Answer:
Note: The full question is attached as picture below
a) Hо : p = 0.71
Ha : p ≠ 0.71
<em>p </em>= x / n
<em>p </em>= 91/110
<em>p </em>= 0.83.
1 - Pо = 1 - 0.71 = 0.29.
b) Test statistic = z
= <em>p </em>- Pо / [√Pо * (1 - Pо ) / n]
= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]
= 0.12 / 0.043265
= 2.77360453
Test statistic = 2.77
c) P-value
P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028
P-value = 0.0056
∝ = 0.01
P-value < ∝
Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.
Answer:
If the children divide evenly onto 3 buses, there will be 11 children on each bus
Step-by-step explanation:
Add all 3 classes to ind total amount of students
Ms. Noels class has 10 students
Mr. Grubbs class has 9 students
Ms. Ackerman's class has 14 students
<u>Total Students:</u> 10 students + 9 students + 14 students
33 total students
Dividing 33 students in 3 busses is 33 / 3 which equals 11
"AC/DF = 5/2" is correct.
5x10^9
Just find out how many zeros there are
