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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 \(99 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=88.3, s=32.1,\) and \(n=15\)

Short Answer

Expert verified
The best point estimate for \(\mu\) is 88.3. The margin of error and the 99% confidence interval for \(\mu\) are obtained using the t-distribution, which should be computed as per the steps outlined. To give the exact number, you need the t-value, which is obtained from a t-distribution table, and this missing information prevents the completion of the calculation for the current task.

Step by step solution

01

Calculate Degrees of Freedom

Degrees of freedom is the number of values in the final calculation that are free to vary. It help to adjust the calculation for the sample size. It can be calculated as follows: Degrees of Freedom (df) = n - 1 = 15 - 1 = 14
02

Find the Critical Value

Because the problem describes a 99% confidence interval, the level of significance (\(\alpha\)) is 1% or 0.01. This is two-tailed, so \(\alpha/2 = 0.005\). Look up this value in the t-table with 14 degrees of freedom and find the critical t-value. Let's denote that value as t*.
03

Calculate the Standard Error

Standard error is a measure of the amount of variability in the sample mean, and it's calculated by dividing the standard deviation by the square root of the sample size, like so: Standard Error (SE) = s/√n = 32.1/√15
04

Calculate the Margin of Error

Margin of error helps us quantify the possible range of the actual mean. It is calculated by multiplying the standard error with the critical t-value determined earlier: Margin of Error (ME) = t* x SE
05

Compute the Confidence Interval

To form the 99% confidence interval, subtract and add the calculated margin of error from/to the sample mean (the best point estimate for \(\mu\)): Confidence Interval = \(\bar{x}\) ± ME = 88.3 ± ME

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Key Concepts

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

t-distribution
When it comes to understanding the t-distribution, it is essential to grasp that it is a type of probability distribution. It closely resembles the normal distribution but has heavier tails, meaning it is more prone to producing values that fall far from its mean. This is particularly useful when dealing with small sample sizes (<30), as it adjusts for the added uncertainty.

The t-distribution is used to estimate population parameters, such as the mean ( mu), when the standard deviation is unknown and the sample size is small. Notably, the shape of the t-distribution changes depending on the degrees of freedom. As a rule of thumb, the fewer the degrees of freedom, the heavier the tails of the distribution. And as the sample size grows, the t-distribution approaches the normal distribution.
Degrees of Freedom
Degrees of freedom (df) are a concept in statistics that represent the number of independent values or quantities that can vary in an analysis without breaking any constraints. It is a foundational element when using the t-distribution. For estimating a single mean, the degrees of freedom are calculated as the sample size ( n) minus one, effectively the number of values that are free to vary while the mean remains fixed.

For the given exercise, with a sample size of 15 individuals, the degrees of freedom is 14 ( n - 1). When consulting a t-table or using statistical software to find critical values, it's vital to use the correct degrees of freedom to ensure accurate confidence intervals.
Standard Error
The standard error (SE) is a measure that describes the fluctuation of the sample mean around the population mean. It basically tells us how far the sample mean is likely to be from the true population mean. The formula, SE = s/√n, where s is the sample standard deviation, and n is the sample size, is used to compute the standard error of the mean.

In the given example, the standard deviation is 32.1, and the sample size is 15, leading to an SE calculation. This value is critical in constructing the confidence interval, as it influences the calculated margin of error.
Margin of Error
The margin of error quantifies the amount by which the estimated value could deviate from the true value. It expresses the range within which we can expect the true population parameter to lie with a certain level of confidence. The margin of error (ME) is calculated by multiplying the standard error by a critical value from the t-distribution, reflecting the desired confidence level.

In the context of the exercise, after calculating the standard error, the next step is to decide the critical t-value related to a 99% confidence level, and then use these two figures to derive the margin of error. This margin is then applied to the sample mean to create the confidence interval.
Critical Value
A critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis, and on the flip side, within which we fail to reject it. For a given confidence interval, the critical value corresponds to the degree of confidence one wishes to have in the interval estimate.

In a t-distribution, the critical value is determined based on the confidence level (like 95% or 99%) and the degrees of freedom. For the 99% confidence interval in this exercise, the critical value is found using 14 degrees of freedom (associated with the sample size) and corresponds to the tails (2.5% on each side in a two-tailed test) of the distribution. This value is crucial as it gauges the extent of the confidence interval around the sample mean.

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Most popular questions from this chapter

A survey \(^{4}\) of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online. (a) Find and interpret a \(90 \%\) confidence interval for the proportion, \(p\), of all US teens who have made a new friend online. (b) Give the best estimate for \(p\) and give the margin of error for the estimate. (c) Use the interval to determine whether we can be \(90 \%\) confident that more than half of US teens have made a new friend online.

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\) A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table:. $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

In Exercises 6.9 and 6.10 , indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, is the sample size large enough so that the sample proportions follow a normal distribution? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

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