In: Math
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 | 91 | 44 | 135 |
B | 96 | 49 | 145 |
C | 81 | 73 | 154 |
Column Total | 268 | 166 | 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 not
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
independent.
H0: Ceremonial ranking and pottery type are
independent.
H1: Ceremonial ranking and pottery type 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?
Yes
No
What sampling distribution will you use?
Student's t
uniform
normal
chi-square
binomial
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 > 0.100
0.050 < p-value < 0.100
0.025 < p-value < 0.050
0.010 < p-value < 0.025
0.005 < p-value < 0.010
p-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.
ANSWER :
From the given data we have to determine the Ceremonial Ranking and Pottery type by using Chi- square test at a level of significance =0.05
H0 : Ceremonial ranking and pottery type
are independent.
H1 : Ceremonial ranking and pottery type
are not independent.
(b) Applying Chi- square test
Expected | Ei = row total x column total / grand total | Cooking | Decorated | Total |
A | 135*268 / 434 = 83.36 | 135*166 / 434 = 51.635 | 135 | |
B | 145*268 /434 = 89.539 | 145*166 / 434 = 55.46 | 145 | |
C | 154*268 / 434 = 95.096 |
154*166 / 434 = 58.90 |
154 | |
Total | 268 | 166 | 434 | |
Chi square X2 |
= ( Oi - Ei )2 / Ei where Oi = observed Frequency Ei = expected Frequency |
Cooking | Decorated | Total |
A | 0.7002 | 1.1257 | 1.8259 | |
B | 0.4662 | 0.7524 | 1.2186 | |
C | 2.0894 | 3.3753 | 5.4647 | |
TOTAL | 3.2558 | 5.2534 | 8.5092 |
Chi - Square statistic for the sample = 8.5092
Expected Frequencies are Greater than 5 : Yes
What sampling distribution will you
use?
answer : I will choose Chi square
Degrees Of Freedom = ( rows-1)* ( columns -1) = (3-1)*(2-1) = 2
(C)
0.05 < p- value < 0.010
Here the sample test statistic value is 8.5092 So at the level of significance where p value it is less than 0.010.
(D) From the above where p - value is very less it can be ignored
Since the P-value ≤ α, we reject the null hypothesis.
(E)
At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.