In: Statistics and Probability
3. A teacher believes that a new method will improve students’ reading ability. For 8 weeks, she teaches a class of 21 students using these new methods. Meanwhile a colleague uses “traditional” methods to teach his 23 students. At the end of the 8 weeks, all the students are given the Degree of Reading Power Test. Here are the scores:
New Method students
24 |
43 |
58 |
71 |
43 |
49 |
61 |
44 |
67 |
49 |
53 |
56 |
59 |
52 |
62 |
54 |
57 |
33 |
46 |
43 |
57 |
Traditional Method students
42 |
43 |
55 |
26 |
62 |
37 |
33 |
41 |
19 |
54 |
20 |
85 |
46 |
10 |
17 |
60 |
53 |
42 |
37 |
42 |
55 |
28 |
48 |
Are we justified in using the t procedures? Explain.
Give a significance test to check the teacher’s theory. Include all relevant components. (you may wish to list some “background” items first)
Give a 99% confidence interval for the mean difference between the two groups of students
Draw an overall conclusion based on parts B and C.
a) To use t procedures for sample size of atleat 15 there should not be a strong skewness in the data.
If skewness is < -1 and > 1 skewness is strong.
Skewness = where is the mean of sample is the observation.
S.No | News Method | Traditional Method |
1 | 24 | 10 |
2 | 33 | 17 |
3 | 43 | 19 |
4 | 43 | 20 |
5 | 43 | 26 |
6 | 44 | 28 |
7 | 46 | 33 |
8 | 49 | 37 |
9 | 49 | 37 |
10 | 52 | 41 |
11 | 53 | 42 |
12 | 54 | 42 |
13 | 56 | 42 |
14 | 57 | 43 |
15 | 57 | 46 |
16 | 58 | 48 |
17 | 59 | 53 |
18 | 61 | 54 |
19 | 62 | 55 |
20 | 67 | 55 |
21 | 71 | 60 |
22 | 62 | |
23 | 85 |
Skewness of new method = -0.626 moderate skewness
Skewness of tradional method = 0.309 approximately symmetric
Since there is no strong skewness in both of the datasets we can use t procedures.
b) Hypothesis
where is the population mean for new method
is the population mean for traditional method
Assuming equal variances of the population i.e., and %
test statistic t0
for the given data
for t0 = 2.265 and 42 degrees of freedom P - Value = 0.014 hence reject null hypothesis.
c) 99% confidence interval
t0.05,42 = 1.682
Substituing the values
d) Since the null hypothesis ie rejected test is statistically significant. The CI does not contain zero and the range is greater than zero hence from this we can conclude that new method of teaching is better than traditional method of teaching.