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Chapter 9: Problem 9

Sale of Air Conditioners (Example 1) The average number of air conditioners sold in 2015 was 3600 per day in a city, and that was larger than the average for any other appliance. Suppose the standard deviation is 1404 and the distribution is right-skewed. Suppose we take a random sample of 81 days in the year. a. What value should we expect for the sample mean? Why? b. What is the standard error for the sample mean?

Short Answer

Expert verified
The expected value for the sample mean is \(3600\) air conditioners per day. The standard error of the sample mean is calculated as \( \frac{1404}{9} = 156\) air conditioners per day.

Step by step solution

01

Calculate The Sample Mean

Since a random sample of 81 days is considered from a normal distribution, which has an average of 3600 per day, the expected value for the sample mean is equivalent to the population mean. Therefore, the expected sample mean is \(3600\) air conditioners per day.
02

Calculate The Standard Error

The standard error of the sample mean can be calculated by dividing the standard deviation by the square root of the sample size (\(n\)). The standard deviation given is \(1404\) and the sample size is \(81\). So the standard error formula is \( \frac{1404}{\sqrt{81}}\). Simplifying \( \sqrt{81} \) results in \(9\), which can be further divided into \(1404\) to obtain the standard error.

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Key Concepts

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

Standard Deviation
Understanding the standard deviation is crucial when analyzing the spread of data around the mean. It provides an insight into how much individual data points deviate from the average, indicating the variability within a set of data. In this context, the standard deviation is given as 1404 air conditioners per day. This means:
  • A higher standard deviation shows more variation from the mean.
  • Conversely, a lower standard deviation would indicate that most data points are close to the mean.

In a positively skewed distribution, like the one described in the exercise, more data points may linger on the lower end of the scale, leading to this higher measure of spread. The understanding of how data spread affects our calculations, especially for errors and deviations, is key when it comes to estimating results with samples.
Sample Mean
The sample mean is an estimate of the population mean, calculated from a subset of the entire population. It helps us understand the average behavior of a dataset over a given period or under certain conditions. Given the information, since our sample of 81 days is drawn from the population鈥檚 normal distribution, and the population mean is 3600 air conditioners per day, the sample mean is expected to be close to this value.
  • An important point to remember is that the larger the sample size, the closer the sample mean will be to the population mean.
  • In this exercise, with a sample size of 81, the expected sample mean remains at 3600 air conditioners.

The accuracy of the sample mean as a reflection of the population mean is reinforced by larger sample sizes, since random fluctuations tend to cancel each other out, bringing the sample mean nearer to the true mean.
Standard Error
The standard error provides an estimate of the variability between the sample mean and the actual population mean. It measures the accuracy with which your sample mean represents the population mean and is derived by the following method:
  • Divide the standard deviation by the square root of the sample size.

In our example, the calculation follows the formula: \[SE = \frac{SD}{\sqrt{n}} = \frac{1404}{\sqrt{81}} = \frac{1404}{9} = 156\]This means the standard error is 156 air conditioners per day.
  • This standard error suggests that we expect an average fluctuation of 156 air conditioners per day in the sample mean from the population mean of 3600.
The smaller the standard error, the closer the sample mean will likely be to the actual population mean. This concept is key when interpreting results and understanding how sample sizes and standard deviations factor into estimates of accuracy.

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Most popular questions from this chapter

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A McDonald's fact sheet says their cones should weigh \(3.18\) ounces (converted from grams). Suppose you take a random sample of four cones, and the weights are \(4.2,3.4,3.9\), and \(4.4\) ounces. Assume that the population distribution is Normal, and, for all three parts, report the alternative hypothesis, the \(t\) -value, the p-value, and your conclusion. The null hypothesis in all three cases is that the population mean is \(3.18\) ounces. a. Test the hypothesis that the cones do not have a population mean of \(3.18\) ounces. b. Test the hypothesis that the cones have a population mean less than \(3.18\) ounces. c. Test the hypothesis that the cones have a population mean greater than 3.18 ounces.

Tomatoes The weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be \(5.1\), \(5.0,5.3\), and \(5.1\) pounds. Assume Normality. a. Find a \(95 \%\) confidence interval for the mean weight of all bags of tomatoes. b. Does the interval capture \(5.0\) pounds? Is there enough evidence to reject a mean weight of \(5.0\) pounds?

Rodents A scientist is interested in studying the effects that applying pesticide to crops has on the local rodent population. The scientist collects 31 rodents from the field where the pesticide is used and finds the mean weight of this sample to be \(84.6\) grams. Assuming the selected rodents are a random sample, he concludes that because the sample mean is an unbiased estimator of the population mean, the mean weight of rodents in the population is also \(84.6\) grams. Explain why this is an incorrect interpretation of what it means to have an unbiased estimator.

Weight A study of all the employees at an office showed a mean weight of \(60.4\) kilograms and a standard deviation of \(1.5\) kilograms. a. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma\) ).
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