Respuesta :
Using the t-distribution, as we have the standard deviation for the samples, it is found that the data does not provide statistical evidence that there is a difference.
What are the hypotheses tested?
At the null hypotheses, it is tested if there is no difference in the means, that is, the subtraction is of 0, hence:
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
At the alternative hypotheses, it is tested if there is a difference, hence:
[tex]H_1: \mu_1 - \mu_2 \neq 0[/tex]
What are the mean and the standard error of the distribution of differences?
For each sample, they are given by:
[tex]\mu_1 = 1.82, s_1 = \frac{1.51}{\sqrt{100}} = 0.151[/tex]
[tex]\mu_2 = 1.65, s_2 = \frac{1.39}{\sqrt{100}} = 0.139[/tex]
Hence, for the distribution of differences, they are given by:
[tex]\overline{x} = \mu_1 - \mu_2 = 1.82 - 1.65 = 0.17[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.151^2 + 0.139^2} = 0.205[/tex]
What is the test statistic?
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Then:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{0.17 - 0}{0.205}[/tex]
[tex]t = 0.83[/tex]
What is the decision?
Considering that it is a two-tailed test, as we are testing if the mean is different of a value, with 100 + 100 - 2 = 198 df and a significance level of 0.02, the critical value is of [tex]|t^{\ast}| = 2.3453[/tex].
Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis, which means that the data does not provide statistical evidence that there is a difference.
More can be learned about the t-distribution at https://brainly.com/question/13873630
