Answer:
a) Statement is true
b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test (right)
c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
Step-by-step explanation:
a) as the significance level is = 0,02 that means α = 0,02 (the chance of error type I ) and the β (chance of type error II is
1 - 0.02 = 0,98
b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right
c) test statistic
Hypothesis test should be:
null hypothesis H₀ = 190
alternative hypothesis H₀ > 190
t(s) = ( μ - μ₀ ) / s/√n ⇒ t(s) = ( 202 - 190 )/(28/√100 )
t(s) = 12*10/28
t(s) = 4.286
That value is far away of any of the values found for 99 degree of fredom and between α ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
If we look table t-student we will find that for 99 degree of freedom and α = 0.02.
Answer:
0.4
Step-by-step explanation:
Let X be the random variable that represents the number of consecutive days in which the parking lot is occupied before it is unoccupied. Then the variable X is a geometric random variable with probability of success p = 2/3, with probability function f (x) = [(2/3)^x] (1/3)
Then the probability of finding him unoccupied after the nine days he has been found unoccupied is:
P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9). For a geometric aeatory variable:
P (X> = 10) = 1 - P (X <10) = 0.00002
P (X> = 9) = 1 - P (X <9) = 0.00005
Thus, P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9) = 0.00002 / 0.00005 = 0.4.
I dont really know. you dont know how big the walls are and how far 1 gallon goes so figure how far 1 gallon goes and then divide it by the 1,162.5
32 quizzes. If you make a proportion of quiz/weeks it would be 16/4 = x/8. Multiply by 2.
