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In Exercises 6.81 to \(6.86,\) use the t-distribution to find a confidence interval for a mean \(\mu\) given the relevant sample results. Give the best point estimate for \(\mu,\) the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A \(95 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=12.7, s=5.6,\) and \(n=30\)

Short Answer

Expert verified
The 95% confidence interval for \( \mu \) using the sample results is \( CI = 12.7 \pm E \), where E is calculated as \( E = 2.045 * \frac{5.6}{\sqrt{30}} \).

Step by step solution

01

Define the Variables

In this problem, the following variables are defined: \( \bar{x} = 12.7 \) (sample mean),\( s = 5.6 \) (sample standard deviation) and \( n = 30 \) (sample size). We need to calculate the confidence interval for the population mean \( \mu \) using these variables.
02

Compute the Standard Error

The standard error of the sample mean(SE) is calculated using the formula: \( SE = \frac{s}{\sqrt{n}} \) where, \( s = 5.6 \) is the sample standard deviation and \( n = 30 \) is the sample size. Substituting these values into the formula, we find that the standard error (SE) is \( \frac{5.6}{\sqrt{30}} \).
03

Determine the T-Score

Using a t-distribution table or calculator, find the t-score associated with the desired confidence level and degrees of freedom \( df = n-1 = 30 - 1 = 29 \). For a 95% confidence level, the t-score (t) is approximately 2.045.
04

Calculate the Margin of Error

The margin of error (E) is calculated as \( E = t*SE \). Substituting our calculated values we get \( E = 2.045 * \frac{5.6}{\sqrt{30}} \).
05

Calculate Confidence Interval

To calculate the 95% confidence interval for \( \mu \), use the formula: \( CI = \bar{x} \pm E \). Substituting the given \( \bar{x} = 12.7 \) and the calculated E gives the range of values for the confidence interval.

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Key Concepts

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

Understanding t-Distribution
The t-distribution, also known as Student's t-distribution, is a type of probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown. Unlike the normal distribution which has a known standard deviation, the t-distribution takes into account additional uncertainty by having fatter tails. This means there's a greater chance of values further from the mean.

The t-distribution is symmetrical and bell-shaped, like the normal distribution, but has heavier tails. This heavier tail feature allows the t-distribution to accommodate the extra variation found in small sample sizes. As the sample size increases, the t-distribution approaches the normal distribution.

To use the t-distribution, one must be aware of the degrees of freedom (df), which in the context of estimating a mean with a confidence interval is typically one less than the sample size (\( n - 1 \)). Degrees of freedom are a crucial concept in statistics that account for the number of values in the final calculation of a statistic that are free to vary.
Calculating Standard Error
Standard error (SE) is a statistical measure that quantifies the amount by which a sample mean, such as sample mean \( \bar{x} \), could differ from the actual population mean \( \mu \). This measure reflects the variability of the sample means across different samples. The formula for standard error of the mean is: \( SE = \frac{s}{\sqrt{n}} \), where \( s \) represents the sample standard deviation and \( n \) is the sample size.

When you calculate the standard error, you're accounting for the fact that you're working with a sample, not the entire population. A larger standard error indicates that the sample mean is less likely to reflect the population mean accurately. The standard error diminishes as the sample size \( n \) increases, reflecting the increased precision that comes with larger samples.
Determining Margin of Error
The margin of error (E) quantifies the range within which the true population parameter, like a mean, is expected to fall. This range reflects the confidence in the sample's representation of the population. The margin of error is dependent on the standard error and the critical value from the selected probability distribution, such as the t-distribution for small samples. The general formula can be expressed as: \( E = t \times SE \), where \( t \) is the t-score at a given confidence level and \( SE \) is the standard error.

To compute the margin of error, you will need a calculated standard error and a critical value, which corresponds to the desired confidence level. In practice, for a 95% confidence level, it means that if you were to take many samples and build confidence intervals, approximately 95 out of 100 of those intervals would contain the true population mean. The margin of error, therefore, provides a way to express this certainty numerically and is used to generate a confidence interval around the sample mean.

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Most popular questions from this chapter

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 ±3 , if the estimated standard deviation is \(\tilde{\sigma}=100\). If the estimated standard deviation is \(\tilde{\sigma}=50\). If the estimated standard deviation is \(\tilde{\sigma}=10 .\) Comment on how the variability in the population influences the sample size needed to reach a desired level of accuracy.

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of \(\pm 2 \%\) with \(95 \%\) confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

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?

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To measure the effectiveness of a new teaching method for math in elementary school, each student in a class getting the new instructional method is matched with a student in a separate class on \(\mathrm{IQ}\), family income, math ability level the previous year, reading level, and all demographic characteristics. At the end of the year, math ability levels are measured again.

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