The Economist collects data each year on the price of a Big Mac in various countries around the world. A
sample of McDonald's restaurants in Europe in July 2021 resulted in the following Big Mac prices (after
conversion to U.S. dollars).
6.27 4.76 2.52 4.75 3.74 3.47 2.88
The mean price of a Big Mac in the United States. in July 2021 was $5.81. For purposes of this exercise,
you can assume it is reasonable to regard the sample as representative of European McDonald's
restaurants. Does the sample provide convincing evidence that the mean July 2021 price of a Big Mac in
Europe is less than the reported U.S. price? Test the relevant hypotheses using alpha = 0.05
State the appropriate null and alternative hypotheses (in dollars).
Find the test statistic. (Round your answer to two decimal places.)
Use technology to find the p-value
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The appropriate null and alternative hypotheses are:
Null hypothesis: The mean July 2021 price of a Big Mac in Europe is equal to or greater than the reported U.S. price ($5.81).
Alternative hypothesis: The mean July 2021 price of a Big Mac in Europe is less than the reported U.S. price ($5.81).
To find the test statistic, we can use the t-test formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / √n)
sample mean = (6.27 + 4.76 + 2.52 + 4.75 + 3.74 + 3.47 + 2.88) / 7 ≈ 4.15
hypothesized mean = $5.81
sample standard deviation = √[(6.27 - 4.15)² + (4.76 - 4.15)² + (2.52 - 4.15)² + (4.75 - 4.15)² + (3.74 - 4.15)² + (3.47 - 4.15)² + (2.88 - 4.15)²] / (7 - 1) ≈ 1.15
n = 7
t = (4.15 - 5.81) / (1.15 / √7) ≈ -3.25 (rounded to two decimal places)
Using technology, we can find the p-value for this test statistic. The p-value represents the probability of observing a test statistic as extreme as the calculated t-value, assuming that the null hypothesis is true.
The p-value for a t-statistic of -3.25 with 6 degrees of freedom (n-1) can be calculated using a t-distribution table or a statistical software. The p-value is approximately 0.012.
Therefore, the p-value for this test is approximately 0.012.
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