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.

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

(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

susie has 7 apples

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !