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A recent Gallup poll asked Americans to disclose the number of books they read during the previous year. Initial survey results indicate that \(s=16.6\) books. (a) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(95 \%\) confidence? (b) How many subjects are needed to estimate the number of books Americans read the previous year within two books with \(95 \%\) confidence? (c) What effect does doubling the required accuracy have on the sample size? (d) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(99 \%\) confidence? Compare this result to part (a). How does increasing the level of confidence in the estimate affect sample size? Why is this reasonable?

Step by step solution

01

Identify the given values

Given values for the problem are: - Standard deviation, s = 16.6 books - Confidence level for part (a) and (b) = 95%- Error margin for part (a) = 4 books- Error margin for part (b) = 2 books- Confidence level for part (d) = 99%- Error margin for part (d) = 4 books

02

Understand the formula for sample size

The formula to calculate the sample size needed when the population standard deviation is known is: \[ n = \left( \frac{z \cdot s}{E} \right)^2 \]Where:- n = required sample size- z = z-value (depends on confidence level)- s = standard deviation- E = margin of error

03

Calculate the z-value for 95% confidence level

For a 95% confidence level, the z-value is approximately 1.96.

04

Calculate the sample size for part (a)

Using the formula for sample size with 95% confidence level and error margin of 4 books:\[ n = \left( \frac{1.96 \cdot 16.6}{4} \right)^2 \]Calculate the value inside the brackets first:\[ \frac{1.96 \cdot 16.6}{4} = 8.127 \]Now square this value:\[ n = 8.127^2 \approx 66 \]So, approximately 66 subjects are needed.

05

Calculate the sample size for part (b)

Using the formula for sample size with 95% confidence level and error margin of 2 books:\[ n = \left( \frac{1.96 \cdot 16.6}{2} \right)^2 \]Calculate the value inside the brackets first:\[ \frac{1.96 \cdot 16.6}{2} = 16.255 \]Now square this value:\[ n = 16.255^2 \approx 264 \]So, approximately 264 subjects are needed.

06

Effect of doubling the accuracy

Doubling the required accuracy (halving the margin of error from 4 to 2 books) results in an increase in the required sample size from 66 to 264. This shows that increasing accuracy requires significantly more subjects.

07

Calculate the z-value for 99% confidence level

For a 99% confidence level, the z-value is approximately 2.576.

08

Calculate the sample size for part (d)

Using the formula for sample size with 99% confidence level and error margin of 4 books:\[ n = \left( \frac{2.576 \cdot 16.6}{4} \right)^2 \]Calculate the value inside the brackets first:\[ \frac{2.576 \cdot 16.6}{4} = 10.687 \]Now square this value:\[ n = 10.687^2 \approx 114 \]So, approximately 114 subjects are needed.

09

Effect of increasing confidence level

Increasing the confidence level from 95% to 99% increases the required sample size from 66 to 114 subjects. This is reasonable because a higher confidence level means more certainty in the estimate, thus requiring more data to support this increased certainty.

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Key Concepts

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

Confidence Intervals

Confidence intervals are used to estimate the range within which a population parameter (like the mean) likely falls. For example, if we want to estimate the average number of books read by Americans in a year, we calculate a confidence interval around that estimate.
The width of this confidence interval depends on our chosen confidence level (such as 95% or 99%), the margin of error, and the variability of the data (standard deviation).
Higher confidence levels (such as 99%) will produce wider intervals as we want to be more sure that the interval contains the true mean.
This means gathering more data. In our exercise, the intervals help gauge how many books Americans might have read within four or two books, providing a clearer picture with more data.

Margin of Error

The margin of error reflects the range within which the true population parameter is expected to lie, measured from the sample estimate. It is calculated as a product of the critical value (z-value) and the standard deviation of the population divided by the square root of the sample size.
Practically, if we wanted a smaller margin of error (like within two books), we would need a larger sample size.
More subjects reduce the uncertainty.
In our example, reducing the error margin from four books to two books significantly increased the sample size needed.

Standard Deviation

Standard deviation (denoted as s) quantifies the variation or dispersion of a dataset. A higher standard deviation indicates that the data points are spread out over a larger range of values.
In sample size estimation, standard deviation helps understand the variability in the number of books read across the population.
A higher standard deviation would generally increase the required sample size because of the greater uncertainty about where the true population mean lies within the sample data.
In the exercise, we used a standard deviation of 16.6 books, meaning the number of books read by Americans varies quite a bit among participants.

Z-value

The z-value (also known as the z-score or critical value) represents the number of standard deviations a data point is from the mean, corresponding to a certain confidence level.
For instance, a 95% confidence level corresponds to a z-value of approximately 1.96, while a 99% confidence level uses a z-value of about 2.576.
These z-values help stretch or shrink the confidence interval based on our desired level of certainty.
In the sample size estimation, higher z-values for increased confidence levels lead to a larger required sample size, as shown by the exercise when shifting from 95% (z = 1.96) to 99% (z = 2.576), resulting in more subjects needed.