Answer:
Let x = number of units produced per hour before the line was modified , y = number of units produced per hour after the line was modified.
The null and alternative hypotheses are :
H0: μbefore = μafter
H1: μbefore ≠ μafter
Step 1: . Calculate the difference (D = y − x) between the two observations on each pair.
See attachment for (D=y-x)
step 2: Calculate ∑D and ∑D2
From table we get ∑D = -10 & ∑D2 = 72
Step 3: Put all the value in test statistic "t"
t = D/√nD^2-D^2/n - 1
t =-10/√6.72-(-10)^2/6-1
t= -10√66.4
= -10/8.1486
= -1.227
Step 4: Compare tcal and ttab
At α = 0.05
t0.05 for 5 d.f. = 3.1634 (two tail test)
Hence | tcal | < t0.05
So, we accept the hypothesis.
So we conclude that the modified (after) layout has not increased worker productivity at 5% level of significance.
Answer:
For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.02 g/mi
Standard Deviation, σ = 0.01 g/mi
Sample size, n = 81
We are given that the distribution of level of nitrogen oxides is a bell shaped distribution that is a normal distribution.
Standard error due to sampling:
Formula:
We have to find the value of x such that the probability is 0.01
P(X > x)
Calculation the value from standard normal z table, we have,
For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.
Question 47= 33 ounces
Question 48= 98 Fahrenheit
Question 49= 127 millimeters
Last question= 2 inches
Twenty four plus (negative nine) minus forty-four plus thirty three minus seventeen.
Let us say that:
K = age of Kristen
B = age of Ben
From the problem, we make the equations:
eqtn 1: K + B = 32
eqtn 2: (K – 4) = 2 (B – 4)
Simplifying eqtn 2:
K – 4 = 2 B – 8
K = 2 B – 4
Plugging in this to eqtn 2:
(2 B – 4) + B = 32
3 B – 4 = 32
3 B = 36
B = 12
From eqtn 2:
K = 2 B – 4 = 2 (12) – 4 = 20
So Kristen is 20 while Ben is 12.
