The table below shows the average scores across a host of standardized tests in

Question

The table below shows the average scores across a host of standardized tests in a random sample of 100 school districts in the U.S. as well as the per-person incomes ("per capita incomes") of the people living inside the borders of each district. Each variable has two valucs - "low" or "high." Using this table, we can answer the research question: Are standardized test scores in U.S school districts related to per capita incomes there? Total Hi Low Hi Total Low 28 24 52 18 30 48 46 54 100 1. State the null and research hypotheses that you can use in answering the research question above. (6 pts 2, we will use the chi-square distribution to test your null hypothesis, and we want to be 95% confident in the outcome of our hypothesis test. Do a chi-square test of independence to test your null hypothesis. Explain what conclusion you can draw about standardized testing and incomes in U.S. school districts as a result of your testing. (Show all work, and use the back of this page or extra paper to answer this question.) (16 pts)

Explanation / Answer

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: no association b/w variables OR variables are independent

alternative, H1: association b/w variables OR variables are are dependent

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =3.8415

since our test is right tailed,reject Ho when ^2 o > 3.8415

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 2.6848

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415

we got | ^2| =2.6848 & | ^2 | =3.8415

make decision

hence value of | ^2 o | < | ^2 | and here we do not reject Ho

^2 p_value =0.1013

ANSWERS

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null, Ho: no association b/w variables OR variables are independent

alternative, H1: association b/w variables OR variables are are dependent

test statistic: 2.6848

critical value: 3.8415

p-value:0.1013

decision: do not reject Ho

we claim that no association b/w variables OR variables are independent

MATRIX col1 col2 TOTALS row 1 28 18 46 row 2 24 30 54 TOTALS 52 48 N = 100

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