5. To assess the usefulness of several learning methods, a cohort of second grad

Question

5. To assess the usefulness of several learning methods, a cohort of second grade students were randomly assigned to learn reading strategies using one of three methods: sight phonics, and write-to-read. Each student was given a pre- and post-test, and improve- ment between the two tests was reported, based on certain educational metrics, as one of no improvement, moderate improvement, or high improvement. The data are given in the following two-way table. (This data set is also available on Blackboard as Reading.txt.) Method Sight Phonics Write-to-Read 43 51 60 None 57 49 38 85 61 71 Improvement Moderate High Determine, at the pendent 0.05 level. if learning method and improvement level are inde-

Explanation / Answer

5.

Given table data is as below

5.

Given table data is as below

MATRIX col1 col2 col3 TOTALS row 1 57 85 43 185 row 2 49 61 51 161 row 3 38 71 60 169 TOTALS 144 217 154 N = 515 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 col3 row 1 row1*col1/N row1*col2/N row1*col3/N row 2 row2*col1/N row2*col2/N row2*col3/N row 3 row3*col1/N row3*col2/N row3*col3/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 row 1 51.728 77.951 55.32 row 2 45.017 67.839 48.144 row 3 47.254 71.21 50.536 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 57 51.728 5.272 27.794 0.537 85 77.951 7.049 49.688 0.637 43 55.32 -12.32 151.782 2.744 49 45.017 3.983 15.864 0.352 61 67.839 -6.839 46.772 0.689 51 48.144 2.856 8.157 0.169 38 47.254 -9.254 85.637 1.812 71 71.21 -0.21 0.044 0.001 60 50.536 9.464 89.567 1.772 ^2 o = 8.713 ------------------------------------------------------------------

set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =9.488
since our test is right tailed,reject Ho when ^2 o > 9.488
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 8.713
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 9.488
we got | ^2| =8.713 & | ^2 | =9.488
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.069


ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 8.713
critical value: 9.488
p-value:0.069
decision: do not reject Ho

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