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Chapter 8: Problem 22

A random sample of \(n=100\) measurements has been selected from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=10 .\) Calculate the width of the confidence interval for \(\mu\) for the confidence levels given. What effect do the changing confidence levels have on the width of the interval? \(95 \%(\alpha=.05)\)

Short Answer

Expert verified
The width of the 95% confidence interval for the population mean 渭 is 3.92. Changing the confidence level affects the width of the interval in such a way that increasing the confidence level results in a larger interval width, whereas decreasing the confidence level results in a smaller interval width. This is because, as the confidence level increases, there is a greater probability of capturing the true population mean 渭 within the interval.

Step by step solution

01

Calculate the standard error

Given the standard deviation \(\sigma=10\) and sample size \(n=100\), we can calculate the standard error (SE) using the formula: \( SE = \frac{\sigma}{\sqrt{n}} \) \( SE = \frac{10}{\sqrt{100}} \) \( SE = \frac{10}{10} \) \( SE = 1 \)
02

Find the critical value for 95% confidence level

For a \(95\%\) confidence level, the \(\alpha = 0.05\). Thus, we want to find \(z_\frac{\alpha}{2}\) corresponding to \(\frac{\alpha}{2} = 0.025\). From a standard normal distribution table or calculator, we find the critical value \(z_\frac{\alpha}{2} = 1.96\).
03

Calculate the width of the 95% confidence interval

Now we can calculate the width of the confidence interval using the formula: \( Width = 2 * z_\frac{\alpha}{2} * SE \) \( Width = 2 * 1.96 * 1 \) \( Width = 3.92 \) The width of the \(95\%\) confidence interval for \(\mu\) is \(3.92\).
04

Analyzing the effect of changing confidence levels on the width of the interval

As the confidence level increases, \(\alpha\) decreases and \(z_\frac{\alpha}{2}\) increases. This means that the confidence interval width will also increase, as it directly depends on \(z_\frac{\alpha}{2}\). In other words, a higher confidence level requires a larger interval width to ensure a greater probability of capturing the true population mean \(\mu\). Conversely, a lower confidence level will have a smaller interval width.

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Key Concepts

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

Standard Error
Understanding the standard error (SE) is crucial when conducting statistical analysis. It measures the variability in your sample mean, indicating how much the sample mean deviates from the true population mean. In the exercise, we calculated the SE by dividing the population standard deviation, \(\sigma=10\), by the square root of the sample size, \(\sqrt{n}\), where the sample size was \(n=100\).To calculate the SE, we used the formula:\[ SE = \frac{\sigma}{\sqrt{n}} \]With \(\sigma=10\) and \(n=100\), the calculation simplified to \(SE = 1\). This low SE implies our sample mean is likely to be a good estimate of the population mean.
Critical Value
The critical value is a factor used to widen your confidence interval to a point where, based on your desired confidence level, you expect it to contain the true population mean. A 95% confidence level was used in our exercise, corresponding to an \(\alpha=0.05\) (5% significance level). We find the critical value, denoted by \(z_\frac{\alpha}{2}\), by looking at a z-table or using statistical software, which corresponds to the point on the standard normal distribution that has \(\frac{\alpha}{2}\) area in the tails.In our exercise, we found the critical value for a 95% confidence level to be 1.96. This value scales the SE to set the width of our confidence interval.
Sample Size
The sample size, denoted by \(n\), plays a key role in determining the precision of our confidence interval. A larger sample size provides a smaller standard error, leading to a narrower confidence interval. It signifies more precision in our estimation of the population mean.In the given problem, we worked with a sample size of \(n=100\). If we were to increase the sample size, we would see a decrease in the standard error and thus a smaller interval width, all else being constant. Smaller sample sizes, on the other hand, lead to greater variability and thereby wider intervals.
Population Mean
The population mean, represented by \(\mu\), is the average value of a variable in the population. It is what we aim to estimate using our sample. Since \(\mu\) is unknown in the exercise, we use the sample mean as an estimator. Confidence intervals provide a range within which we can confidently say the population mean lies.Our confidence interval is constructed around the sample mean and extends \(z_\frac{\alpha}{2} * SE\) in both directions. The actual value of \(\mu\) remains unknown but with the confidence interval, we estimate it with a stated level of confidence.
Standard Deviation
Standard deviation, symbolized as \(\sigma\), is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. In our case, the standard deviation of the population is known and is set at 10. This value is used to calculate the standard error, which is integral to finding the width of the confidence interval.The smaller the standard deviation of a given data set, the closer the sample means are to the population mean, making our estimation more precise.

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Most popular questions from this chapter

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. $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 500 & 500 \\\\\text { Number of Successes } & 120 & 147 \\\\\hline\end{array}$$

Will there be a time when hiring decisions will be done using a computer algorithm, and without any human involvement? A Pew Research survey found that \(67 \%\) of the adults surveyed were worried about the development of algorithms that can evaluate and hire job candidates. \({ }^{12}\) To check this claim, a random sample of \(n=100\) U.S. adults was selected, with 58 saying they were worried about this issue. a. Construct a \(98 \%\) confidence interval for the true proportion of adults that are worried about being "hired by a robot". b. Does the confidence interval constructed in part a confirm the claim of the Pew Research survey? Why or why not?

As Americans become more conscious of the importance of good nutrition, some researchers believe that we may be eating less red meat than we used to eat. To test this theory, a researcher selects two groups of 400 subjects each and collects the following sample information on the annual beef consumption now and 10 years ago: $$\begin{array}{lcc}\hline & \text { Ten Years Ago } & \text { This Year } \\\\\hline \text { Sample Mean } & 73 & 63 \\\\\text { Sample Standard Deviation } & 25 & 28 \\\\\hline\end{array}$$ a. The researcher would like to show that per-capita beef consumption has decreased in the last 10 years, so she needs to show that the difference in the averages is greater than 0 . Find a \(99 \%\) lower confidence bound for the difference in the average per-capita beef consumptions for the two groups. b. What conclusions can the researcher draw using the confidence bound from part a?

In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24 -hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error.

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 the difference between two means with a margin of error equal to 卤5 . Assume that the sample sizes will be equal and that \(\sigma_{1} \approx \sigma_{2} \approx 24.5\).
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