Using the t-distribution, it is found that since the absolute value of the test statistic is less than the critical value for the two-tailed test, there is not enough evidence to conclude that the price of the gas in Anne's city is different of $4.06.
At the null hypothesis, we test if the average price is of $4.06 per gallon, hence:
[tex]H_0: \mu = 4.06[/tex]
At the alternative hypothesis, we test if the average is different of $4.06 per gallon, hence:
[tex]H_1: \mu \neq 4.06[/tex]
We have 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:
In this problem, considering the table, we have that the values of the parameters are: [tex]\overline{x} = 4.038, \mu = 4.06, s = 0.0533, n = 10[/tex].
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{4.038 - 4.06}{\frac{0.0533}{\sqrt{10}}}[/tex]
[tex]t = -1.31[/tex]
Using a t-distribution calculator, considering a two-tailed test, as we are testing if the mean is different of a value, 10 - 1 = 9 df and a standard significance level of 0.05, the critical value is [tex]|t^{\ast}| = 2.2622[/tex]
Since the absolute value of the test statistic is less than the critical value for the two-tailed test, there is not enough evidence to conclude that the price of the gas in Anne's city is different of $4.06.
You can learn more about the t-distribution at https://brainly.com/question/13873630