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91Ó°ÊÓ

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 a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

Short Answer

Expert verified
The 99% confidence interval for the population mean \(\mu\) of life insurance policy premiums is \((654.76, 715.24)\)$.

Step by step solution

01

Identify Given Information

The sample size (n) is 25, while the sample mean (\(\bar{x}\)) is $685 and the standard deviation (s) is $74. The confidence level is 99%.
02

Calculate the Standard Error

The standard error (SE) can be calculated with the formula \(SE = \frac{s}{\sqrt{n}}\). So, \(SE = \frac{74}{\sqrt{25}} = 14.8\)
03

Determine the Z-Score for Desired Confidence Level

For a 99% confidence level, the Z-Score (Z) is approximately 2.58 based on a z-table or online z-value calculator.
04

Calculate Confidence Interval

The formula for calculating confidence interval is \(\bar{x} \pm Z*SE\). Therefore, the confidence interval = 685 \pm 2.58 * 14.8.
05

Solve for Confidence Interval Range

Solving the equation gives us the range for the confidence interval: (654.76, 715.24).

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Key Concepts

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

Population Mean
In statistics, the population mean is a critical concept to understand. The population mean, often represented by the Greek letter \( \mu \), refers to the average of a set of values for an entire population. In the context of our life insurance problem, if we had data for every policyholder, the population mean would indicate the average premium paid across all policyholders.

However, it is usually impractical to collect data from an entire population, especially if the population is large. Therefore, we use a sample, which is a smaller, manageable subset of the population, to estimate the population mean. Using the sample mean \( \bar{x} \) (which is \( 685 \) in our problem), we can estimate \( \mu \) and provide insights into the population's characteristics. Recall, even though the sample mean can provide an estimate of the population mean, they aren't necessarily the same unless the entire population is surveyed.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In our exercise, the standard deviation is \( 74 \). This number tells us how much the annual premiums of the life insurance policyholders vary from the average premium within our sample. The higher the standard deviation, the more spread out the values are in the dataset.

Why is understanding standard deviation important? It provides valuable insight into the risk associated with data variation. In financial contexts, like with insurance premiums, knowing the standard deviation helps insurers understand the range within which most premium amounts fall. This information is crucial when insurers set policy terms and determine coverage levels.
  • Helps in assessing variability.
  • Indicates the data's distribution around the mean.
  • Used to calculate other statistical measures, like the standard error in confidence intervals.
Normal Distribution
A normal distribution is a common way to represent data. It's defined by its bell-shaped curve, with most of the observations clustering around the central peak—and symmetrically tapering off toward both ends. In our exercise, we're assuming the life insurance premiums follow a normal distribution.

Understanding normal distribution is pivotal because it allows us to apply statistical tools like the confidence interval. Many statistical analyses assume data is normally distributed because this form of distribution enables simpler predictions and inferences about the population.
  • Mean, median, and mode of a normal distribution are equal.
  • Empirical rule: 68% of data falls within one standard deviation, 95% within two, and 99.7% within three in a normally distributed dataset.
  • Normal distribution simplifies the calculations of statistical properties.
Z-Score
A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. In our exercise, the Z-score is used in calculating confidence intervals to account for the level of confidence we desire in our estimates.

In context, a Z-score helps us understand how far from the mean our sample mean falls, standardized to account for the sample size and variability. For a 99% confidence level, a Z-score of 2.58 tells us that we're looking for an interval where the true population mean falls within these standardized boundaries in 99% of cases.
  • A high absolute Z-score indicates an observation is far away from the mean.
  • Critical for confidence interval calculations to determine reliability.
  • Z-tables or calculators provide Z-scores corresponding to different confidence levels.

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Most popular questions from this chapter

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the amounts of milk in all such cartons is always equal to \(.15\) ounce. The quality control department takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(32.15\) ounces or the lower limit of this confidence interval is less than \(31.85\) ounces, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.94\) ounces. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the amounts of milk put in all such cartons have a normal distribution.

a. A sample of 100 observations taken from a population produced a sample mean equal to \(55.32\) and a standard deviation equal to \(8.4 .\) Make a \(90 \%\) confidence interval for \(\mu .\) b. Another sample of 100 observations taken from the same population produced a sample mean equal to \(57.40\) and a standard deviation equal to \(7.5 .\) Make a \(90 \%\) confidence interval for \(\mu .\) c. A third sample of 100 observations taken from the same population produced a sample mean equal to \(56.25\) and a standard deviation equal to \(7.9 .\) Make a \(90 \%\) confidence interval for \(\mu .\) d. The true population mean for this population is \(55.80 .\) Which of the confidence intervals constructed in parts a through c cover this population mean and which do not?

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a \(99 \%\) confidence level. Assume that the population standard deviation for such times is \(5.5\) minutes.

A sample of 20 managers was taken, and they were asked whether or not they usually take work home. The responses of these managers are given below, where yes indicates they usually take work home and no means they do not. \(\begin{array}{lllllllll}\text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { No } \\ \text { Yes } & \text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { No } & \text { Yes }\end{array}\) Make a \(99 \%\) confidence interval for the percentage of all managers who take work home.

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