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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 99% confidence interval for \(\mu\) with given sample results is approximately \(88.3 \pm 2.576 \times \frac{32.1}{\sqrt{15}}\). Remember to evaluate this expression to get the final answer for the confidence interval.

Step by step solution

01

Find the Z score corresponding to the given confidence level

Using a standard normal distribution table, or Z-table, find the alpha level (100 - the given confidence level) and find the 'Z' value that corresponds to this alpha level. Given a 99% confidence interval, the alpha level is 1% divided by 2 (two tails of the distribution), or 0.005. A 'Z' score that corresponds to an alpha level of 0.005 is approximately 2.576.
02

Calculate the standard error

The standard error can be calculated by dividing the standard deviation of the sample by the square root of the size of the sample. Substituting the given values, \(SE = \frac{s}{\sqrt{n}} = \frac{32.1}{\sqrt{15}}\).
03

Find the confidence interval

The confidence interval is found by multiplying the standard error by the 'Z' score and then subtracting this value from and adding it to the sample mean. The formula is \(\mu = \bar{x} \pm Z(SE)\). Substituting the given sample mean and the calculated standard error and 'Z' score, we find the solution.

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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 statistical term that measures the accuracy with which a sample represents a population. In other words, it quantifies the variability of the sampling distribution of a statistic, most commonly of the mean. The formula to calculate the standard error is:
\[\begin{equation}SE = \frac{\sigma}{\sqrt{n}}\end{equation}\]
where \(\sigma\) is the standard deviation of the population and \(n\) is the sample size. When the population standard deviation is unknown, as is often the case, the sample standard deviation (denoted as \(s\)) is used instead, yielding the formula:
\[\begin{equation}SE = \frac{s}{\sqrt{n}}\end{equation}\]
In the given exercise, the standard error is calculated using this formula with the provided sample standard deviation and size, which helps to determine how much the sample mean \(\bar{x}\) is expected to fluctuate from the true population mean \(\mu\).
Z-score
The Z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean of a distribution. It is a pivotal concept when working with normal distributions. The Z-score can be calculated by the formula:
\[\begin{equation}Z = \frac{(X - \mu)}{\sigma}\end{equation}\]
where \(X\) is the value from the distribution, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. In the context of confidence intervals, the Z-score provides a critical value that determines the range around the sample mean within which the true population mean is likely to lie with a given level of confidence. Higher confidence levels correspond to larger Z-scores. For our example, a 99% confidence level leads to a Z-score of approximately 2.576, indicating that the interval extends 2.576 standard errors from the sample mean on both sides to capture the true population mean.
Normal Distribution
The normal distribution, commonly associated with the bell curve, is a continuous probability distribution that is symmetrical around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In real-life situations, the normal distribution can be used to describe characteristics such as test scores, measurement errors, and many other phenomena.When constructing confidence intervals, the normal distribution assumes that the means of the samples follow a normal distribution around the true population mean. This allows us to make probabilistic statements about where the true mean lies based on our sample data. A 99% confidence interval indicates that we are 99% certain that the true population mean falls within our calculated range. The

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Most popular questions from this chapter

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cancer of any kind? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{N}-\bar{x}_{E},\) where \(\bar{x}_{N}\) represents the mean Mathematics score for a sample of 100 people for whom the native language is not English and \(\bar{x}_{E}\) represents the mean Mathematics score for a sample of 100 people whose native language is English, is centered at 10 with a standard deviation of 17.41 . Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, non-native English speakers or native English speakers? Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.224 and \(6.225,\) use StatKey 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.

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from a population with proportion 0.08

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the mean Critical Reading score for a sample of 50 males and \(\bar{x}_{f}\) represents the mean Critical Reading score for a sample of 50 females, is centered at 5 with a standard deviation of \(22.5 .\) Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, males or females?

A sample with \(n=18, \bar{x}=87.9,\) and \(s=10.6\)

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