/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Catalog mail-order sales A compa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Catalog mail-order sales A company that sells its products through mail-order catalogs wants information about the success of its most recent catalog. The company decides to estimate the mean dollar amount of items ordered from those who received the catalog. For a random sample of 100 customers from their files, only 5 made an order, so 95 of the response values were \(\$ 0 .\) The overall mean of all 100 orders was \(\$ 10,\) with a standard deviation of \(\$ 10\) a. Is it plausible that the population distribution is normal? Explain and discuss how much this affects the validity of a confidence interval for the mean. b. Find a \(95 \%\) confidence interval for the mean dollar order for the population of all customers who received this catalog. Normally, the mean of their sales per catalog is about \(\$ 15\). Can we conclude that it declined with this catalog? Explain.

Short Answer

Expert verified
The distribution is not normal. The confidence interval is $(8.016, 11.984)$; sales declined.

Step by step solution

01

Evaluate Normality of Distribution

The question asks whether a normal distribution is plausible. In this case, with 95 out of 100 customers ordering nothing, the data is heavily skewed towards zero, not resembling a normal distribution at all. Thus, assuming normality is not reasonable, affecting the validity of a standard confidence interval that assumes a normal distribution.
02

Calculate Sample Mean and Standard Deviation

The mean of the sample of sales values is provided as $10. The standard deviation of the sample is also given as $10. These values will be used to construct the confidence interval.
03

Determine Confidence Interval Formula

Since the assumption of normality is not met, using the standard method (a t-distribution or z-distribution-based interval) may not be appropriate without caution. However, for illustrative purposes, let's proceed as if it were normal. Since the population standard deviation is unknown and n=100 (a large sample size), use:\[\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}\]Where \(\bar{x} = 10\), \(s = 10\), \(n = 100\), and \(t_{\alpha/2, n-1}\) is the t-value for 95% confidence and 99 degrees of freedom.
04

Find the Critical T-Value

Look up the critical t-value for 95% confidence with 99 degrees of freedom. For simplicity, approximate using t = 1.984, which is accurate for large samples where it converges to the Z-distribution value.
05

Calculate Confidence Interval

Substitute the values into the formula:\[10 \pm 1.984 \times \frac{10}{\sqrt{100}} = 10 \pm 1.984 \times 1 = 10 \pm 1.984\]Therefore, the confidence interval is:\[ 8.016 \text{ to } 11.984 \]
06

Compare to Previous Mean Sales

Previously, the mean sales per catalog were $15. The calculated confidence interval for mean sales with this catalog is $(8.016, 11.984)$, which does not include $15. Thus, this suggests that the mean sales have declined and it is statistically significantly lower under these assumptions.

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Key Concepts

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

Normal Distribution
The normal distribution is a key concept in statistics, often referred to as the "bell curve" due to its shape. It is symmetrical around the mean, meaning that the left and right sides of the graph are mirror images. This distribution plays a crucial role in statistics because it allows statisticians to make inferences about populations based on sample data. However, in the scenario provided by our exercise, where 95% of the customers did not make any orders (resulting in $0 sales), the data set is heavily skewed towards zero. This skewness indicates that the distribution of the data is far from normal.

The assumption of normality is significant as it impacts the validity of statistical procedures, such as calculating confidence intervals. If the data does not follow a normal distribution, standard methods for confidence interval estimation may not be valid or may require modifications. When working with non-normally distributed data, alternative methods or adjustments might be necessary to obtain accurate statistical inferences.
Sample Mean
The sample mean is a measure of the central tendency of a data set, calculated as the sum of all observed values divided by the number of observations. In our exercise, the sample mean is given as $10. This value tells us the average order amount across all customers sampled.

