Question

TUDENT NAME TAKE IT HOME Students take many fmn, which is given to students in grades 1 -8. Student scores are reported on a scale that runs achievenent tests. One test is the lowa Test of Basic Skills (ITBS) for Reading from o to 13. Higher scores indicate better performance. The North Snowshoe Community Schools is a school district in lowa. Imagine that the district is evaluating the reading program for students in the 3"d and 4" grades. The district has scores for ten randomly selected srudents who took the ITBS Reading Comprehension test in third grade and again in the fourth grade. The scores of the ten students for each year are listed below. The grade that students were in when they took the test is provided on the left. Assume that the population of differences in reading test scores is normally distributed Student# | 1 | 2 | 3 | 4 | S | 6 | 7 | 8 |9|10 3rd Grade Score 85 7.26.7 79 756.8 6.6 6.5 4.4 9.5 4th Grade Score 9.1 7.56.6 88 8.265 7.4 6.9 4.39.7 Differences Is this an example of paired data? Explain why it is or is not an example of paired data. 2 Suppose that a student's reading test score improved between 3* and 4h grade. Would you expect the difference (4th grade score minus 3 grade score) to be zero, positive, or negative? Explain why you chose your answer. 3 Determine the difference for each student (4h grade score minus 3" grade score). ill in those differences in the last row of the table. 4 Determine the mean of the sample differences. Statway Version 3.0, 2016 The Carnegie Foundation for the Advancement of Teaching 9.2.3 18 231

Answer 1

1:

yes it is paired data set.

2:

The difference should eb positive because we supposed that reading score increases for 4th graders.

3:

Following is the completed data:

3rd grade score	4th grade score	d=4th grade score- 3rd garde score
8.5	9.1	0.6
7.2	7.5	0.3
6.7	6.6	-0.1
7.9	8.8	0.9
7.5	8.2	0.7
6.8	6.5	-0.3
6.6	7.4	0.8
6.5	6.9	0.4
4.4	4.3	-0.1
9.5	9.7	0.2
Total		3.4

4:

The mean difference:

$\bar{d}=\frac{3.4}{10}=0.34$

5:

Following table shows the calcualtions:

3rd grade score	4th grade score	d=4th grade score- 3rd garde score	(d-mean)^2
8.5	9.1	0.6	0.0676
7.2	7.5	0.3	0.0016
6.7	6.6	-0.1	0.1936
7.9	8.8	0.9	0.3136
7.5	8.2	0.7	0.1296
6.8	6.5	-0.3	0.4096
6.6	7.4	0.8	0.2116
6.5	6.9	0.4	0.0036
4.4	4.3	-0.1	0.1936
9.5	9.7	0.2	0.0196
Total		3.4	1.544

$s_{d}=\sqrt{\frac{\sum \left ( d-\bar{d} \right )^{2}}{n-1}}=0.4142$

6)

The degree of freedom: df=n-1=9

The critical value for 95% confidence interval is:2.262

The margin of error is

$ME=t_{c}\cdot \frac{s_{d}}{\sqrt{n}}=2.262\cdot \frac{0.4142}{\sqrt{10}}=0.2963$

7:

The confidence interval is:

$\bar{d}\pm ME=0.34\pm 0.2963=(0.0437, 0.6363)$

8:

Since confidence interval does not zero so the difference is significant.