Question

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during 2005 . 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 the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) 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) The standard deviation can be larger than the mean if the data points are widely spread out from the mean, typically due to certain very high or low values. b) It's hard to definitively say if the distribution is normal with just mean and standard deviation. However, the data could be skewed because the standard deviation is much larger than the mean. c) The t-interval could be used despite a quite large sample size as the distribution may not be normal. d) Using the t-interval, a 99% confidence interval can be calculated. However, due to potential non-normality, this interval should be interpreted with caution.

Step-by-step solution

Explain why the standard deviation may exceed the mean

Standard deviation measures the dispersion or spread of data from its mean. If the standard deviation is larger than the mean, it suggests that the data points are widely spread out from the mean. This can happen when there are extreme values (very high or low) in the data, causing a higher variance and thus a larger standard deviation.

Discuss the likelihood of normal distribution

Given just the mean and standard deviation, it is difficult to definitively say if the data follows a normal distribution. However, considering practical factors, it's reasonable to think that the distribution of the anticipated Halloween spending may not be normal, particularly because the standard deviation is much larger than the mean, suggesting a possible skewed distribution. However, without additional information such as histograms or Q-Q plots, it's hard to make a definitive determination.

Consider the use of t-interval

The t-confidence interval is often used when the sample size is relatively small, and the population standard deviation is unknown. Since the sample size in this case is 1000, which is large, and we know the standard deviation, usually the z-interval would be more appropriate. However, considering we don't know if the distribution is normal or not, the t-interval could still be used as it is more robust against non-normality for large sample sizes.

Compute and interpret the confidence interval

The formula to calculate the t-confidence interval is: \[ \overline{X} \pm t_{\frac{a}{2},df} \cdot \frac{\sigma}{\sqrt{n}} \] where \(\overline{X}\) is the sample mean, \( \sigma \) is the standard deviation, \(n\) is the sample size, and \( df \) is the degree of freedom which is \(n-1\). For a 99% confidence interval and df=999, \( t_{\frac{a}{2},df} \) is approximately 2.617. However, given Step 3, the confidence interval would usually be offered with a caution that it might not accurately reflect the parameters of the population.