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Chapter 8: Problem 42

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 2000 seniors who pay for thein medicines showed that they spent an average of \(\$ 4600\) last year on medicines with a standard deviation of \(\$ 800\). Make a \(98 \%\) confidence interval for the corresponding population mean.

Short Answer

Expert verified
The 98% confidence interval for the population mean is calculated by following the steps above. It will give a range of values that is likely to include the actual population mean with 98% confidence.

Step by step solution

01

Find the standard error

Firstly, calculate the standard error (SE) of sample. The formula for standard error is \(SE = \frac{s}{\sqrt{n}}\), where s is the standard deviation of the sample and n is the sample size. Substituting the given values into the formula gives \[SE = \frac{800}{\sqrt{2000}}\]
02

Determine the Z-value

Next, determine the Z-value corresponding to a 98% confidence interval. This Z-value is a standard score that measures the number of standard deviations an element is from the mean in a standard normal distribution. For a 98% confidence interval, the Z-value is approximately 2.33.
03

Calculate the confidence interval

Finally, calculate the confidence interval using the formula \[\bar{x} \pm Z * SE\]Here, \(\bar{x}\) represents the sample mean, Z is the Z-value, and SE is the standard error. Substituting the given values gives \[4600 \pm 2.33 * SE\]. Simplify this to get the final confidence interval.

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Key Concepts

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

Standard Error
The concept of "standard error" is essential in statistics when estimating population parameters. To put it simply, the standard error (SE) measures how much the sample mean (\(\bar{x}\)) is expected to vary due to randomness in sampling. It gives us an idea of the precision of our sample mean estimate.
  • To calculate SE, use the formula: \[SE = \frac{s}{\sqrt{n}}\]where \(s\) is the standard deviation of the sample, and \(n\) represents the sample size.
  • The smaller the standard error, the more reliable our estimate of the population mean is likely to be.
In the exercise, the sample size \(n\) was 2000, and the standard deviation \(s\) was 800. Using the formula, we calculated:\[SE = \frac{800}{\sqrt{2000}}\].
This result helps us understand how spread out our sample mean might be if we were to take multiple samples.
Z-value
The Z-value is an integral part of calculating confidence intervals as it reflects how many standard deviations away from the mean the data point of interest falls within a standard normal distribution. For different confidence levels, different Z-values apply, indicating how wide or narrow the confidence interval should be.
  • In general, the Z-value is determined by the desired confidence level. For a 98% confidence interval, the Z-value is approximately 2.33.
  • This means that if you were to draw an infinite number of samples, about 98% of the sample means would fall within 卤2.33 standard errors of the true population mean.
Bringing this back to our exercise, the Z-value used was 2.33, as it corresponds to the 98% confidence interval we aimed to construct.
Sample Mean
When trying to understand what a population mean might be, it is common to take a sample and calculate the sample mean. The sample mean, symbolized as \(\bar{x}\), gives us an estimation of the average value of an entire population based on the collected data.
  • To find the sample mean, sum up all the data points in a sample and divide by the number of data points (sample size \(n\)).
  • The sample mean is used as an unbiased estimator of the population mean, meaning it accurately reflects the central tendency of the population.
In the provided exercise, the average expenditure on medicines by sampled seniors was given as \(\$4600\), and this is the sample mean used to help estimate the population mean of all seniors' expenditures on medicines.
Standard Deviation
Standard deviation is a fundamental concept in statistics, measuring the amount of variation or dispersion in a set of values. It is a key component in calculating the standard error and ultimately constructing confidence intervals.
  • A small standard deviation means data points tend to be close to the mean, while a large standard deviation indicates that the data is spread out over a wider range of values.
  • The standard deviation for a sample is calculated using: \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]where \(x_i\) represents each data point.
In this exercise example, the standard deviation was provided as \(\$800\). This measure helped us understand how expenses varied among seniors in the sample, influencing the calculation of our standard error.

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Most popular questions from this chapter

A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of \(\$ 1575\) per month for their mortgages. The population standard deviation of all such mortgages is \(\$ 215\). a. Find a \(97 \%\) confidence interval for the mean amount of mortgage paid per month by all homeowners in this area. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

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

A businesswoman is considering whether to open a coffee shop in a local shopping center. Before making this decision, she wants to know how much money, on average, people spend per week at coffee shops in that area. She took a random sample of 26 customers from the area who visit coffee shops and asked them to record the amount of money (in dollars) they would spend during the next week at coffee shops. At the end of the week, she obtained the following data (in dollars) from these 26 customers: \(\begin{array}{rrrrrrrrr}16.96 & 38.83 & 15.28 & 14.84 & 5.99 & 64.50 & 12.15 & 14.68 & 33.37 \\ 37.10 & 18.15 & 67.89 & 12.17 & 40.13 & 5.51 & 8.80 & 34.53 & 35.54 \\ 8.51 & 37.18 & 41.52 & 13.83 & 12.96 & 22.78 & 5.29 & 9.09 & \end{array}\) Assume that the distribution of weekly expenditures at coffee shops by all customers who visit coffee shops in this area is approximately normal. a. What is the point estimate of the corresponding population mean? b. Make a \(95 \%\) confidence interval for the average amount of money spent per week at coffee shops by all customers who visit coffee shops in this area.

A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of . 75 hour. Find a \(98 \%\) confidence interval for the population mean.

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 all such times is \(5.5\) minutes.
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