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91Ó°ÊÓ

A survey of 500 randomly selected adult men showed that the mean time they spend per week watching sports on television is \(9.75\) hours with a standard deviation of \(2.2\) hours. Construct a \(90 \%\) confidence interval for the population mean, \(\mu\).

Short Answer

Expert verified
The 90% confidence interval for the population mean is \(sample mean - E to sample mean + E).

Step by step solution

01

Compute the Standard Error

The standard error (SE) can be calculated using the formula: SE = \(s/\sqrt{n}\), where \(s\) is the sample standard deviation and \(n\) is the sample size. So, SE = \(2.2/\sqrt{500}\).
02

Look Up the Z-Score

For a 90% confidence interval, we need to look up the Z-score that corresponds to the area to the right of 0.05 (since the total area to the right of the mean in a standard normal distribution is 0.5, and we want 5% on each side of our distribution for a 90% confidence interval in the middle). Using a standard normal distribution table or a Z-score calculator, we find that Z = 1.645.
03

Compute the Margin of Error

The margin of error (E) can be calculated using the formula: E = Z * SE, where Z is the Z-score and SE is the standard error. So, E = 1.645 * SE.
04

Construct the Confidence Interval

The final step is to construct the confidence interval for the population mean. We have the lower and upper limits calculated as, lower limit = sample mean - E and upper limit = sample mean + E. These two values give the lower and upper bounds of our 90% confidence interval.

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Key Concepts

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

Standard Error
The standard error (SE) is a key statistical concept used to measure the precision of the sample mean as an estimate of the population mean.
It tells us how much variability we can expect between different samples drawn from the same population.
The standard error is calculated using the formula:
  • SE = \(\frac{s}{\sqrt{n}}\)
Where:
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
By dividing the standard deviation by the square root of the sample size, we adjust for larger sample sizes, acknowledging that more data will tend to give a more precise estimate of the population mean.
A smaller standard error indicates that the sample mean is a more accurate reflection of the population mean.
Z-Score
A Z-Score is a statistical measure that describes a value's position relative to the mean of a group of values.
In simpler terms, it tells us how many standard deviations away a data point is from the mean.
For constructing confidence intervals, the Z-score corresponds to the amount of data within a certain level of confidence. In our scenario, a 90% confidence interval leaves 5% in each tail of a standard normal distribution.
This is because 100% minus 90% leaves 10%, with 5% in each tail.
To find the Z-score, you can consult a Z-table or use a calculator, where for 90% confidence, the Z-score is typically 1.645.
This Z-score helps in determining the range in which we expect to find the true population mean with 90% confidence.
Margin of Error
The margin of error (E) gives us an idea of how accurate our confidence interval is, or how much we are likely to be wrong.
It is computed as follows:
  • E = Z * SE
Where:
  • Z is the Z-score found for the desired confidence level.
  • SE is the standard error of the sample mean.
The margin of error reflects uncertainty, showing how far off the estimate of the population mean might be.
A smaller margin of error indicates a more reliable confidence interval.
It essentially adds and subtracts around the sample mean to establish the interval lower and upper bounds.
Population Mean
The population mean (\( \mu \)) is the average of a set of characteristics for the entire population we are interested in.
However, since calculating the population mean directly is often challenging due to the size of populations, we use sample data to estimate it.In a survey like ours, the sample mean provides an estimate of the population mean.
By using confidence intervals, we get a range that is likely to contain the true population mean.
With a 90% confidence interval, there is a 90% chance the interval includes the true population mean.
This means we accept a 10% risk of the interval not capturing \( \mu \). This statistical estimation ensures that conclusions drawn about population means are backed by a degree of certainty.

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Most popular questions from this chapter

Determine the sample size for the estimation of the population proportion for the following, where \(\hat{p}\) is the sample proportion based on a preliminary sample. a. \(E=.025, \hat{p}=.16, \quad\) confidence level \(=99 \%\) b. \(E=.05, \quad \hat{p}=.85, \quad\) confidence level \(=95 \%\) c. \(E=.015, \quad \hat{p}=.97, \quad\) confidence level \(=90 \%\)

A sample selected from a population gave a sample proportion equal to .31. a. Make a \(95 \%\) confidence interval for \(p\) assuming \(n=1200\). b. Construct a \(95 \%\) confidence interval for \(p\) assuming \(n=500\). c. Make a \(95 \%\) confidence interval for \(p\) assuming \(n=80\). d. Does the width of the confidence intervals constructed in parts a through \(c\) increase as the sample size decreases? If yes, explain why.

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

The U.S. Senate just passed a bill by a vote of \(55-45\) (with all 100 senators voting). A student who took an elementary statistics course last semester says, "We can use these data to make a confidence interval about \(p\). We have \(n=100\) and \(\hat{p}=55 / 100=.55\) " Hence, according to him, a \(95 \%\) confidence interval for \(p\) is $$ \hat{p} \pm z \sigma_{\hat{p}}=.55 \pm 1.96 \sqrt{\frac{(.55)(.45)}{100}}=.55 \pm .098=.452 \text { to } .648 $$ Does this make sense? If not, what is wrong with the student's reasoning?

When calculating a confidence interval for the population mean \(\mu\) with a known population standard deviation \(\sigma\), describe the effects of the following two changes on the confidence interval: (1) doubling the sample size, (2) quadrupling (multiplying by 4) the sample size. Give two reasons why this relationship does not hold true if you are calculating a confidence interval for the population mean \(\mu\) with an unknown population standard deviation.

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