Answer:
H0: μm − μw = 0
against the claim
Ha: μm − μw ≠ 0
Since the calculated value of z= 0.6177 does not lie in the critical region the null hypothesis is accepted that men and women have equal success in challenging calls.
Step-by-step explanation:
1) Let the null and alternate hypothesis be
H0: μm − μw = 0
against the claim
Ha: μm − μw ≠ 0
2) The significance level is set at 0.05
3) The critical region is z > + 1.96 and z< -1.96
4) The test statistic
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
Here p1= 411/ 1390= 0.2956 and p2= 213/753=0.2829
pc = 411+ 213/1390+753
pc=624/2143
pc= 0.2912
qc= 1-pc= 1-0.2912=0.7088
5) Calculations
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
z= 0.2956-0.2829/√ 0.2912*0.7088( 1/1390+ 1/753)
z= 0.0127/ √0.2064 (0.00204)
z= 0.0127/0.02056
z= 0.6177
6) Conclusion
Since the calculated value of z= 0.6177 does not lie in the critical region the null hypothesis is accepted that men and women have equal success in challenging calls.
As we can see from the table there is a pattern to the whole data in the table. A closer look reveals that when we multiply the number of students, n by the cost per student, c we get a constant number, 72. Let us see what we mean here:
Thus, the function which models the data is the 
.
Now, let us move on to the next part of the question:
If 12 students go on the trip then the cost per student as per this model will be:
dollars=$6
Therefore, out of the given options, the option the correct option is number B.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data :
Technician __Shutdown
Taylor, T___4
Rousche, R _ 3
Hurley, H__ 3
Huang, Hu___2
Gupta, ___ 5
The Numbe of samples of 2 possible from the 5 technicians :
We use combination :
nCr = n! ÷ (n-r)!r!
5C2 = 5!(3!)2!
5C2 = (5*4)/2 = 10
POSSIBLE COMBINATIONS :
TR, TH, THu, TG, RH, RHu, RG , HHu, HG, HuG
Sample means :
TR = (4+3)/2 = 3.5
TH = (4+3)/2 = 3.5
THu = (4+2) = 6/2 = 3
TG = (4 + 5) = 9/2 = 4.5
RH = (3+3) = 6/2 = 3
RHu = (3+2) /2 = 2.5
RG = (3 + 5) = 8/2 = 4
HHu = (3+2) = 2.5
HG = (3+5) = 8/2 = 4
HuG = (2+5) / 2 = 3.5
Mean of sample mean (3.5+3.5+3+4.5+3+2.5+4+2.5+4+3.5) / 10 = 3.4
Population mean :
(4 + 3 + 3 + 2 + 5) / 5 = 17 /5 = 3.4
Population Mean and mean of sample means are the same.
This distribution should be approximately normal.
Answer:
Step-by-step explanation:
Rent, wages and car insurance don't vary much. Grocery expenses depend upon the particular groceries purchased each time one goes to the grocery store, and are least likely to be steady / constant from week to week.
