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.
Answer:
Step-by-step explanation:
2915 can stay as it is; 15.458 should be rounded down to 15. Then we have
2915-15, or 2900 (which is a very reasonable estimate).
Answer:
Left 3, down 2
Step-by-step explanation:
(6,4) --> (3,2)
-3, -2
Left 3, down 2
8.1x - 4.5y = 5.4.....subtract 8.1x from both sides
-4.5y = -8.1x + 5.4...divide both sides by -4.5
y = (- 8.1/-4.5)x + 5.4 / -4.5...simplify
y = 1.8x - 1.2 <==
