To test the results of a conventional versus a new approach to the teaching of

reading, 12 pupils were selected and matched according IQ, age, and present

reading ability (there are 6 pairs!). One from each of the pairs was assigned to

the conventional reading program and the other to the new reading program. At

the end of 6 weeks, their progress was measured by a reading test. Do the data

present sufficient evidence to indicate that the new approach is better than the

conventional approach at the alpha=0.05 level?
Pair Conventional New Difference

1 78 83 5

2 65 69 4

3 88 87 -1

4 91 93 2

5 72 78 6

6 59 59 0

To determine if the new approach to teaching reading is better than the conventional approach, we need to conduct a paired t-test. This test compares the differences between pairs of observations (in this case, the improvement in reading scores) to determine if there is a significant difference between the two approaches.

To perform the paired t-test, we first need to calculate the mean difference, the standard deviation of the differences, and the standard error of the mean difference.

Step 1: Calculate the mean difference.
- Add up all the differences: 5 + 4 + (-1) + 2 + 6 + 0 = 16
- Divide the sum by the number of pairs: 16 / 6 = 2.67

Step 2: Calculate the standard deviation of the differences.
- Subtract the mean difference from each difference, square the result, and sum them up:
[(5 - 2.67)^2 + (4 - 2.67)^2 + (-1 - 2.67)^2 + (2 - 2.67)^2 + (6 - 2.67)^2 + (0 - 2.67)^2] = 60.67
- Divide the sum by the number of pairs minus 1 (n-1): 60.67 / (6-1) = 12.13
- Take the square root: √12.13 = 3.48

Step 3: Calculate the standard error of the mean difference.
- Divide the standard deviation of the differences by the square root of the number of pairs: 3.48 / √6 ≈ 3.48 / 2.45 = 1.42

Finally, we can calculate the t-value using the formula:
t = (mean difference - hypothesized mean difference) / standard error

In this case, the hypothesized mean difference is 0 because we want to test if there is any difference between the approaches. Therefore:
t = 2.67 - 0 / 1.42 ≈ 1.88

Consulting the t-distribution table with (6-1) degrees of freedom and a significance level of 0.05 (alpha = 0.05 for 95% confidence), we find that the critical t-value is approximately 2.571.

Since our calculated t-value of 1.88 is less than the critical t-value of 2.571, we fail to reject the null hypothesis. This means that there is not enough evidence to indicate that the new approach is significantly better than the conventional approach at the 0.05 level of significance.

Therefore, based on the data provided, we do not have sufficient evidence to conclude that the new reading program is better than the conventional program.