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According to a 2017 Pew Research report, \(40 \%\) of millennials have a BA degree. Suppose we take a random sample of 500 millennials and find the proportion who have a driver's license. Find the probability that at most \(35 \%\) of the sample has a BA degree. Begin by verifying that the conditions for the Central Limit Theorem for Sample Proportions have been met.

Short Answer

Expert verified
The probability that at most 35% of the sample has a BA degree can be found by calculating the z-score for 0.35 and looking up the probability corresponding to this z-score in the z-table. The exact value depends on the result of the z-score and the specific values in the z-table, which are standard.

Step by step solution

01

Making sure the conditions for the CLT are met

The conditions for the Central Limit Theorem for Sample Proportions are that the sample is random, most likely stated in the problem. The sample is assumed to be independent since one person having a BA degree should not affect whether another person has a BA degree and the sample size, 500, is large enough as it is more than 30.
02

Calculate the mean and standard deviation of the proportion

The mean of a proportion is simply the proportion in the population that have a BA degree, in this case 0.40. The standard deviation of the proportion can be calculated with the formula \( \sqrt {(P(1-P)/n)} \), where P is the population proportion (0.40) and n is the sample size (500). Hence, standard deviation is \( \sqrt {(0.40*0.60/500)} \)
03

Convert the proportion of interest to a z-score

The proportion of interest is 0.35. To convert this to a z-score, we'll use the formula \( Z = (x - mean) / standard deviation \). In this case, \( Z = ((0.35 - 0.40) / \sqrt {(0.40*0.60/500)}) \)
04

Find the probability using the Z-table

A Z-table can be used to find the probability that the z-score is less than or equal to the calculated z-score. This gives us the probability that at most 35% of the sample has a BA degree. Hence, the answer is the value from the Z-table corresponding to the calculated Z value in the previous step.

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Key Concepts

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

Sample Proportions
When we talk about sample proportions, we are referring to the fraction of a certain characteristic within a sample. In the given exercise, this proportion would be the percentage of millennials in a sample of 500 who have a BA degree. If 40% of millennials in the population have a BA degree, and we are pondering over a sample, sample proportions help us understand how this trait would distribute within that smaller group.

Key points about sample proportions include:
  • The proportion can vary from sample to sample due to variability.
  • The sample proportion (\( \hat{p} \)) can be used to make estimates about a population proportion.
  • This is where the Central Limit Theorem is crucial as it helps in predicting the behavior of this sample proportion when taken repeatedly from the population.
Knowing the expected behavior of sample proportions allows researchers to understand how a sampled statistic relates to the actual population parameter, which in this case is 40%.
Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. For sample proportions, it measures how much these individual sample proportions would differ from the expected population proportion. In a population where the proportion is known, the standard deviation is computed using the formula \[\sqrt{\frac{P(1-P)}{n}}\]where:
  • \(P\) is the population proportion (in our exercise, 0.40).
  • \(n\) is the sample size (here, 500).
This standard deviation assists in understanding how much the sample proportion \(\hat{p}\) might differ from \(P\). It indicates the typical spread of \(\hat{p}\) around \(P\), and in our case helps ascertain how likely it is that a certain percentage of the sample will deviate from the 40% seen in the larger population.
Z-score
The Z-score is a measure that describes how many standard deviations an element is from the mean. It is used in statistics to determine the probability of a sample proportion being a particular value or less. To calculate a Z-score for a sample proportion, \(x\), you use the formula:
  • \[Z = \frac{x - \text{mean}}{\text{standard deviation}}\]
In our scenario, we need to find the Z-score to determine the probability that at most 35% of the sample has a BA degree. By substituting the values of 0.35 for \(x\) and calculating using the mean of 0.40 and the standard deviation from earlier, we can compute a Z value. This Z-score allows us to look up the probability in a Z-table, showing how likely it is for this proportion to be at the specified level or lower.

