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91Ó°ÊÓ

The accompanying data consists of prices \((\$)\) for one sample of California cabernet sauvignon wines that received ratings of 93 or higher in the May 2013 issue of Wine Spectator and another sample of California cabernets that received ratings of 89 or lower in the same issue. \(\begin{array}{rrrrrrrr}\geq 93: & 100 & 100 & 60 & 135 & 195 & 195 & \\ & 125 & 135 & 95 & 42 & 75 & 72 & \\ \leq 89: & 80 & 75 & 75 & 85 & 75 & 35 & 85 \\ & 65 & 45 & 100 & 28 & 38 & 50 & 28\end{array}\) Assume that these are both random samples of prices from the population of all wines recently reviewed that received ratings of at least 93 and at most 89 , respectively. a. Investigate the plausibility of assuming that both sampled populations are normal. b. Construct a comparative boxplot. What does it suggest about the difference in true average prices? c. Calculate a confidence interval at the \(95 \%\) confidence level to estimate the difference between \(\mu_{1}\), the mean price in the higher rating population, and \(\mu_{2}\), the mean price in the lower rating population. Is the interval consistent with the statement "Price rarely equates to quality" made by a columnist in the cited issue of the magazine?

Short Answer

Expert verified
The data seems plausible as normal; boxplots indicate higher medians for better rated wines; the confidence interval likely excludes zero, suggesting price differs with quality.

Step by step solution

01

Check Normality Assumption

For part (a), we usually check normality using visual tools like Q-Q plots or statistical tests like the Shapiro-Wilk test. Given that we have small sample sizes, a visual check can be a preliminary approach. Plotting a Q-Q plot for each sample can help determine if the distribution of data follows a straight line, suggesting normality. Alternatively, completeness of normality can be further checked using the Shapiro-Wilk test for each dataset, where a p-value greater than 0.05 would indicate no significant deviation from normality.
02

Draw the Comparative Boxplot

For part (b), construct a boxplot for both datasets (wines rated ≥ 93 and ≤ 89). In a boxplot, we will display the median, quartiles, and potential outliers. By comparing the two boxplots, we can visually assess the differences in median prices, variability (interquartile range), and presence of any outliers. Note how the center of each boxplot (median) compares and whether one boxplot consistently lies above the other.
03

Calculate the 95% Confidence Interval

For part (c), calculate a 95% confidence interval for the difference in means between the two populations. Using the formula for confidence interval between two means: \(CI = \bar{x}_1 - \bar{x}_2 \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \( \bar{x}_1 \) and \( \bar{x}_2 \) are sample means, \( s_1^2 \) and \( s_2^2 \) are sample variances, \( n_1 \) and \( n_2 \) are sample sizes, and \( t^* \) is the critical value from the t-distribution for 95% confidence level. Calculate and interpret whether this interval includes zero, indicating negligible difference in true average prices.

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Key Concepts

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

Normality Assumption
When dealing with statistical analysis, one assumption that often comes up is the normality of the data. The normality assumption refers to the belief that data follows a normal distribution, which is a symmetric, bell-shaped curve. This assumption is crucial in many statistical tests because it simplifies the calculations involved and validates the use of certain statistical measures. To verify this, tools like Q-Q plots or the Shapiro-Wilk test are commonly used.

In our wine price scenario, checking normality helps us know if the prices from the two ratings groups (93 or higher and 89 or lower) can be compared using parametric tests. With small sample sizes, a Q-Q plot is a quick way to visually inspect if the sample data aligns with a normal distribution. In simpler terms, if the plot points follow a straight line, normality might be assumed. If instead you see a lot of deviation from this line, the data probably isn't normal.

The Shapiro-Wilk test, on the other hand, provides a more formal statistical test of normality. Here, a p-value greater than 0.05 suggests the data does not significantly deviate from normality, supporting the assumption. It’s important to note, however, that even if the data doesn't follow a perfect normal distribution, other methods or transformations can still make the analysis viable.
Comparative Boxplot
A comparative boxplot is a robust way of visually presenting and comparing the distribution of two or more data sets. It effectively showcases the data's center, spread, and potential outliers, providing crucial insights at a glance. So when comparing wine prices between different ratings, boxplots highlight these differences through several visual cues.

Here's what you can look for in a boxplot:
  • The line inside the box indicates the median, showing the central tendency of the data.
  • The box itself captures the interquartile range (IQR), which is the middle 50% of data, highlighting data variability.
  • Whiskers extend to the smallest and largest non-outlier data points, while individual points outside the whiskers are considered potential outliers.

In our exercise, drawing a comparative boxplot for the two wine rating groups allows us to quickly assess which group generally has higher prices. If one boxplot is noticeably higher across all sections compared to the other, it suggests that the group may have higher median prices. Examination of outliers can also reveal the extent to which certain wines deviate significantly from others in terms of price.
Confidence Interval
Confidence intervals are essential for estimating the range within which a population parameter lies, with a specific level of certainty. In our wine price example, constructing a confidence interval (CI) helps gauge the difference between the average prices of the two wine rating groups, rated 93 or higher and 89 or lower.

To construct a 95% confidence interval for the difference in mean prices, we use the formula:\[ CI = \bar{x}_1 - \bar{x}_2 \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for each group.
  • \(s_1^2\) and \(s_2^2\) are the sample variances for each group.
  • \(n_1\) and \(n_2\) are the sample sizes.
  • \(t^*\) is the critical value from the t-distribution for the 95% confidence level.

The resulting interval provides a range in which the true difference in means likely resides. If this interval includes zero, it suggests there may be no significant difference in average prices, offering insight into whether price actually reflects wine quality, as speculated in the magazine. If zero isn't within the interval, it implies a significant price difference, prompting us to explore further underlying causes.

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Most popular questions from this chapter

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