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What sample size is needed to give the desired margin of error in estimating a population proportion with the indicated level of confidence? A margin of error within \(\pm 2 \%\) with \(95 \%\) confidence. An initial small sample has \(\hat{p}=0.78\).

Short Answer

Expert verified
To achieve a margin of error within \(2\%\) at a confidence level of \(95\%\), with an initial estimate sample proportion of \(0.78\), approximately 1696 samples are needed.

Step by step solution

01

Determine the Z-Score for Desired Confidence Level

The z-score is a measure of how many standard deviations an element is from the mean. For a confidence level of \(95\%\), the z-score is \(1.96\). This value can be found in standard statistical tables for normal distribution, correlating to the area (probability) to the left of the z-score.
02

Convert Margin of Error to Decimal

The margin of error is given in percentage. To use it in calculations, we need to convert it into decimal. Margin of error within \( \pm 2\% \) = \( \pm 0.02 \) in decimal.
03

Apply the Formula for Sample Size

The formula to calculate the sample size(n) given a margin of error(E), confidence level(z), and the population proportion(p) is \( n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \). Plug in the given values into the formula: \( n = \left( \frac{(1.96)^2 \cdot 0.78 \cdot (1-0.78)}{(0.02)^2} \right) \)
04

Compute the Sample Size

By performing above calculation, you'll get n as floating point number. Rounded to the nearest whole number this will represent the sample size needed to achieve the desired margin of error. Always round this final answer up to ensure adequate sample size.

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Key Concepts

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

Margin of Error
The margin of error is a crucial concept in statistics that tells us how much we can expect estimated values to differ from the true population value. It gives researchers an idea of the accuracy of their sample estimates. Essentially, the margin of error defines an interval around the sample statistic that is likely to contain the actual population parameter.

When you hear that the margin of error is \(\pm 2\%\), it means that the estimate could be 2% higher or lower than the calculated sample proportion. It's vital for ensuring that results are statistically significant and reliable.

Calculating it involves a few components, most importantly:
  • The standard deviation of the sample.
  • The size of the sample.
  • The desired level of confidence.
In simple terms, for a smaller margin of error, you need a larger sample size. This ensures that your results will be more precise and closer to the true population value.
Population Proportion
Population proportion is the ratio of members in the population with a particular attribute to the total number of members in the population. For example, if we're trying to find out the proportion of people who prefer tea over coffee in a given city, the population proportion would be the number of tea lovers divided by the total number of people surveyed in that city.

Understanding population proportion helps researchers to make informed predictions and decisions based on the data they have. In the given problem, the initial survey tells us that \(\hat{p}=0.78\), indicating that 78% of the small sample prefer tea. This is an estimate of the true population proportion.

Visualization of population proportion

As the population proportion affects the sample size calculation, precise estimates are required. Larger deviations in the proportion may require more extensive research or larger sample surveys to refine the calculated statistics further.
Confidence Level
Confidence level reflects how certain we are about our estimation. For instance, a \(95\%\) confidence level signifies that if we were to take 100 different samples and compute an interval estimate for each sample, we expect about 95 of those intervals to contain the true population parameter.

The confidence level is directly tied to the z-score in statistics. It doesn't say that there's a 95% probability that the actual parameter lives within the interval, but rather reflects the long-term reliability of the estimation method.
  • A higher confidence level means a wider confidence interval.
  • A lower confidence level gives a narrower interval but with more uncertainty.
In practice, high confidence levels are generally preferred to ensure the reliability of the results, especially in critical decision-making scenarios.

In our exercise, the desired confidence is \(95\%\), corresponding to a z-score of 1.96 in statistical tables. This value is pivotal because it influences how accurately we can pinpoint the sample size needed to estimate the population parameter precisely.

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Most popular questions from this chapter

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\) A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table:. $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=5.2, s_{1}=2.7, n_{1}=10\) and \(\bar{x}_{2}=4.9, s_{2}=2.8, n_{2}=8 .\)

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether males are less likely than females to support a ballot initiative, if \(24 \%\) of a random sample of 50 males plan to vote yes on the initiative and \(32 \%\) of a random sample of 50 females plan to vote yes.

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. A study investigating the effect of exercise on brain activity recruits sets of identical twins in middle age, in which one twin is randomly assigned to engage in regular exercise and the other doesn't exercise.

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