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.

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, = 10.06 and s = 0.2459. We assume that X is normally distributed.

We have the following null and alternative hypothesis

vs (two-tailed alternative)

We will use the test statistic  

because we have a small sample size. And the observed value is

if 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.

i just took the test, the correct answer is .. 31.5 in.