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A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

Short Answer

Expert verified
For a 95% confidence interval, use the z-value of 1.96, substitute it into the Margin of Error formula, and then use this value to calculate the confidence interval using the formula: \(\bar{x} \pm E\).

Step by step solution

01

Understand the Statistical Terms

Here, \(n\) is the sample size, or the number of observations. The term \(\bar{x}\) represents the sample mean, or the average of all metrics. And \(s\) is the standard deviation, which is a measure of how much variance there is in a set of numbers compared to the average (mean) of the numbers.
02

Calculate the Margin of Error

The margin of error (\(E\)) for a sample mean is calculated by using the formula \(E = z \times \frac{s}{\sqrt{n}}\), where \(z\) is the value from the z-table, which corresponds to the desired confidence level. Let's say we want a 95% confidence interval, the z-value is 1.96.
03

Calculate the Confidence Interval

The confidence interval for the mean can be calculated using the formula: \(\bar{x} \pm E\). Substituting the given values and the calculated margin of error into this formula gives the confidence interval.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that is likely to contain the true population mean. It's like a margin of error that tells us how much we can expect the actual value to differ from our observed sample mean. Imagine it as a net that tries to capture the real average by giving us upper and lower bounds based on our sample data.

  • The confidence level (like 95% or 99%) indicates how sure we are that the interval contains the true mean.
  • A higher confidence level will give us a wider interval because we want to be more certain.
The process involves calculating the margin of error using a z-score, which corresponds to the chosen confidence level. Once we find the margin of error, the confidence interval is simply the sample mean plus and minus this error. For example, if the sample mean is 508.5 and the margin of error is found to be 13, your confidence interval would be from 495.5 to 521.5.
Sample Mean
The sample mean, represented as \(\bar{x}\), is the average of all data points in a sample. It is a central value that summarizes the data well. When you have a set of observations, you simply add them up and divide by the total number of observations, \(n\).

  • Formula: \(\bar{x} = \frac{\sum x_i}{n}\)
  • In our given example, \(\bar{x}\) is 508.5
The sample mean is crucial because it serves as the best estimate of the population mean, assuming our sample size is representative. However, remember that it is still just an approximation, as we only have a subset of the entire population. With larger sample sizes, this estimate tends to be more accurate.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A small standard deviation means that the values tend to be close to the mean of the set, while a large standard deviation means the values are spread out over a wider range.

  • It is denoted by \(s\) in the sample standard deviation.
  • Calculated using the formula: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\)
  • In our example, the standard deviation is 21.5
Standard deviation is important in the context of statistical inference because it affects the width of the confidence interval. It tells us how much we can expect our data to vary and, thus, how precise our estimate of the mean is. In summary, understanding this variability helps statisticians make more informed conclusions about their data.

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Most popular questions from this chapter

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

In Exercise 6.93 on page \(430,\) we see that the average number of close confidants in a random sample of 2006 US adults is 2.2 with a standard deviation of \(1.4 .\) If we want to estimate the number of close confidants with a margin of error within ±0.05 and with \(99 \%\) confidence, how large a sample is needed?

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

In Exercises 6.109 to 6.111 , we examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within ±10 . Within ±5 . Within ±1 . Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the margin of error.

Exercises 6.192 and 6.193 examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DEHP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=83.6\) with \(s_{F}=194.7\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=59.1\) with \(s_{N}=152.1\)

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