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Find the z-values needed to calculate large-sample confidence intervals for the confidence levels given. A \(95 \%\) confidence interval

Short Answer

Expert verified
Answer: The z-value needed to calculate a 95% confidence interval for a large-sample estimate is 1.96.

Step by step solution

01

Identify the confidence level

The given confidence level is \(95\%\). This means that we want to find the z-value such that \(95\%\) of the area under the normal curve is between -\(z\) and \(z\).
02

Calculate the area in each tail

Since we are calculating a two-sided confidence interval, we need to divide the remaining area outside of the interval (100% - 95%) by 2 to find the area within each tail. The area in each tail can be calculated as follows: \(\frac{100\% - 95\%}{2} = \frac{5\%}{2} = 2.5\%\)
03

Find the z-value using a standard normal table

We will use a standard normal table to find the z-value corresponding to the cumulative probability of \(1 - 2.5\% = 97.5\%\). In the table, look for the closest value to \(0.975\) and identify its corresponding z-score. The closest value in the table is usually \(0.9750\) or something else extremely close, and the corresponding z-score is \(1.96\).
04

Interpret the z-value

The z-value of \(1.96\) represents the number of standard deviations away from the mean to capture \(95\%\) of the area under the normal curve. This means that a \(95\%\) confidence interval for a large-sample estimate will extend \(1.96\) standard deviations away from the sample mean in both directions.

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Key Concepts

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

Z-Value
The z-value plays an essential role in statistical analysis, particularly when it comes to understanding the concept of confidence intervals. In basic terms, the z-value is a measure of how many standard deviations away an element is from the mean. When constructing a confidence interval, the z-value determines how far we extend from the mean to include a certain percentage of the data under the standard normal curve.

For instance, if we talk about a 95% confidence interval, we would want to find the z-values that capture 95% of the data within that interval centered around the mean. This is achieved by calculating the amount of data falling into the tails (the extreme ends of the curve) and identifying the z-value that corresponds to the cut-off point where this tail begins. In practical terms, this is often done using a z-table or statistical software to find the exact z-value, which, for a 95% confidence level, happens to be approximately 1.96.
Standard Normal Distribution
The standard normal distribution, which is sometimes referred to as the Z-distribution, is a special case of the normal distribution with a mean of zero and a standard deviation of one. It's a foundational concept in statistics that allows us to compare scores from different normal distributions or to standardize a single distribution to simplify calculations.

When we talk about z-values, we are actually referring to a position on the standard normal distribution curve. Because this curve is symmetrical, we can easily use it to determine the proportion of data within a certain number of standard deviations from the mean. This property is particularly useful for finding confidence intervals, as it allows us to describe the expected variability in terms of the percentage of data within these intervals. The area under the standard normal curve to the left of a z-value corresponds to the cumulative probability linked to that z-value.
Statistical Significance
Statistical significance is a determination of whether an observed effect is likely due to chance or some factor of interest. When we perform hypothesis testing, for example, we are essentially testing whether the results we observe could be due to random fluctuations or if they are significant enough to be attributed to the specific factor we are studying.

To determine statistical significance, we set a threshold known as the significance level, often denoted by \( \alpha \). This level represents the probability of rejecting the null hypothesis if it is true—basically, it's the risk we are willing to take to incorrectly claim a significant effect. A commonly used significance level is 5%, which corresponds to a 95% confidence interval. If our test statistic, such as a z-value, falls outside the range defined by this interval, we declare the finding statistically significant.
Confidence Level
The confidence level is a measurement of certainty or how confident we can be in the process of estimating a population parameter based on a sample statistic. It quantifies the idea that if we were to take many samples and calculate many confidence intervals, a specific percentage of these intervals would contain the true population parameter.

For instance, a 95% confidence level implies that if we were to take 100 random samples and compute a confidence interval for each sample, we would expect about 95 of these intervals to include the true population parameter. It's important to remember that the confidence level does not reflect the probability of a particular interval containing the parameter; rather, it describes a long-term proportion of intervals that will capture the parameter. Therefore, when you see a 95% confidence interval in a statistical analysis, it provides a range for the estimate that is supported by data with a certain level of confidence, based on the chosen confidence level.

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Most popular questions from this chapter

An entomologist wishes to estimate the average development time of the citrus red mite, a small spider-like insect that causes damage to leaves and fruit, correct to within . 5 day. From previous experiments it is known that \(\sigma\) is approximately 4 days. How large a sample should the entomologist take to be \(95 \%\) confident of her estimate?

Refer to Exercise 18 (Section 8.5). The percentage of people catching a cold when exposed to a cold virus is shown in the table, for a group of people with only a few social contacts and a group with six or more activities. \(^{21}\) $$\begin{array}{lcc}\hline & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \% \\\\\hline\end{array}$$ Construct a \(95 \%\) lower confidence bound for the difference in population proportions. Does it appear that when exposed to a cold virus, a greater proportion of those with fewer social contacts contracted a cold? Is this counter-intuitive?

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(11-12 .\) Construct a \(98 \%\) lower confidence bound for the difference in the population proportions. $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 800 & 640 \\\\\text { Number of Successes } & 337 & 374 \\\\\hline\end{array}$$

Use the information given to find the necessary confidence interval for the binomial proportion \(p .\) Interpret the interval that you have constructed. A \(95 \%\) confidence interval for \(p,\) based on a random sample of 500 trials of a binomial experiment which produced 27 successes.

Catching a Cold Do well-rounded people get fewer colds? A study in the Chronicle of Higher Education found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities. \({ }^{21}\) Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: \begin{tabular}{lcc} \hline & Few Social Outlets & Many Social Outlets \\ \hline Sample Size & 96 & 105 \\ Percent with Colds & \(62 \%\) & \(35 \%\) \\ \hline \end{tabular} a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

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