A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random samp
1 answer:
Answer:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Step-by-step explanation:
Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
We find mean = 11.015
Sample std deviation = 3.157
a)
(Right tailed test)
Mean difference /std error = test statistic
p value =0.174
Since p >0.01, our alpha, fail to reject H0
Conclusion:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
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a. The first part asks for how many ways they can be seated together in a row. Therefore we want the permutations of the set of 6 people, or 6 factorial,
6! = 6 5
= 30 4
= 360 2 = 720 possible ways to order 6 people in a row
b. There are two cases to consider here. If the doctor were to sit in the left - most seat, or the right - most seat. In either case there would be 5 people remaining, and hence 5! possible ways to arrange themselves.
5! = 5 4
= 20 3
= 120 1 = 120 possible ways to arrange themselves if the doctor were to sit in either the left - most or right - most seat.
In either case there are 120 ways, so 120 + 120 = Total of 240 arrangements among the 6 people if the doctor sits in the aisle seat ( leftmost or rightmost seat )
c. With each husband on the left, there are 3 people left, all women, that we have to consider here.
3! = 3 2 6 ways to arrange 3 couples in a row, the husband always to the left
See, pls, the suggested decision, if it's possible check the arithmetic part.
