binomials are quite skewed.

01:17 Is this reasonably symmetric or is it skewed? well, let's have a look at the standard deviation of this binomial, np or minus p.

01:27 If it's large enough, we can consider it to be approximately normal.

01:32 This is 10.

01:37 If this is true, there'll be enough successes and failures that it'll be roughly symmetric and we can take a good normal approximation.

01:46 Let's start through here.

01:47 Let's see, 0 .6 times 0 .4 is 0 .24.

01:53 Multiply it by 50.

01:55 Yes, it's above 10.

01:57 So both of these are true.

01:58 So option a is going to be the correct answer.

02:01 And if we look at this normal approximation we have now, so the binomial variable is x, the number of people in the sample or number of anything in the sample that meets this criteria.

02:15 The normal distribution is also a distribution of x.

02:20 But what we then do is we just divide the entire distribution by n.

02:24 And now it's a distribution of p hat, the sample proportions.

02:29 The mean of the binomial was np.

02:33 The standard deviation of the binomial was root np one minus p.

02:37 Just the mean and standard deviation of any binomial.

02:40 These were also parameters of the initial normal distribution.

02:44 And now i've gone and divided them by n to make sure they match up with this sampling distribution.

02:50 So how does that change them? well, mu becomes mu p hat.

02:53 That is p.

02:56 Sigma becomes sigma p hat, also called the standard error.

03:00 That is p minus p over n, all in this square root...