A simple random sample of size n=1460 is obtained from a...

A simple random sample of size $n=1460$ is obtained from a population whose size is $N=1,500,000$ and whose population proportion with a specified characteristic is $p=0.42$ (a) Describe the sampling distribution of $\hat{p}$. (b) What is the probability of obtaining $x=657$ or more individuals with the characteristic? (c) What is the probability of obtaining $x=584$ or fewer individuals with the characteristic?

A simple random sample of size $n=1460$ is obtained from a population whose size is $N=1,500,000$ and whose population proportion with a specified characteristic is $p=0.42$
(a) Describe the sampling distribution of $\hat{p}$.
(b) What is the probability of obtaining $x=657$ or more individuals with the characteristic?
(c) What is the probability of obtaining $x=584$ or fewer individuals with the characteristic?

Key Concepts

Sampling Distribution of a Sample Proportion

This concept refers to the probability distribution of a sample proportion (p-hat) taken from a population. It describes how p-hat varies from sample to sample and is centered around the true population proportion (p). For sufficiently large samples, the sampling distribution is approximately normal, which allows us to perform inference about the population proportion.

Central Limit Theorem

The Central Limit Theorem states that, for large enough sample sizes, the distribution of the sample mean (or sample proportion) approximates a normal distribution regardless of the population's distribution, as long as the samples are independent. This theorem is fundamental in justifying the use of normal probability models when estimating probabilities about the sample proportion.

Normal Approximation to the Binomial Distribution

When dealing with proportions and a large sample size, the binomial distribution of the number of successes can be approximated by a normal distribution. This approximation is used for calculating probabilities (such as the probability of obtaining a number of successes above or below a certain threshold) and is valid when the conditions np and n(1-p) are both sufficiently large.

Standard Error of a Proportion

The standard error quantifies the variability of the sample proportion around the population proportion. It is calculated as the square root of [p(1-p)/n], reflecting the spread or dispersion in the sampling distribution. A smaller standard error indicates that the sample proportion is likely to be closer to the population proportion.

Continuity Correction

When a discrete distribution (like the binomial) is approximated by a continuous distribution (like the normal), a continuity correction is applied to adjust for the differences in distribution type. By adding or subtracting 0.5, the approximation becomes more accurate for probability calculations involving specific counts.

Z-score Calculation

A z-score is a standardized value that indicates how many standard deviations an observation is from the mean of its distribution. In the context of a sample proportion, it involves subtracting the population proportion from the sample proportion and dividing by the standard error. This conversion allows us to easily calculate probabilities using the standard normal distribution.

Transcript

00:01 So we're dealing with a binomial distribution problem here.

00:04 We're told 29 % of adults do not own a credit card.

00:08 Part a says to suppose that we take a random sample of 500 adults and we ask them if they own a credit card.

00:14 We're then asked to describe the sampling distribution of p hat.

00:18 So first thing we need to do is define our variables.

00:21 So our variables are n, p, and q.

00:24 N is the sample size, which we're told as 500.

00:28 P we're also given, which is 29%, because p is the probability of a success.

00:34 So here we'll call a success not owning a credit card.

00:37 So again, that's 29 % or as a decimal.

00:41 0 .29.

00:42 And then q is the probability of failure.

00:44 So this would be that they own a credit card, and this would be 1 minus p.

00:49 So 1 minus 0 .29 is equal to 0 .71.

00:53 Now to find our sampling distribution in part a, we need to find a mean and standard deviation.

00:58 Which i have the equations for on the screen here.

01:01 We'll begin with the mean, which is equal to n times p.

01:04 So we'll take 500 times 0 .29, and we'll get 145.

01:11 Then we'll find the standard deviation, which is equal to the square root of n times p times q.

01:16 That's the square root of 500 times 0 .29 times 0 .71.

01:22 And this comes out to approximately 10 .146.

01:26 So our distribution here is approximately normally distributed with a mean of 145 and a standard deviation of about 10 .146.

01:37 We'll use this now in part b.

01:39 Because in part b, we're asked to find the probability that a random sample of 500 adults more than 30 % do not own a credit card.

01:50 So first thing we need to do is find what 30 % of 500 is.

01:54 To do that, we'll simply multiply 500 times 0 .3, and we get 150.

02:01 So now it's a little bit easier to understand.

02:03 So we're finding the probability that more than 150 do not own a credit card.

02:10 So that would be the probability of x being greater than 150...

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