Question

Test the hypothesis that the average content of containers of a particular lubricant is 10.110.1 liters if the contents of a random sample of 1010 containers are 10.110.1​, 9.79.7​, 10.510.5​, 10.410.4​, 9.89.8​, 9.79.7​, 9.49.4​, 9.99.9​, 10.710.7​, and 9.79.7 liters. Use a 0.100.10 level of significance and assume that the distribution of contents is normal.

Answer 1

Answer:

Step-by-step explanation:

Given is a sample of contents of a random sample of 10 containers as follows:

10.1 9.7​, 10.5​, 10.4​, 9.8​, 9.7​, 9.4​, 9.9​, 10.7​, and 9.7 liters.

To check whether the mean is 10.1 litres.

Set up hypothesis as

$H_0: \bar x = 10.1\\H_a:\bar x \neq 10.1$

(two tailed test at 10% level of significance)

Sample mean =9.9

Sample std dev=0.420185

Std error =0.1329

n = 10

df = 9

Mean difference = 9.9-10.1 =-0.2

Test statistic t = -0.2/0.1329 =-1.505

p value= 0.1666

Since p value is >0.10 we accept null hypothesis

Conclusion:

the average content of containers of a particular lubricant is 10.1 liters can be taken as true at 90% level of confidence