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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

Radio and television stations often air controversial issues during broadcast time. A poll is then conducted, asking viewers who agree with the issue to call a certain 900 telephone number and those who disagree to call a second 900 telephone number. All respondents pay a fee for their calls. a. Does this polling technique result in a random sample? b. What can be said about the validity of the results of such a survey? Do you need to worry about a margin of error in this case?

Chicken Feed An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true dif ference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?

Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating \(\mu\) to within 1.6 with probability .95. Prior experience suggests that \(\sigma=12.7\).

To determine whether there is a significant difference in the average weights of boys and girls beginning kindergarten, random samples of 50 5-year-old boys and 50 5-year-old girls produced the following information. \(^{17}\) $$\begin{array}{llcc}\hline & & \text { Standard } & \text { Sample } \\\& \text { Mean } & \text { Deviation } & \text { Size } \\\\\hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\\\\text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \\\\\hline\end{array}$$ Find a \(99 \%\) lower confidence bound for the difference in the average weights of 5 -year-old boys and girls. Can you conclude that on average 5 -year-old boys weigh more than 5-year-old girls?

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?

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