Answer:
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
Step-by-step explanation:
Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.
Let five years ago be group I X and as of now be group II Y
(Two tailed test at 5% level of significance)
Group I Group II combined p
n 270 300 570
favor 120 140 260
p 0.4444 0.4667 0.4561
Std error for differene = 
p difference = -0.0223
Z statistic = p diff/std error = -1.066
p value =0.2864
Since p value >0.05, we accept null hypothesis.
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
Answer:
SITE A
Step-by-step explanation:
Given :
proposed-site Area-Served
1 2 3 4
A 5.2 4.4 3.6 6.5
B 6.0 7.4 3.4 4.0
C 5.8 5.9 5.9 5.8
D 4.3 4.8 6.5 5.1
Area 1 2 3 4
Number-runs 150 65 175 92
Computing the weighted average for the 4 sites :
Site A:
((150*5.2) + (65*4.4) + (175*3.6) + (92*6.5)) / (150 + 65 + 175 + 92)
= 2294 / 482
= 4.7593
Site B:
((150*6.0) + (65*7.4) + (175*3.4) + (92*4.0)) / (150 + 65 + 175 + 92)
= 2344/ 482
= 4.863
Site C:
((150*5.8) + (65*5.9) + (175*5.9) + (92*5.8)) / (150 + 65 + 175 + 92)
= 2819.6/ 482
= 5.850
Site D:
((150*4.3) + (65*4.8) + (175*6.5) + (92*5.1)) / (150 + 65 + 175 + 92)
= 2563.7/ 482
= 5.319
From the weighted average response time computed for the different sites ;
The best location for the emergency facility would be one with the least average response time; which is SITE A.
Answer: 2/3
Step-by-step explanation:
5/6 divided 5/4
5 times 4 and 6 times 5
equals 20/30
simplify 20/30 =2/3
Answer: 0,2 or 20%
Step-by-step explanation:
