Answer:
Part a:
Since the calculated z= 8.1649 falls in the critical region z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.
Part b:
P (Type II Error)= 0.6628
Step-by-step explanation:
The null and alternate hypothesis are
H0: μ1 ≥ μ2 against the claim Ha: μ1 < μ2
This is left tailed test
The critical region for this test at 0.10 ∝ is z < ± 1.28
The test statistic
z= x1-x2/ s/√n
z= 66-61/ 3/√24
z= 5/0.61238
z= 8.16486
Since the calculated z= 8.1649 falls in the critical region z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.
Part b:
P (Type II Error) = P ( accept H0/ H0 is false)
z ( ∝)= X-u/s/√n
1.28= X-61/0.61238
X= 61.7838
So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 61.7838
The probability of a Type II error is given by
P (z> 61.7838-66/24)
P (z>- 0.42161)
From the tables
P (z>- 0.42161)= 0.6628
For b. They are all correct, you can justify because all pairs are different versions of the same angle
Answer: Stephen gets 42 lb
Step-by-step explanation:
1 = Richard
3 = Stephen
14:?
1 x 14 = 14
3 x 14 = 42
For this case we have the following table:
x f(x)
<span><span><span>0 2
</span><span>1 5
</span><span>2 10
</span><span>3 17
</span></span></span> The equation that best fits the data in the table, for this case, is given by a quadratic function.
<span><span><span> </span></span></span>The quadratic function in its standard form is:
f (x) = x2 + 2x + 2
Answer:
f (x) = x2 + 2x + 2
Answer:
no
Step-by-step explanation:
No since the divisor and dividend don't change by the same factor of 10.