Understanding the Z-score is essential for students as it is a gateway into probability in hypothesis testing and inference making based on sample data.
Pew Research Report
The Pew Research Center is a widely respected source of quantitative social science research. Reports like the one referenced here provide snapshots of societal trends and are invaluable to researchers and policymakers. The 2017 Pew Research report cited gives critical insights into the educational attainment of millennials, stating that 40% hold a BA degree. This context serves as the baseline for statistical analysis in our exercise.

Understanding research reports such as those by Pew is crucial in academic and societal decision-making:
  • They offer data that can be used to establish population parameters, as seen in our exercise.
  • These insights help in formulating hypotheses and setting expectations for demographic-based studies or polls.
  • They often serve as benchmarks for understanding changes over time or differences across populations.
Thus, reports like these are not just informative but form a foundational element in statistical studies and real-world application of statistical principles.

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Most popular questions from this chapter

In a 2018 survey conducted by Northeastern University, \(28 \%\) of working adults with education levels less than a bachelor's degree worried that their job would be eliminated due to new technology or automation. This was based on a \(95 \%\) confidence interval with a margin of error of 3 percentage points. a. Report the confidence interval for the proportion of adults with education level less than a bachelor's degree who are worried about job loss due to new technology or automation. b. If the sample size were smaller and the sample proportion stayed the same, would the resulting interval be wider or narrower than the one obtained in part a? c. If the confidence level were \(99 \%\) rather than \(95 \%\) and the sample proportion stayed the same, would the interval be wider or narrower than the one obtained in part a? d. In 2018 the population of the United States was roughly 327 million. If 50 million people were added to the population what effect, if any, would this have on the intervals obtained in this problem?

\(1,3,5,7\), and 9 are odd and 0,2, 4,6, and 8 are even. Consider a 30 -digit line from a random number table. a. How many of the 30 digits would you expect to be odd on average? b. If you actually counted, would you get exactly the number you predicted in part a? Explain.

A double-blind study using random assignment was done of pregnant women in Denmark. Women were given fish oil or a placebo during pregnancy. Their children were followed during the first 5 years of life to see if they developed asthma. The results are summarized in the table. (Bisgaard et al., "Fish Oil-Derived Fatty Acids in Pregnancy and Wheeze and Asthma in Offspring," New England Journal of Medicine, vol. 375: 2530-2539. doi: 10.1056/NEJMoa1503734) $$ \begin{array}{|lcc|} \hline \text { Developed asthma } & \text { Fish Oil } & \text { Placebo } \\ \hline \text { Yes } & 58 & 83 \\ \hline \text { No } & 288 & 266 \\ \hline \end{array} $$ a. Calculate and compare the percentages of children who developed asthma in the fish oil group and in the placebo group. b. Check that the conditions for using a two-population confidence interval hold. c. Find the \(95 \%\) confidence interval for the difference in the proportion of children who develop asthma in the two groups. Based on your confidence interval, can we conclude that there is a difference in the population proportions?

A large collection of one-digit random numbers should have about \(50 \%\) odd and \(50 \%\) even digits, because five of the ten digits are odd \((1,3,5,7\), and 9\()\) and five are even \((0,2,4,6\), and 8\()\). a. Find the proportion of odd-numbered digits in the following lines from a random number table. Count carefully. $$ \begin{array}{lll} 57.283 \mathrm{pt} & 74834 & 81172 \\ \hline 89281 & 48134 & 71185 \end{array} $$ b. Does the proportion found in part a represent \(\hat{p}\) (the sample proportion) or \(p\) (the population proportion)? c. Find the error in this estimate, the difference between \(\hat{p}\) and \(p\) (or \(\hat{p}-p)\).

According to a 2017 Pew Research survey, \(60 \%\) of young Americans aged 18 to 29 say the primary way they watch television is through streaming services on the Internet. Suppose a random sample of 200 Americans from this age group is selected. a. What percentage of the sample would we expect to watch television primarily through streaming services? b. Verify that the conditions for the Central Limit Theorem are met. c. Would it be surprising to find that 125 people in the sample watched television primarily through streaming services? Why or why not? d. Would it be surprising to find that more than \(74 \%\) of the sample watched television primarily through streaming services? Why or why not?

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