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The Economist collects data each year on the price of a Big Mac in various countries around the world. The price of a Big Mac for a sample of McDonald's restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llllll}3.80 & 5.89 & 4.92 & 3.88 & 2.65 & 5.57\end{array}\) \(\begin{array}{ll}6.39 & 3.24\end{array}\) The mean price of a Big Mac in the U.S. in May 2009 was \(\$ 3.57\). For purposes of this exercise, assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using \(\alpha=.05\).

Short Answer

Expert verified
A t-test was conducted to compare the mean prices of a Big Mac in Europe and the U.S. The calculated t-value, and hence the result, would determine whether we reject or fail to reject our null hypothesis.

Step by step solution

01

Set up the Hypotheses

First, the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)) need to be established. For this exercise: \(H_0: \mu = \$3.57 - this signifies that the mean price in Europe equals the U.S. mean price. \) \(H_1: \mu > \$3.57 - Represents that the mean price in Europe is greater than the U.S. mean price.\)
02

Calculate Sample Mean and Sample Standard Deviation

Add up all the given European Big Mac prices and divide by the total number of observations to get the sample mean (\(\overline{x}\)). The sample standard deviation (s) can be obtained by calculating the square root of the variance, which itself is the average of the squared deviations from the mean. For this data set: \(\overline{x} = (\$3.80+\$5.89+\$4.92+\$3.88+\$2.65+\$5.57+\$6.39+\$3.24)/8 = \$4.505 \) For standard deviation, use a standard deviation calculator inputting all the European data set values to get \(s = \$1.21\)
03

Conduct the Hypothesis Test

Since the sample size is smaller than 30, therefore a t-test will be used. The test statistic, t, can be calculated as: \(t = (\overline{x} - \mu_{0}) / (s / \sqrt{n}) \) Where \(n\) is the sample size, \(\mu_{0}\) is the population mean, in this case the U.S. mean price, \(\$3.57\). Calculation yields: \(t = (\$4.505 - \$3.57) / (\$1.21 / \sqrt{8}) = 2.44 \) Then, compare this calculated t-value to the t critical value from the t-Distribution table given \(\alpha = 0.05\) and degrees of freedom = \(n - 1 = 7\). If the calculated t-value is larger than the standard t-value (one-tailed test), reject the null hypothesis.
04

Make a Conclusion

If the calculated t-value is higher than the table value, we reject the null hypothesis and conclude that the mean price of a Big Mac in Europe is significantly larger than in the U.S. On the other hand, if the table value is higher, we fail to reject the null hypothesis, suggesting there's not enough evidence to support that the mean price of the Big Mac in Europe is larger than that in the U.S. We would have to look up the t-table to finalize the conclusion.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is crucial for hypothesis testing. It's a statement suggesting that there is no effect or no difference, and it serves as a starting point for statistical analysis. In our Big Mac price scenario, the null hypothesis (\( H_0 \)) posits that the mean price of a Big Mac in Europe is equal to the mean price in the United States, specifically \( \$3.57 \). This is the benchmark against which we test the possibility of a difference in pricing.
Alternative Hypothesis
Complementary to the null hypothesis is the alternative hypothesis (\( H_1 \) or \( H_a \)), which represents what we aim to support with our data. For the Big Mac exercise, the alternative hypothesis claims that the average price in Europe is greater than in the U.S., or \( \mu > \$3.57 \). Demonstrating the alternative hypothesis typically requires showing that the observed difference is unlikely to have occurred due to chance if the null hypothesis were true.
T-Test
When we have a smaller sample size (less than 30), or we don’t know the population standard deviation, we use the t-test. It's a statistical tool to determine if there is a significant difference between the means of two groups. In our case, we are using a one-sample t-test to compare the sample mean of the Big Mac prices in Europe with the known U.S. mean price.
Sample Mean
The sample mean (\( \overline{x} \) is the average value of our sample data. It’s calculated by adding up all the sample values and dividing by the number of observations. For our Big Mac price data, the sample mean calculated from the sample of European prices is \( \$4.505 \), which we use to compare against the U.S. mean price.
Sample Standard Deviation
Sample standard deviation (s) measures the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. Our calculated sample standard deviation for European Big Mac prices is \( \$1.21 \) and reflects the variability within the sample prices.
Degrees of Freedom
Degrees of freedom (df) are an essential concept in statistical tests, including the t-test. They represent the number of values in the final calculation of a statistic that are free to vary. In the context of the t-test, the degrees of freedom is the sample size minus one (\( n - 1 \)). For our set of Big Mac prices, we have a sample size of 8, so the degrees of freedom we use is 7.
Statistical Significance
Statistical significance is a determination about the null hypothesis. In hypothesis testing, we use a significance level (\( \alpha \)), typically set at 0.05 or 5%, to decide whether to reject the null hypothesis. It represents the probability threshold under which we determine our test results are significant. If our test statistic is past this threshold, we have grounds to reject the null hypothesis, suggesting our findings are not due to random chance.
P-Value
The p-value is a metric used to summarize the strength of the evidence against the null hypothesis provided by our sample. It represents the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. A small p-value (\( p < \alpha \) indicates strong evidence against the null hypothesis, so it is rejected, and the alternative hypothesis is accepted. In our example, we would need to compute or look up the p-value associated with the t-statistic to decide on the hypothesis.

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Most popular questions from this chapter

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(p\) be the proportion of all apartments that prohibit children. If the city council is convinced that \(p\) is greater than 0.75 , it will consider appropriate legislation. a. If 102 of the 125 sampled apartments exclude renters with children, would a level .05 test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(p=.8\) and \(\alpha=.05 ?\)

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases. It is therefore important that the information on packages be accurate. A random sample of \(n=12\) frozen dinners of a certain type was selected from production during a particular period, and the calorie content of each one was determined. (This determination entails destroying the product, so a census would certainly not be desirable!) Here are the resulting observations, along with a boxplot and normal probability plot: \(\begin{array}{llllllll}255 & 244 & 239 & 242 & 265 & 245 & 259 & 248\end{array}\) \(\begin{array}{llll}225 & 226 & 251 & 233\end{array}\) a. Is it reasonable to test hypotheses about mean calorie content \(\mu\) by using a \(t\) test? Explain why or why not. b. The stated calorie content is \(240 .\) Does the boxplot suggest that true average content differs from the stated value? Explain your reasoning. c. Carry out a formal test of the hypotheses suggested in Part (b).

The power of a test is influenced by the sample size and the choice of significance level. a. Explain how increasing the sample size affects the power (when significance level is held fixed). b. Explain how increasing the significance level affects the power (when sample size is held fixed).

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10 .\) a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate? b. If \(t=-1.83\) and \(\alpha=.01\) is selected, what conclusion is appropriate? c. If \(t=0.47,\) what conclusion is appropriate?

According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net wealth in their lifetime ("One in Five Believe Path to Riches Is the Lottery," San Luis Obispo Tribune, January 11, 2006 ). Although the article does not describe how the sample was selected, for purposes of this exercise, assume that the sample can be regarded as a random sample of adult Americans. Is there convincing evidence that more than \(20 \%\) of adult Americans believe that playing the lottery is the best strategy for accumulating \(\$ 200,000\) in net wealth?

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