91Ó°ÊÓ

The study described in the paper "Marketing Actions Can Modulate Neural Representation of Experienced Pleasantness" (Proceedings of the National Academy of Science [2008]\(: 1050-1054)\) investigated whether price affects people's judgment. Twenty people each tasted six cabernet sauvignon wines and rated how they liked them on a scale of 1 to 6 . Prior to tasting each wine, participants were told the price of the wine. Of the six wines tasted, two were actually the same wine, but for one tasting the participant was told that the wine cost \(\$ 10\) per bottle and for the other tasting the participant was told that the wine cost \(\$ 90\) per bottle. The participants were randomly assigned either to taste the \(\$ 90\) wine first and the \(\$ 10\) wine second, or the \(\$ 10\) wine first and the \(\$ 90\) wine second. Differences (computed by subtracting the rating for the tasting in which the participant thought the wine cost \(\$ 10\) from the rating for the tasting in which the participant thought the wine cost \(\$ 90\) ) were computed. The differences that follow are consistent with summary quantities given in the paper. Difference \((\$ 90-\$ 10)\) \(\begin{array}{cccccccccccccccccccc}2 & 4 & 1 & 2 & 1 & 0 & 0 & 3 & 0 & 2 & 1 & 3 & 3 & 1 & 4 & 1 & 2 & 2 & 1 & -1\end{array}\) Carry out a hypothesis test to determine if the mean rating assigned to the wine when the cost is described as \(\$ 90\) is greater than the mean rating assigned to the wine when the cost is described as \(\$ 10 .\) Use \(\alpha=.01\)

Step by step solution

01

Formulate the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) : The mean rating difference is zero, i.e., the price does not affect the rating.\(\mu = 0\)\n\nThe alternative hypothesis (\(H_a\)) : The mean rating difference is greater than zero, which would indicate that the \(\$ 90\) wine has a higher rating. \(\mu > 0\)

02

Calculate Sample Mean and Standard Deviation

Use the given differences to calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s). The differences are: \[2, 4, 1, 2, 1, 0, 0, 3, 0, 2, 1, 3, 3, 1, 4, 1, 2, 2, 1, -1\]\n\nCalculate the sample mean (\(\bar{x}\)) as the sum of all differences divided by the number of differences. Then calculate the sample standard deviation (s) by taking the square root of the sum of the squared difference between each difference and the sample mean, divided by the number of differences minus 1.

03

Calculate Test Statistic

The test statistic for the t-test is calculated as follows: \n\n\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]\n\nwhere \(\mu\) is the population mean (our null hypothesis), \(\bar{x}\) is the sample mean, s is the sample standard deviation, and n is the number of observations (differences).

04

Find the Critical Value

From the t-distribution table, find the critical value (\(t_{crit}\)) for a one-tailed test with degrees of freedom = n-1 and significance level (\(\alpha\)) of 0.01.

05

Reject or Fail to Reject the Null Hypothesis

If the calculated test statistic (t) is greater than the critical value (\(t_{crit}\)), then reject the null hypothesis in favor of the alternative hypothesis. This would mean that the price does affect the wine's rating. Otherwise, fail to reject the null hypothesis, which means that the price does not significantly affect the wine's rating.

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Key Concepts

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

Null Hypothesis

In the realm of hypothesis testing, the null hypothesis (\(H_0\)) stands as a statement that indicates no effect or no difference. It serves as a baseline or a default position that a researcher seeks to challenge. As exemplified by the wine study, the null hypothesis suggested that price has no impact on how participants rated the wine, essentially stating that the mean rating difference due to price is zero (\( \bar{x} = 0 \)). Critically, it is the assumption that any observed difference in data, such as the ratings of the wines, is due to chance rather than a systematic effect.

Alternative Hypothesis

The alternative hypothesis (\(H_a\) or \(H_1\) represents the statement investigators aim to support, which proposes that there is a true effect or difference. In our study, the alternative hypothesis suggests there is an effect of the price on wine ratings—specifically, that the wine described as costing \$90 is rated higher than when it is described as costing \$10. The formulation of \(H_a: \bar{x} > 0 \) sets the stage for a one-tailed test, where the direction of the effect is specified as higher ratings for the more expensive wine label.

Statistical Significance

Statistical significance acts as a threshold for determining whether the observed effect is unlikely to have occurred by chance. This concept relies upon a predefined alpha level (\(\alpha\) which represents the probability of rejecting the null hypothesis when it is indeed true (Type I error). In this scenario, \(\alpha = 0.01\) implies that there is only a 1% probability that the findings could be due to random variation, as opposed to a genuine difference in ratings based on the perceived price of the wine.

t-test

A t-test is a statistical examination utilized to compare the mean of a sample to a known value (often the population mean), or to compare the means of two samples. In explicit terms for our study, we apply a one-sample t-test to determine if the mean difference in wine ratings (\(\bar{x}\) differs significantly from zero under the null hypothesis. The t-test calculates a t-value that serves as a test statistic, which we then compare with the critical value to draw conclusions about our hypotheses.

Sample Mean

The sample mean (\(\bar{x}\) is a measure that captures the average rating difference observed in the study sample. It is computed by summing all the rating differences provided by participants and dividing by the total number of observations. The study's sample mean is crucial since it functions as an estimator of the population mean difference, which we are looking to compare against our null hypothesis of zero effect.

Sample Standard Deviation

Sample standard deviation (s) quantifies the extent of variation or dispersion in the rating differences from the sample mean. It is essential for calculating the standard error, which adjusts for the fact that we are dealing with a sample rather than an entire population. Sample standard deviation is found by taking the root of the sum of squared deviations of each sample data point from the sample mean divided by one less than the number of observations.

Critical Value

The critical value, which is found using statistical tables or computational tools, is a cut-off point that determines the boundary for rejecting the null hypothesis. Depending on the chosen significance level and the degrees of freedom (usually \(n - 1\) for a sample of size n), the critical value forms the threshold for how extreme the test statistic needs to be to deem the result statistically significant. In hypothesis testing, comparing the test statistic against the critical value decides whether the null hypothesis can be rejected.

p-value

The p-value is the probability of obtaining a result at least as extreme as the one actually observed, given that the null hypothesis is true. It quantifies the evidence against the null hypothesis where a small p-value (\(p \< \alpha\) indicates strong evidence to reject the null hypothesis. In contrast, a larger p-value suggests insufficient evidence to do so, leading to retaining the null hypothesis. The p-value complements the decision-making process by providing a scale of the evidence rather than just a binary outcome.