Question

In: Statistics and Probability

3. A teacher believes that a new method will improve students’ reading ability. For 8 weeks,...

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.

Solutions

Expert Solution

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.


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