Respuesta :
Using the t-distribution, it is found that since the test statistic is greater than the critical value, the sample information supports the suspicion that the average strength of corrugated fiberboards from Firm 1 is three pounds per square inch higher than that from Firm 2.
At the null hypothesis, we test if the average strength of corrugated fiberboards from Firm 1 is not three pounds per square inch higher than that from Firm 2, that is, we test if the subtraction of the means is at most 3, hence:
[tex]H_0: \mu_1 - \mu_2 \leq 3[/tex]
At the alternative hypothesis, we test if the difference is more than 3, that is:
[tex]H_1: \mu_1 - \mu_2 > 3[/tex]
The mean and the standard deviation for both tests are:
[tex]\mu_1 = 366, s_1 = 11.67[/tex]
[tex]\mu_2 = 357.4, s_2 = 4.27[/tex]
The standard errors are:
[tex]s_1 = \frac{11.67}{\sqrt{9}} = 2.9175[/tex]
[tex]s_2 = \frac{4.27}{\sqrt{10}} = 1.35[/tex]
The distribution of the difference has mean and standard deviation given by:
[tex]\overline{x} = \mu_1 - \mu_2 = 366 - 357.4 = 8.6[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.9175^2 + 1.35^2} = 3.2147[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 3[/tex] is the value tested at the null hypothesis.
It's value is:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{8.6 - 3}{3.2147}[/tex]
[tex]t = 1.742[/tex]
The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with 9 + 10 - 2 = 17 df and a significance level of 0.05 is [tex]t^{\ast} = 1.7396[/tex]
Since the test statistic is greater than the critical value, the sample information supports the suspicion that the average strength of corrugated fiberboards from Firm 1 is three pounds per square inch higher than that from Firm 2.
A similar problem is given at https://brainly.com/question/25562298
