Question

A high school science teacher decided to give a series of lectures on current events. To determine if the lectures had any effect on awareness of current events, an exam was given to the class before the lectures, and a similar exam was given after the lectures. The scores follow. Use a 0.05 level of significance to test the claim that the lectures made no difference against the claim that the lectures did make some difference (one way or the other). Student 1 2 3 4 5 6 7 8 9 After Lectures 105 116 123 73 89 52 73 80 72 Before Lectures 116 119 96 71 86 55 78 78 86 Student 10 11 12 13 14 15 16 17 18 After Lectures 48 117 134 94 94 97 81 108 116 Before Lectures 48 114 104 114 92 91 77 104 104 (a) What is the level of significance? State the null and alternate hypotheses. Ho: Distributions are different. H1: Distributions are the same. Ho: Distributions are different. H1: Distributions are different. Ho: Distributions are the same. H1: Distributions are different. Ho: Distributions are the same. H1: Distributions are the same. (b) Compute the sample test statistic. (Use 2 decimal places.) What sampling distribution will you use? uniform normal chi-square Student's t (c) Find the P-value of the sample test statistic. (Use 4 decimal places.)

Answer 1

(a)

Level of significance = $\alpha$= 0.05

H0: Differences are the same.

H1:Differences are different

(b)

From the given data,values of d = After Lectures - Before Lectures are calculated as follows:

Student	After Lectures	Before Lectures	d = After Lectures
1	105	116	- 11
2	116	119	- 3
3	123	96	27
4	73	71	2
5	89	86	3
6	52	55	- 3
7	73	78	- 5
8	80	78	2
9	72	86	- 14
10	48	48	0
11	117	114	3
12	134	104	30
13	94	114	- 20
14	94	92	2
15	97	91	6
16	81	77	4
17	108	104	4
18	116	104	12

From the d values,the following statistics are calculated

n = 18

$\bar{d}$=39/18= 2.1667

s_d =12.2390

SE = s_d/$\sqrt{n}$

= 12.2390/$\sqrt{18}$

=2.8848

(i)

Test statistic is given by:

t = 2.1667/2.8848

= 0.75

So,

The Sample test statistic is:

t = 0.75

(ii)

Distribution used:

Student's t

(c)

ndf =n - 1 = 18 - 1= 17

By Technology, P - Value = 0.4635

So,

P - Value = 0.4635

Since P - Value is greater than $\alpha$, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that the lectures did make some difference (one wy or the other).