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 0.1 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 103 119 128 72 84 51 74 80 79

Before Lectures 118 112 94 73 83 58 74 78 87

Student 10 11 12 13 14 15 16 17 18

After Lectures 46 117 138 96 97 96 85 105 119

Before Lectures 42 114 106 116 94 99 79 107 104

a) What sampling distribution will you use?

b) How do you put in in your TI-84 Calculator to find the sample test statistic?

c) How do you put in in your TI-84 Calculator to find the P-value?

Answer 1

Claim: The lectures did not make some difference.

The null and alternative hypothesis is

$H0:\mu_{d}=0$

$H1:\mu_{d}\neq 0$

Level of significance = 0.1

a) Here data is given a before and after type so we have to us paired t-test.

T distribution we have to use as a sampling distribution.

b) By using the TI-84 calculator we have to find the sample test statistic.

Before	After	Difference(d)
118	103	15
112	119	-7
94	128	-34
73	72	1
83	84	-1
58	51	7
74	74	0
78	80	-2
87	79	8
42	46	-4
114	117	-3
106	138	-32
116	96	20
94	97	-3
99	96	3
79	85	-6
107	105	2
104	119	-15

First, enter difference(d) data into TI-84 calculator.

Click on STAT------>Edit ----->Enter Before values into L1.

Then click on STAT -------> TESTS -------> T-Test --------> Data ------>

$\mu0$ : 0

List: L1

Freq: 1

$\mu:\neq \mu0$

Calculate

We get test statistic = t = - 0.8849

C) See below test statistic value we get P-value = 0.3885

P-value > 0.1 we fail to reject null hypothesis.

Conclusion: The lectures did make some difference.