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A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have an approximate normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

Short Answer

Expert verified
The \(99 %\) confidence interval for the population mean of life insurance policy premiums is \( \$ 685 \pm ME \), where ME is the calculated margin of error.

Step by step solution

01

Identifying the values from the problem

First, identify the needed values from the exercise: the sample mean \( x \) is $685, the sample standard deviation \( s \) is $74, the sample size \( n \) is 25, and the desired confidence level is 99 %.
02

Calculating the standard error

Next, calculate the standard error (SE) of the sample. The standard error is given by the formula \( SE = \frac{s}{\sqrt{n}} \), or in this case, \( SE = \frac{74}{\sqrt{25}} \). Calculate this to get the standard error.
03

Finding the Z-score

Find the Z-score that corresponds to the desired 99% confidence level. For a 99% confidence interval, the Z-score is typically ±2.576.
04

Calculating the Margin of Error

Calculate the margin of error (ME) for the given confidence interval. The margin of error is calculated by multiplying the Z-score by the standard error, or \( ME = Z \times SE \).
05

Calculating the Confidence Interval

Finally, the 99% confidence interval for the population mean is derived by adding and subtracting the margin of error from the sample mean. So, the confidence interval is \( CI = x \pm ME \), or in this case, \( CI = 685 \pm ME \).

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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 crucial statistic in understanding the precision of sample data in estimating the population parameter. It essentially measures how spread out the sample mean is likely to be from the actual population mean. In this context, the standard error is calculated using the formula: - \( SE = \frac{s}{\sqrt{n}} \) where: - \( s \) is the sample standard deviation - \( n \) is the sample size
  • For our exercise, the standard deviation is $74.
  • The sample size is 25 policyholders.
Thus, the SE guides us in understanding that smaller standard errors indicate more reliable and precise estimates of the population mean. A lower SE suggests that the sample mean is a good estimate of the population mean. While a higher SE means more variability in your estimate, implying more uncertainty.
Sample Mean
The sample mean is the average value calculated from the sample data. It serves as a point estimate of the population mean. The sample mean acts as our best guess of the population mean based on the data we have. In the exercise given, the sample mean is:- \( \bar{x} = 685 \)
  • This means the average premium paid by the sampled life insurance policyholders is $685.
Understanding the sample mean is fundamental, as it provides a single summary value from the entire dataset, encapsulating a central measure of tendency. Hence, it functions as a backbone for further statistical calculations, like constructing confidence intervals.
Z-score
A Z-score is a statistical measurement that explains how many standard deviations an element is from the mean. In creating confidence intervals, the Z-score tells us the multiplier of the standard error that is needed to achieve a certain level of confidence.
  • For a 99% confidence level, the Z-score is typically ±2.576.
This means we expect 99% of sample means to fall within values that are ±2.576 standard errors from the population mean. Thus, understanding Z-scores allows you to contextualize your confidence interval and the certainty of your sample estimate.
Margin of Error
The margin of error (ME) quantifies the uncertainty and ensures that the actual population parameter lies within this limit around the sample statistic. It is key in forming a confidence interval. Calculating the margin of error involves multiplying the Z-score with the standard error:- \( ME = Z \times SE \)
  • Here, with an SE calculated from the given data and a Z-score from statistical tables, you can find the margin of error.
The resulting ME reflects the potential range of uncertainty around the sample mean, implying that the true population mean likely falls within this zone. The smaller the margin, the more confident you can be in your estimates, representing less variability.

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Most popular questions from this chapter

The management of a health insurance company wants to know the percentage of its policyholders who have tried alternative treatments (such as acupuncture, herbal therapy, etc.). A random sample of 24 of the company's policyholders were asked whether or not they have ever tried such treatments. The following are their responses. \(\begin{array}{lllllll}\text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } & \text { No } \\ \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } \\ \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No }\end{array}\) a. What is the point estimate of the corresponding population proportion? b. Construct a \(99 \%\) confidence interval for the percentage of this company's policyholders who have tried alternative treatments.

You are working for a bank. The bank manager wants to know the mean waiting time for all customers who visit this bank. She has asked you to estimate this mean by taking a sample. Briefly explain how you will conduct this study. Collect data on the waiting times for 45 customers who visit a bank. Then estimate the population mean. Choose your own confidence level.

Yunan Corporation produces bolts that are supplied to other companies. These bolts are supposed to be 4 inches long. The machine that makes these bolts does not produce each bolt exactly 4 inches long but the length of each bolt varies slightly. It is known that when the machine is working properly, the mean length of the bolts made on this machine is 4 inches. The standard deviation of the lengths of all bolts produced on this machine is always equal to \(.04\) inch. The quality control department takes a random sample of 20 such bolts every week, calculates the mean length of these bolts, and makes a \(98 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(4.02\) inches or the lower limit of this confidence interval is less than \(3.98\) inches, the machine is stopped and adjusted. A recent such sample of 20 bolts produced a mean length of \(3.99\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the population distribution is approximately normal.

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10 .\) b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025 .\) c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

Briefly explain the difference between a confidence level and a confidence interval.

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