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Which would be hardest for you to give up: Your computer or your television? In a recent survey of 1677 U.S. Internet users, \(74 \%\) of the young tech elite (average age of 22 ) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48\% say their television would be very hard to give up. a. Develop a \(95 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer. b. Develop a \(99 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their television. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

Short Answer

Expert verified
The margin of error is larger in part (b) because it uses a higher Z-score for the 99% confidence interval.

Step by step solution

01

Identify the parameters for part (a)

For part (a), we need to calculate the 95% confidence interval for the proportion of young tech elite who would find it hard to give up their computer. We are given the sample proportion \( p = 0.74 \) and the sample size \( n = 1677 \).
02

Compute the standard error for part (a)

The standard error \( SE \) for a proportion is calculated using the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \). Substitute in the known values: \( SE = \sqrt{\frac{0.74(1-0.74)}{1677}} \).
03

Calculate the Z-score for 95% confidence interval

For a 95% confidence interval, the Z-score is approximately 1.96 based on the standard normal distribution.
04

Determine the confidence interval for part (a)

The confidence interval is calculated as \( p \pm Z \times SE \). Substitute \( p = 0.74 \), \( Z = 1.96 \), and \( SE \) from Step 2 into the formula to find the interval.
05

Identify the parameters for part (b)

For part (b), we need to develop a 99% confidence interval for the proportion of young tech elite that would find it hard to give up their television. Here, \( p = 0.48 \) and the sample size is still \( n = 1677 \).
06

Compute the standard error for part (b)

The standard error \( SE \) is calculated as \( SE = \sqrt{\frac{p(1-p)}{n}} \) for \( p = 0.48 \). Calculate \( SE = \sqrt{\frac{0.48(1-0.48)}{1677}} \).
07

Calculate the Z-score for 99% confidence interval

For a 99% confidence interval, the Z-score is approximately 2.576 based on the standard normal distribution.
08

Determine the confidence interval for part (b)

The confidence interval is calculated as \( p \pm Z \times SE \). Substitute \( p = 0.48 \), \( Z = 2.576 \), and \( SE \) from Step 6 into the formula to find the interval.
09

Compare the margins of error for parts (a) and (b)

The margin of error is larger in part (b) due to the higher Z-score for the 99% confidence interval compared to the 95% confidence interval in part (a). That is because higher confidence levels result in wider intervals to account for more uncertainty.

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Key Concepts

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

Understanding Sample Proportion
When conducting a survey or a study, the sample proportion is a crucial statistic that helps us understand the makeup of our sample. In this context, the sample proportion is used to estimate the true proportion of a larger population. It's basically the proportion of individuals in a sample with a certain characteristic, which we use to represent the whole.

In the given exercise, two specific examples are provided: the proportion who find it hard to give up their computer, and the proportion who would find it hard to give up their television. These proportions are respectively 0.74 and 0.48. To arrive at these proportions, we divide the number of respondents with the characteristic by the total sample size.

Using the responses, out of 1677 participants, 74% find it hard to give up their computer. We can express this as:
  • Sample Proportion = 0.74
  • Sample Size = 1677
Understanding sample proportion is fundamental because it provides the basis for further statistical work, such as computing confidence intervals.
Calculating Standard Error
The standard error of a sample proportion is an important part of understanding the variability or dispersion of the sample data around the true population proportion. It tells us how much the sample proportion can be expected to fluctuate from the actual population proportion.

To calculate standard error (SE), we use the formula:\[SE = \sqrt{\frac{p(1-p)}{n}}\], where:
  • \(p\)is the sample proportion.
  • \(n\)is the sample size.
For example, in the exercise for part (a), the standard error computation for those who can't give up their computers (\(p = 0.74\)) is:\[SE = \sqrt{\frac{0.74 \times (1-0.74)}{1677}}\]From this, we can grasp the spread of our data, which aids in constructing confidence intervals. Smaller standard error indicates the sample proportion is a more accurate reflection of the true population proportion.
Understanding Margin of Error
The margin of error is the range of values below and above the sample proportion that represents uncertainty. It accounts for any error due to sampling variability, allowing us to express that we are a certain percentage confident the true proportion lies within this range.

To calculate the margin of error, we multiply the Z-score corresponding to our desired confidence level by the standard error (SE) of our sample proportion. For instance, in a 95% confidence interval, we use a Z-score of 1.96, and for a 99% confidence interval, a Z-score of 2.576.

In the exercise:
  • Part (a): Margin of Error = 1.96 \(\times\) SE
  • Part (b): Margin of Error = 2.576 \(\times\) SE
This means that in part (b), increasing the confidence level requires a larger margin of error. This is because higher confidence levels widen the interval to ensure the true proportion is captured, acknowledging more uncertainty.

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Most popular questions from this chapter

Consumption of alcoholic beverages by young women of drinking age has been increasing in the United Kingdom, the United States, and Europe (The Wall Street Journal, February 15 2006 ). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women. $$\begin{array}{rrrrr}266 & 82 & 199 & 174 & 97 \\\170 & 222 & 115 & 130 & 169 \\\164 & 102 & 113 & 171 & 0 \\\93 & 0 & 93 & 110 & 1300\end{array}$$ Assuming the population is roughly symmetric, construct a \(95 \%\) confidence interval for the mean annual consumption of alcoholic beverages by European young women.

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In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a \(95 \%\) level of confidence? How large a sample should be taken for a \(99 \%\) level of confidence? Use a planning value for the population standard deviation of eight minutes.

A simple random sample of 400 individuals provides 100 Yes responses. a. What is the point estimate of the proportion of the population that would provide Yes responses? b. What is your estimate of the standard error of the proportion, \(\sigma_{\bar{p}} ?\) c. Compute the \(95 \%\) confidence interval for the population proportion.

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

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