The Cincinnati Enquirer reported that, in the United States,...

The Cincinnati Enquirer reported that, in the United States, $66 \%$ of adults and $87 \%$ of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February 7,2006 ). Use the reported numbers as the population proportions and assume that samples of 300 adults and 300 youths will be used to learn about attitudes toward Internet security. a. Show the sampling distribution of $\bar{p}$ where $\bar{p}$ is the sample proportion of adults using the Internet. b. What is the probability that the sample proportion of adults using the Internet will be within ±.04 of the population proportion? c. What is the probability that the sample proportion of youths using the Internet will be within ±.04 of the population proportion? d. Is the probability different in parts (b) and (c)? If so, why? e. Answer part (b) for a sample of size $600 .$ Is the probability smaller? Why?

The Cincinnati Enquirer reported that, in the United States, $66 \%$ of adults and $87 \%$ of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February 7,2006 ). Use the reported numbers as the population proportions and assume that samples of 300 adults and
300 youths will be used to learn about attitudes toward Internet security.
a. Show the sampling distribution of $\bar{p}$ where $\bar{p}$ is the sample proportion of adults using the Internet.
b. What is the probability that the sample proportion of adults using the Internet will be within ±.04 of the population proportion?
c. What is the probability that the sample proportion of youths using the Internet will be within ±.04 of the population proportion?
d. Is the probability different in parts (b) and (c)? If so, why?
e. Answer part (b) for a sample of size $600 .$ Is the probability smaller? Why?

Key Concepts

Sampling Distribution of a Sample Proportion

This concept refers to the probability distribution of the sample proportion (p-hat) obtained from repeatedly taking samples of the same size from a population. It describes how p-hat varies from sample to sample, with its center at the true population proportion (p) and its spread measured by the standard error. Understanding this distribution is crucial for making inferences about the population based on sample data.

Standard Error of the Proportion

The standard error quantifies the variability of the sample proportion and is calculated using the formula sqrt[p(1 - p)/n], where p is the population proportion and n is the sample size. It provides a measure of how much the sample proportion is expected to fluctuate around the true proportion and plays a key role in assessing the precision of estimates and the calculation of probabilities for specified intervals.

Normal Approximation to the Sampling Distribution

When the sample size is large, the Central Limit Theorem assures that the sampling distribution of the sample proportion can be approximated by a normal distribution. This normal approximation allows us to use z-scores and standard normal probability calculations to determine the likelihood that the sample proportion falls within a certain range of the population proportion.

Margin of Error

The margin of error in the context of a sample proportion represents the maximum expected difference between the sample proportion and the true population proportion due to random sampling variability. It is often used to express the reliability of an estimate, and determining the probability of the sample proportion lying within a specific margin (for example, ±0.04) is directly linked to the concepts of the standard error and normal approximation.

Impact of Sample Size

Sample size plays a critical role in determining the accuracy and reliability of the sample proportion. An increase in sample size reduces the standard error, leading to a narrower sampling distribution. This, in turn, increases the likelihood that the sample proportion will lie within any fixed margin of error around the population proportion, thereby enhancing the precision of statistical estimates.

Transcript

00:01 So we have the cincinnati enquirer reported that in the us, 66 % of adults and 87 % of youths use the internet.

00:12 And we are assuming that these population proportions are coming from, or the sample proportions are coming from samples of size 300.

00:20 And it says in part a, show the sampling distribution of p -hat, where p -hat is a sample proportion for adults.

00:28 So the mean of those values would end up being approximately .66, and the standard deviation of those would be the square root of .66 times .34 divided by that 300.

00:45 And that value comes out to be, let me quickly calculate that, square root of .66 times .34 divided by 300 gives us a value of approximately .0273.

01:01 And it would be approximately normal, because n times p and n times 1 minus p would be greater than or equal to 10.

01:10 Part b says, what is the probability that the sample proportion of adults is within .04 of the population proportion? so we want to find the likelihood of being within .04.

01:25 So we'd have a negative .04 as a difference between it and the sample proportion and the population proportion.

01:35 And then to find it, convert it to a z value, we are going to divide it by that standard error, or that standard deviation of the p -hats of that sample size.

01:47 And so this will give us our value.

01:51 So i'm going to take that .04 divided by, and that's the second answer in my calculator, and that z value comes out to be 1 .46, negative 1 .46.

02:01 And then the probability of being in between those values, we can look it up in a table, or we can use our normal cdf.

02:10 And i'm just going to quickly put it into my normal cdf and find the values without looking it up in the table.

02:18 So i'm finding that that is .8557.

02:22 Now on part c, we want to find what the probability is that the youth are within .04.

02:30 And for the youth, we're going to have the negative .04 and the positive .04, and then we're going to convert that to a z value...

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