The following table shows site type and type of pottery for a random sample of 6

Question

The following table shows site type and type of pottery for a random sample of 628 sherds at an archaeological location.

Use a chi-square test to determine if site type and pottery type are independent at the 0.01 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H0: Site type and pottery are not independent.
H1: Site type and pottery are not independent.H0: Site type and pottery are independent.
H1: Site type and pottery are not independent.    H0: Site type and pottery are not independent.
H1: Site type and pottery are independent.H0: Site type and pottery are independent.
H1: Site type and pottery are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

chi-squareuniform    Student's tbinomialnormal

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > , we fail to reject the null hypothesis.Since the P-value > , we reject the null hypothesis.    Since the P-value , we reject the null hypothesis.Since the P-value , we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is sufficient evidence to conclude that site and pottery type are not independent.At the 1% level of significance, there is insufficient evidence to conclude that site and pottery type are not independent.   

Pottery Type Site Type Mesa Verde
Black-on-White McElmo
Black-on-White Mancos
Black-on-White Row Total Mesa Top 80 59 50 189 Cliff-Talus 82 75 56 213 Canyon Bench 89 68 69 226 Column Total 251 202 175 628

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 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 = 2.269
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| =2.269 & | ^2 | =9.488
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.686
ANSWERS
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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: 2.269
critical value: 9.488
p-value:0.686
decision: do not reject Ho                                      

col1 col2 col3 row 1 80 59 50 189 row 2 82 75 56 213 row 3 89 68 69 226 TOTALS 251 202 175 N = 628

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