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Do you drink the cereal milk? AUSA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 67\(\%\) said that they drink it. Suppose that 70\(\%\) of U.S. adults actually drink the cereal milk. Let \(\hat{p}\) be the proportion of people in the sample who drink the cereal milk. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p} .\) Check to see if the 10\(\%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Normal condition is met. (d) Find the probability of obtaining a sample of 1012 adults in which 67\(\%\) or fewer say they drink the cereal milk. Do you have any doubts about the result of this poll?

Short Answer

Expert verified
(a) 0.70; (b) approximately 0.0141; (c) Yes, conditions met; (d) Probability is 0.0166, unusual result suggests doubts.

Step by step solution

01

Mean of the Sampling Distribution

The mean of the sampling distribution of the sample proportion \( \hat{p} \) is equal to the actual proportion of the population \( p \). Given that \( p = 0.70 \) (since 70\(\%\) of U.S. adults actually drink the cereal milk), the mean of the sampling distribution of \( \hat{p} \) is \( 0.70 \). This is because the expected value of the sample proportion equals the true population proportion.
02

Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of the sample proportion \( \hat{p} \) is given by the formula:\[\sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} }.\]Substituting the values, \( p = 0.70 \) and \( n = 1012 \), we get:\[\sigma_{\hat{p}} = \sqrt{ \frac{0.70(1-0.70)}{1012} } \approx 0.0141.\]To check the 10\(\%\) condition: our sample size \( n = 1012 \) is much less than 10\(\%\) of the total U.S. adult population, so the condition is met.
03

Normal Condition Verification

For the sampling distribution to be approximately Normal, the sample must satisfy the conditions: \( np \geq 10 \) and \( n(1-p) \geq 10 \). We calculate these as follows:- \( np = 1012 \times 0.70 = 708.4 \geq 10 \)- \( n(1-p) = 1012 \times 0.30 = 303.6 \geq 10 \)Both conditions are satisfied, indicating that the sampling distribution of \( \hat{p} \) is approximately Normal.
04

Probability Calculation

We want to calculate the probability that \( \hat{p} \leq 0.67 \). First, calculate the Z-score:\[ z = \frac{ \hat{p} - p }{ \sigma_{\hat{p}} } = \frac{ 0.67 - 0.70 }{ 0.0141 } \approx -2.13 \]Using the Standard Normal Distribution table, the probability \( P(Z \leq -2.13) \) is approximately 0.0166.Since this probability is very low, it suggests that obtaining a sample proportion of 67\% is unusual if the true proportion is 70\%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
When we talk about the sample proportion, denoted as \( \hat{p} \), we're referring to the fraction of the sample that has a particular characteristic. In our cereal milk example, 67% of the 1012 US adults polled said they drink the milk left in their cereal bowl. So, \( \hat{p} = 0.67 \). This value gives us an idea of what fraction of our sample exhibits the behavior we're interested in. We use this estimate because sometimes it's not feasible to gather data from an entire population.

The sample proportion is crucial in understanding how close our estimate is to the true population proportion, which in this case is 70% or \( p = 0.70 \). In statistical studies, the sample proportion helps us infer or predict population behaviors without surveying everyone. It acts as a best guess based on available data.
Standard Deviation of Sampling Distribution
The standard deviation of the sampling distribution of a sample proportion provides us with insight into the variability of the sample's proportion from the actual population proportion. It tells us how much the sample proportion \( \hat{p} \) would "wobble" if we took multiple samples. Calculation involves the formula: \[ \sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} }. \] For our cereal scenario, \( p = 0.70 \) and \( n = 1012 \), so substituting gives: \[ \sigma_{\hat{p}} = \sqrt{ \frac{0.70 \times (1-0.70)}{1012} } \approx 0.0141. \]

This tells us that typically the sample proportion will differ from the true population proportion by around 0.014 or 1.4%. The smaller this standard deviation, the closer our sample estimate can be expected to be to the true population proportion. It's like how wide a net is cast when estimating population characteristics based on sample data.
Normal Condition
The Normal condition refers to the requirements that must be met for a sampling distribution to be approximately Normal. In practical terms, this means we can use normal probability models to make inferences from our sample data. We require at least:
  • \( np \geq 10 \)
  • \( n(1-p) \geq 10 \)
where \( n \) is the sample size, and \( p \) is the true population proportion.

For our example:
  • \( np = 1012 \times 0.70 = 708.4 \) which is greater than 10.
  • \( n(1-p) = 1012 \times 0.30 = 303.6 \) which is also greater than 10.
Since both values exceed 10, the sampling distribution of \( \hat{p} \) can be considered approximately Normal. This is important because it allows us to use reliable statistical techniques like Z-tests to make conclusions about the population.
10% Condition
The 10% condition is a check to ensure our sample is truly random and independently chosen. Essentially, it verifies that the sample size \( n \) does not exceed 10% of the total population. Meeting this condition implies the "infinite population" assumption can be applied, simplifying calculations using binomial approximations.

Since we're dealing with the adult population in the US, our sample size of 1012 is definitely less than 10% of all U.S. adults. If we assume the U.S. adult population is in the range of hundreds of millions, 1012 is just a tiny fraction. Hence, the 10% condition is comfortably met.

Meeting this condition ensures that each individual's response remains independent of others' in the sample. It safeguards our calculations, giving more confidence that our sample results can generalize to the entire population.

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Most popular questions from this chapter

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