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A sample of 352 subscribers to Wired magazine shows the mean time spent using the Internet is 13.4 hours per week, with a sample standard deviation of 6.8 hours. Find the 95 percent confidence interval for the mean time Wired subscribers spend on the Internet.

Short Answer

Expert verified
The 95% confidence interval is approximately (12.69, 14.11).

Step by step solution

01

Identify the Sample Statistics

To find the confidence interval, identify the sample mean \( \bar{x} = 13.4 \) hours per week, the sample standard deviation \( s = 6.8 \) hours, and the sample size \( n = 352 \).
02

Determine the Confidence Level and Corresponding Z-Score

We are looking for a 95% confidence interval. For this confidence level, the corresponding Z-score is approximately 1.96 (from the standard normal distribution table).
03

Compute the Standard Error

The standard error (SE) is calculated using the formula:\[ \text{SE} = \frac{s}{\sqrt{n}} \]Plugging in the values, we get\[ \text{SE} = \frac{6.8}{\sqrt{352}} \approx 0.3623 \]
04

Calculate the Margin of Error

The margin of error (ME) is given by:\[ \text{ME} = Z \times \text{SE} \]Using our Z-score of 1.96, the calculation becomes:\[ \text{ME} = 1.96 \times 0.3623 \approx 0.7101 \]
05

Determine the Confidence Interval

Compute the lower and upper bounds of the confidence interval using the formulas:- Lower bound: \( \bar{x} - \text{ME} = 13.4 - 0.7101 \approx 12.6899 \)- Upper bound: \( \bar{x} + \text{ME} = 13.4 + 0.7101 \approx 14.1101 \) Thus, the 95% confidence interval is approximately (12.69, 14.11).

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Key Concepts

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

Sample Statistics
Sample statistics are incredibly useful when we want to understand a larger population based on a smaller group.
In this exercise, we start by looking at a sample of Wired magazine subscribers.
We found that, on average, they spend 13.4 hours per week browsing the Internet. This average is called the sample mean and is denoted by \( \bar{x} \).

Additionally, we need to consider the sample's spread, measured by the standard deviation, which here is 6.8 hours.
This tells us how varied the internet usage times are within our sample group.
Finally, the sample size \( n \), which is 352 subscribers, helps us understand how reliable our sample mean might be as an estimate of the entire population's mean.
  • Sample Mean: 13.4 hours per week
  • Standard Deviation: 6.8 hours
  • Sample Size: 352
By focusing on sample statistics, we can make well-informed assumptions about the entire population.
Standard Error
The standard error (SE) is a key statistic for measuring the precision of a sample mean.
The SE gives us an indication of how much our sample mean might differ from the true population mean.

To calculate the standard error, we divide the sample standard deviation by the square root of the sample size: \[ \text{SE} = \frac{s}{\sqrt{n}} \]Substitute \( s = 6.8 \) and \( n = 352 \) to get: \[ \text{SE} = \frac{6.8}{\sqrt{352}} \approx 0.3623 \]A smaller SE indicates that the sample mean is a more accurate estimate of the population mean.
In our case, an SE of approximately 0.3623 shows a reasonably precise estimate.
  • Calculation: Showcases reliability of the sample mean
  • Indicates precision: Smaller SE means greater precision
This statistic allows us to confidently progress toward calculating the confidence interval.
Margin of Error
The margin of error (ME) quantifies the amount of random sampling error in our estimate of the population mean.
It helps create a range within which the true population mean is likely to fall.

We derive the margin of error by multiplying the Z-score, which reflects our desired confidence level, by the standard error:\[ \text{ME} = Z \times \text{SE} \]For a 95% confidence interval, the Z-score is approximately 1.96. Thus, using our earlier SE value of 0.3623: \[ \text{ME} = 1.96 \times 0.3623 \approx 0.7101 \]This margin of error tells us that our estimate of the mean internet usage time will typically be within 0.71 hours of the true mean.
  • Purpose: Defines precision level
  • Dependence: Relies on SE and Z-score
  • Outcome: Helps form a confidence interval
By knowing the margin of error, we can proceed to determine the full confidence interval.
Z-Score
The Z-score plays a crucial role in confidence intervals.
It represents how many standard deviations an element is from the mean.

For a 95% confidence level, a Z-score of 1.96 is used. This value reflects that we are looking to have 95% of the expected sample means within our range.
  • Z-score of 1.96: Matches a 95% confidence level
Z-scores are critical not just for confidence intervals, but for any analysis involving normal distribution, standardizing data comparisons, and hypothesis testing. They translate our desired confidence level into quantifiable numbers for further calculations.
Understanding the Z-score aids the calculation of the margin of error and, ultimately, the creation of a precise confidence interval.

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Most popular questions from this chapter

The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season is \(70 .\) A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the population mean. b. Does the 99 percent confidence interval include the value suggested by the NCAA? Interpret this result. c. Suppose you decided to switch from a 99 to a 95 percent confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?

You are to conduct a sample survey to determine the mean family income in a rural area of central Florida. The question is, how many families should be sampled? In a pilot sample of 10 families, the standard deviation of the sample was \(\$ 500 .\) The sponsor of the survey wants you to use the 95 percent confidence level. The estimate is to be within \(\$ 100 .\) How many families should be interviewed?

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 49 steady smokers revealed that \(\bar{X}=\$ 20\) and \(s=\$ 5\) a. What is the point estimate of the population mean? Explain what it indicates. b. Using the 95 percent level of confidence, determine the confidence interval for \(\mu .\) Explain what it indicates.

The attendance at the Savannah Colts minor league baseball game last night was \(400 .\) A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumed per person was 1.86 with a standard deviation of \(0.50 .\) Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 95 percent confidence level and state that the estimated proportion must be within 2 percent of the population proportion. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?

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