Respuesta :
Using the t-distribution, it is found that since the p-value of the test is 0.0428 < 0.05, the sample provides convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price.
At the null hypothesis, it is tested if the mean is the same, that is:
[tex]H_0: \mu = 3.57[/tex]
At the alternative hypothesis, it is tested if it is greater, that is:
[tex]H_1: \mu > 3.57[/tex]
We can find the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem, two of the values of the parameters are: [tex]n = 8, \mu = 3.57[/tex]
Additionally, with the help of a calculator: [tex]\overline{x} = 4.5225, s = 1.3468[/tex]
The value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{4.5225 - 3.57}{\frac{1.3468}{\sqrt{8}}}[/tex]
[tex]t = 2[/tex]
The p-value of the test is found using a right-tailed test, as we are testing if the mean is greater than a value, with t = 2 and 8 - 1 = 7 df.
- Using a t-distribution calculator, this p-value is of 0.0428.
Since the p-value of the test is 0.0428 < 0.05, the sample provides convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price.
A similar problem is given at https://brainly.com/question/13873630
