Question

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24,2005\()\) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\), respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the one-sample \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

Short answer

a. Higher standard deviation than the mean indicates a large spread in the expenses. b. Whether the distribution is normal cannot be definitively stated from the information given. c. One-sample t confidence interval is appropriate if the sample size is large enough (by Central Limit Theorem). d. With the given values, it would be possible to calculate the 99% confidence interval assuming normality and large enough sample size.

Step-by-step solution

Understanding Variance

The standard deviation of the anticipated Halloween expense can be larger than the mean anticipated expense if the expenses are spread out a lot. In other words, if there's a huge variation in the amount people plan to spend, some spending a lot and others a little, the standard deviation can be quite large compared to the mean.

Assessing Normality

To determine if the distribution of anticipated Halloween expense is approximately normal, we typically look at the sample size and graphical representations (such as a histogram or a box plot). In the absence of these, we can't make a definitive statement. However, given the large variation (high standard deviation compared to mean), the distribution may not be strictly normal.

Suitability of One-sample t Confidence Interval

Whether a one-sample t confidence interval is appropriate to estimate the mean anticipated Halloween expense largely depends on whether the distribution of the anticipated expense is normal or not. It's also important to consider whether the sample size is large enough. If the sample size is larger than 30, the Central Limit Theorem says the distribution of sample means will be approximately normal regardless of the shape of the population distribution.

Constructing 99% Confidence Interval

Assuming the sample size is large enough, and that the distribution is approximately normal, the 99% confidence interval for the mean can be calculated using the one-sample t-test formula: \[ X̄ ± t * (s/√n)\]. Assuming a large sample size, and degrees of freedom df=n-1 we can look up t for a 99% confidence interval in a t-table, which is approximately 2.626. Plugging the values, we have: \[ \$46.65 ± 2.626 * \$83.70/√1000\]. This will give the interval.

Key concepts

Standard Deviation

Standard deviation is a measure of how spread out numbers are in a dataset. Think of it as an average distance of every data point from the mean (average) of the data. If each individual's Halloween spending is consistently around \(46.65, then the standard deviation would be low, indicating that most people spend a similar amount. However, the high standard deviation of \)83.70 tells us that spending habits vary greatly—from people who spend very little to those who may splurge on Halloween. This can occur in populations where there are extremes, for example, a few individuals planning to host lavish Halloween parties while others may not spend much at all.

In this context, it's quite possible for the standard deviation to be larger than the mean—this suggests that the data isn't clustered close to the mean but widely scattered, with significant variations in spending habits among Canadians.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve that is symmetric about the mean. Most values are close to the mean, with fewer and fewer appearing as you move away. In real-life data, perfect normality is rare, but many processes are approximately normal, especially with larger sample sizes.

When we talk about anticipated Halloween spending, we may expect that the majority of people would spend an amount near the average, with fewer spending much more or much less. However, with a standard deviation much larger than the mean, as in this case, it suggests that the spending is not clustered symmetrically around the mean. This could result from a few spenders who plan to purchase expensive items skewing the distribution. The normality assumption might not hold in this situation without further graphical analysis and normality tests.

One-Sample T Confidence Interval

A one-sample t confidence interval is a statistical tool used to estimate the mean of a population. It’s based on the sample mean, sample standard deviation, and the sample size. The t-distribution is used when the sample size is small or the population standard deviation is unknown—common conditions in real-world research.

To apply this method to the Halloween spending data, we would need to assume that the underlying distribution of expenses is approximately normal. This is often reasonable with large sample sizes due to the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size grows, regardless of the population's distribution. Assuming that the sample size is sufficient (large, typically over 30), and there's no evidence of extreme skewness or outliers, the one-sample t confidence interval can provide a reliable estimate for Canadian residents' mean anticipated Halloween expense. The final confidence interval, based on the exercise solution provided, would offer an estimated range in which we expect the true population mean to lie with 99% confidence.