3,952 x 20=79,140
P.S. Not be rude but do you have a caculator?
-Nana
Answer:
Value of test statistics = 2.30
Step-by-step explanation:
We are given that for a sample of 35 items from a population for which the standard deviation is 20.5, the sample mean is 458.
Also, Null Hypothesis, 
: 
= 450
Alternate Hypothesis, 
: 
450
The test statistics that will be used here is One sample t-test statistics;
T.S. = 
~ 
where, 
= sample mean = 458
s = sample standard deviation = 20.5
n = sample of items = 35
So, test statistics = 
~ 
= 2.30
Therefore, the value of test statistics is 2.30 .
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.
8.5 cm I’m guessing so pls if I’m wrong I’m rlly Srry :(