The sample mean serves as an unbiased estimator of the population mean, especially in large samples. This means that, on average, the sample mean will equal the population mean if you take many samples. However, with the heavy skewness observed in the data in the exercise, the sample mean alone may not be a fully adequate summary without considering its variability and the shape of the distribution. In contexts like this, complementing the sample mean with additional analyses or descriptive statistics can provide a more comprehensive understanding of the dataset.
Standard Deviation
The standard deviation is another critical concept in statistical analysis. It measures the amount of variation or dispersion in a set of values. A low standard deviation indicates the data points are close to the mean, while a high standard deviation suggests they are spread out over a broader range.

In our scenario, the standard deviation is also $10. This value, combined with the sample mean, gives insight into how our sales values deviate from the center point. Even when data is not normally distributed, as is the case here, the standard deviation still acts as a helpful measure of spread. It's important to understand that while the mean tells us where the center of our data is, the standard deviation gives us a picture of how varied the data is around that center, offering a fuller understanding of data variability.
T-Distribution
In statistics, the t-distribution is used to estimate population parameters when the sample size is small, or when the population standard deviation is unknown. It is similar to the normal distribution but has thicker tails. This means there is more probability in the tail ends, which accounts for greater uncertainty in estimates from small samples.

In the exercise, we encounter the t-distribution because we don't know the true population standard deviation, and although we have a large sample size (n=100), the heavily skewed data calls for caution. The t-distribution is utilized to derive a 95% confidence interval for the mean under the given conditions. This adjustment helps to achieve reliable estimates even if the data doesn't perfectly adhere to normal distribution assumptions. The critical t-value used (approximately 1.984 for 99 degrees of freedom) accounts for sampling variability and provides an interval thought to include the true population mean with 95% confidence.

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Most popular questions from this chapter

Employment percentage in the United States According to the U.S. Bureau of Labor Statistics, \(80.3 \%\) out of the nation's 81.4 million families had at least one employed member in 2015 (Source: http://www.bls.gov/news. release/famee.nr0.htm). What should be the sample size needed to estimate the proportion of families having at least one employed member in 2015 within an accuracy of 3 percentage points at a \(95 \%\) level of confidence?

In 2014, news reports worldwide alleged that the U.S. government had hacked German chancellor Angela Merkel's cell phone. A Pew Research Center survey of German citizens at about that time asked whether they find it acceptable or unacceptable for the U.S. government to monitor communications from their country's leaders. Results from the survey show that of 1000 citizens interviewed, 900 found it unacceptable. (Source: Pew Research Center, July 2014, "Global Opposition to U.S. Surveillance and Drones, But Limited Harm to America's Image") a. Find the point estimate of the population proportion of German citizens who find spying unacceptable. b. The Pew Research Center reports a margin of error at the \(95 \%\) confidence level of \(4.5 \%\) for this survey. Explain what this means.

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Effect of confidence level Find the margin of error for estimating the population mean when the sample standard deviation equals 100 for a sample size of \(25,\) using confidence levels (i) \(95 \%\) and (ii) \(99 \%\). What is the effect of the choice of confidence level? (You can use Table \(\mathrm{B}\) in the back to find the appropriate \(t\) -scores.)

Wage discrimination? According to a union agreement, the mean income for all senior-level assembly-line workers in a large company equals \(\$ 500\) per week. A representative of a women's group decides to analyze whether the mean income for female employees matches this norm. For a random sample of nine female employees, using software, she obtains a \(95 \%\) confidence interval of ( 371 , 509 ). Explain what is wrong with each of the following interpretations of this interval. a. We infer that \(95 \%\) of the women in the population have income between \(\$ 371\) and \(\$ 509\) per week. b. If random samples of nine women were repeatedly selected, then \(95 \%\) of the time the sample mean income would be between \(\$ 371\) and \(\$ 509\). c. We can be \(95 \%\) confident that \(\bar{x}\) is between \(\$ 371\) and \(\$ 509 .\) d. If we repeatedly sampled the entire population, then \(95 \%\) of the time the population mean would be between \(\$ 371\) and \(\$ 509 .\)

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