Answer:
t= 0.4933
t ≥ t ( 0.025 ,8 ) = 2.306
Since the calculated value of t= 0.4933 is less than t ( 0.025 ,8 ) = 2.306 therefore we accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the book had no effect on their scores.
Step-by-step explanation:
We state our null and alternative hypotheses as
H0: ud= 0 Ha: ud≠0
The significance level is set at ∝= 0.05
The test statistic under H0 is
t= d`/ sd/√n
which has t distribution with n-1 degrees of freedom
The critical region is t ≥ t ( 0.025 ,8 ) = 2.306
Computations
Student Scores before Scores after Difference d²
reading book ( after minus before)
1 720 740 20 400
2 860 860 0 0
3 850 840 -10 100
4 880 920 40 1600
5 860 890 30 900
6 710 720 10 100
7 850 840 -10 100
8 1200 1240 40 1600
<u>9 950 970 20 40</u>
<u>∑ 6930 8020 140 4840</u>
d`= ∑d/n= 140/9= 15.566
sd²= 1/8( 4840- 140²/9) = 1/8 (4840 - 2177.778) = 2662.22/8= 332.775
sd= 18.2422
t= 3/ 18.2422/ √9
t= 0.4933
Since the calculated value of t= 0.4933 is less than t ( 0.025 ,8 ) = 2.306 therefore we accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the book had no effect on their scores.
Solution is 10. How?
I dont know if it helps but tell me if it does
Three days,
1 1/4 = 5/4
1/2= 2/4
2/4+1/4= 3/4 the first day
5/4-3/4= 2/4
2/4 = 1/4 and 1/4 = two days
the first day+ two days= 3 days
9514 1404 393
Answer:
C 8 in
Step-by-step explanation:
The area is given by the formula ...
A = 1/2bh
Solving for h gives ...
h = 2A/b
Filling in the given values, we find h to be ...
h = 2(36 in²)/(9 in) = 72/9 in
h = 8 in
