In: Statistics and Probability
The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.
Ceremonial Ranking | Cooking Jar Sherds | Decorated Jar Sherds (Noncooking) | Row Total |
A | 89 | 46 | 135 |
B | 96 | 49 | 145 |
C | 78 | 76 | 154 |
Column Total | 263 | 171 | 434 |
Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Ceremonial ranking and pottery type are
not independent.
H1: Ceremonial ranking and pottery type are
independent.H0: Ceremonial ranking and pottery
type are independent.
H1: Ceremonial ranking and pottery type are
independent. H0:
Ceremonial ranking and pottery type are independent.
H1: Ceremonial ranking and pottery type are not
independent.H0: Ceremonial ranking and pottery
type are not independent.
H1: Ceremonial ranking and pottery type are not
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 normalStudent's tbinomial
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.)
p-value > 0.1000.050 < p-value < 0.100 0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005
(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 > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.
(a) The level of significance = 0.05
The null and alternate hypotheses.
H0: Ceremonial ranking and pottery type are independent.
H1: Ceremonial ranking and pottery type are not independent.
(b)
Ceremonial Ranking | Cooking Jar Sherds | Decorated Jar Sherds (Noncooking) | Row Total | |
A | 89 | 46 | 135 | |
B | 96 | 49 | 145 | |
C | 78 | 76 | 154 | |
Column Total | 263 | 171 | 434 | |
Expected Frq= (RT*CT)/GT | O= observed | E=Expected | ||
RT= | Row Total | |||
CT= | Column Total | |||
GT= | 434 | |||
E1= | 135*263/434 | |||
E1= | 81.80875576 | |||
Calculation table for Chi-Square | ||||
O | E | O-E | (O-E)2/E | |
89 | 81.8087558 | 7.19124424 | 0.63213275 | |
46 | 53.1912442 | -7.19124424 | 0.972227562 | |
96 | 87.8686636 | 8.13133641 | 0.75247112 | |
49 | 57.1313364 | -8.13133641 | 1.157309384 | |
78 | 93.3225806 | -15.3225806 | 2.515805671 | |
76 | 60.6774194 | 15.3225806 | 3.869338547 | |
SUM= | 434 | 434 | 9.899285034 | |
Chi-Square = | sum of | (O-E)2/E | ||
Chi-Square = | 9.899 |
Test Statistics = 9.899285034
Are all the expected frequencies greater than 5?
Yes
What sampling distribution will you use?
Chi-Square
Degrees of freedom = (r-1)(c-1)
(3-1)(2-1)
= 2
The P-Value = 0.00708
0.005 < p-value < 0.010
Since the P-value ≤ α, we reject the null hypothesis
At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.