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Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating \(p\) to within .04 with probability .95. You suspect that \(p\) is equal to some value between .1 and \(.3 .\)

Short Answer

Expert verified
Answer: The required sample size is approximately \(n = 369\).

Step by step solution

01

Identify the margin of error, confidence level, and suspected range of \(p\).

The margin of error is 0.04, the probability (confidence level) is 0.95, and the suspected value of \(p\) lies between 0.1 and 0.3.
02

Determine the value of \(Z_{\alpha / 2}\).

For a 95% confidence interval, we will need to find the z-score, \(Z_{\alpha / 2}\), which corresponds to 0.975 in the standard normal distribution table. The z-score for a 95% confidence interval is approximately 1.96.
03

Choose the highest variance for \(p\).

Since the variance of a proportion is given by \(p(1-p)\), we need to choose the value which will result in the highest variance within the given suspected range of 0.1 to 0.3. In this case, the highest variance will occur when \(p=0.3\) because \((0.3)(1-0.3) > (0.1)(1-0.1)\).
04

Calculate the sample size, \(n\).

We will now use the formula \(n \geq \left(\frac{Z_{\alpha / 2} \times \sqrt{p(1-p)}}{E}\right)^2\) and the values we've identified to calculate the sample size required: \(n \geq \left(\frac{1.96 \times \sqrt{(0.3)(1-0.3)}}{0.04}\right)^2\) \(n \geq \left(\frac{1.96 \times \sqrt{(0.3)(0.7)}}{0.04}\right)^2\) \(n \approx 368.99\) Since we cannot have a decimal as our sample size, we round up to the next whole number to ensure the criteria are met:
05

Finalize the sample size.

The required sample size to estimate \(p\) to within 0.04 with a probability of 0.95 is approximately \(n = 369\).

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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, especially when you're conducting polls or surveys. It tells you how much the results from your sample might differ from the true value in the whole population. Think of it as a buffer zone; if you estimate that 60% of voters favor a certain candidate with a margin of error of 4%, the real support could lie somewhere between 56% and 64%.

To achieve a small margin of error, you often need a larger sample size. But remember, a small margin of error increases precision but also requires more resources like time and money, so it's all about finding a balance. Sample size calculations, like the one in our exercise, usually start by deciding upon an acceptable margin of error for your estimate.
Confidence Interval
A confidence interval gives you a range of values that, with a certain level of confidence, includes the true value of the parameter you're estimating. It's like saying 'I'm 95% sure the true average height of sunflowers in my garden is between 150 and 160 cm.'

The width of this interval depends on your chosen confidence level and the variability of your data. Higher confidence levels tend to produce wider intervals. In our exercise, we're using a 95% confidence level, which is pretty standard, to say that we're 95% confident that the true proportion lies within our calculated interval.
Proportion Variance
When we talk about proportion variance in statistics, we're dealing with the spread or dispersion of our data in terms of proportions. In the context of a binary outcome, like a 'yes' or 'no' answer, the variance is highest when the proportion is at 0.5 - when outcomes are most uncertain. In our example, we are estimating a proportion, so we calculate variance as \(p(1-p)\).

When determining sample size, we want to account for the greatest variability to be on the safe side, which is why we chose \(p = 0.3\) as it gives us the maximum variance within the suspected range of 0.1 - 0.3. This ensures our sample size is sufficient even in the scenario of greatest possible variability.
Z-Score
A z-score is like a scoreboard for individual scores comparing them to the group average. It tells you how many standard deviations a point is away from the mean. In the field of inferential statistics, we use the z-score to find out how improbable our sample statistics are if the null hypothesis was true.

For our confidence interval, we pick the z-score corresponding to our desired confidence level by looking in the standard normal distribution table. A high z-score indicates a rare event. In our example, we use a z-score of 1.96 for a 95% confidence level, meaning that we're looking at the points 1.96 standard deviations on either side of the mean, capturing the middle 95% of the data according to our normal distribution.

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Most popular questions from this chapter

Auto Accidents A recent year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 3000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc}\hline & \text { Under } \$ 3000 & \$ 3000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 3000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 3000\) and those with damage of \(\$ 3000\) or more. Use a \(95 \%\) confidence interval.

Calculate the margin of error in estimating a population mean \(\mu\) for the values given in Exercises \(7-10 .\) Comment on how an increased sample size affects the margin of error: \(n=100, s^{2}=4\)

Does It Pay to Haggle? In Exercise \(16,\) a survey done by Consumer Reports indicates that you should always try to negotiate for a better deal when shopping or paying for services. \({ }^{20}\) In fact, based on the survey, \(37 \%\) of the people under age 34 were more likely to "haggle," while only \(13 \%\) of those 65 and older. Suppose that this group included 72 people under the age of 34 and 55 people who are 65 or older. a. What are the values of \(\hat{p}_{1}\) and \(\hat{p}_{2}\) for the two groups in this survey? b. Find a \(95 \%\) confidence interval for the difference in the proportion of people who are more likely to "haggle" in the "under 34 " versus " 65 and older" age groups. c. What conclusions can you draw regarding the groups compared in part b?

Calculate the margin of error in estimating a binomial proportion \(p\) for the sample sizes given in Exercises \(11-14\). Use \(p=.5\) to calculate the standard error of the estimator, and comment on how an increased sample size affects the margin of error. \(n=400\)

A pediatrician randomly selected 50 six-month-old boys from her practice's database and recorded an average weight of 8.0 kilograms with a standard deviation of 0.30 kilogram. She also recorded an average length of 67.3 centimeters with a standard deviation of 0.64 centimeter. a. Find a \(95 \%\) confidence interval for the average weight of all six-month- old boys. b. Find a \(99 \%\) confidence interval for the average length of all six-month- old boys. c. What do you have to assume about the pediatrician's database in order to make inferences about all sixmonth-old boys?

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