Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, and 9.8 liters. Use a 0.01 level of significance and assume that the distribution of contents is normal.

Accepted Solution

Answer:We fail to reject the null hypothesis that the average content of containers of the lubricant is 10 liters, this at the significance level of 0.01 Step-by-step explanation:Let X be the random variable that represents the content of a container of the lubricant. We have observed n = 10 values, [tex]\bar{x}[/tex] = 10.06 and s = 0.2459. We assume that X is normally distributed. We have the following null and alternative hypothesis [tex]H_{0}: \mu = 10[/tex] vs [tex]H_{1}: \mu \neq 10[/tex] (two-tailed alternative) We will use the test statistic Β [tex]T = \frac{\bar{X}-10}{S/\sqrt{10}}[/tex] because we have a small sample size. And the observed value is [tex]t = \frac{10.06-10}{0.2459/\sqrt{10}} = 0.7716[/tex] if [tex]H_{0}[/tex] is true, then T has a t distribution with n-1 = 9 degrees of freedom. The rejection region for a two-tailed alternative and a significance level of 0.01 is given by RR = {t | t < -3.2498 or t > 3.2498}, where 3.2498 is the value such that there is an area of 0.005 above this number and under the density of the t distribution with 9 df. Because the observed value 0.7716 does not fall inside RR, we fail to reject the null hypothesis.