Answer:
The 90% confidence interval is (493.1903, 550.2097), and the critical value to construct the confidence interval is 2.0150
Step-by-step explanation:
Let X be the random variable that represents a measurement of helium gas detected in the waste disposal facility. We have observed n = 6 values,
= 521.7 and S = 31.6368. We will use
as the pivotal quantity. T has a
distribution with 5 degrees of freedom. Then, as we want a 90% confidence interval for the mean level of helium gas present in the facility, we should find the 5th quantile of the t distribution with 5 degrees of freedom, i.e.,
, this value is -2.0150. Therefore the 90% confidence interval is given by
, i.e.,
(493.1903, 550.2097)
To find the 5th quantile of the t distribution with 5 degrees of freedom, you can use a table from a book or the next instruction in the R statistical programming language
qt(0.05, df = 5)
The manager can do both but if he does 10 he will have 25 groups but if he does 5 he will have 50 groups so the simple way is 10 so he will have 25 groups
There's two ways, you can either multiply 11 and 7 to get 77 and then divide by 2, OR, you can divide 11 by 2 and get 5.5 and multiply that by 7, hope this makes sense, if you don't understand just comment and I'll explain it more :)
Answer:
Step-by-step explanation: okay each hour she makes 120 cookies right then 2 hours would be 240 and three hours would be 360 and 4 would be 480 and 5 would be 600 hundred