After a large fund drive to help the Boston City Library, the following information was obtained from a random sample of contributors to the library fund. Using a 1% level of significance, test the claim that the amount contributed to the library fund is independent of ethnic group. Number of People Making Contribution
Ethnic Group $1-50 $51-100 $101-150 $151-200 Over $200 Row Total
A 77 63 54 39 18 251
B 90 48 76 26 24 264
C 73 67 59 37 30 266
D 108 82 70 50 30 340
Column Total 348 260 259 152 102 1121

Required:
a. What is the level of significance?
b. Find the value of the chi-square statistic for the sample.
c. Find or estimate the P-value of the sample test statistic.
d. Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

α = 0.01 ; 2 - tailed

χ² = 16.998

Pvalue = 0.1497

Fail to reject H0

Step-by-step explanation:

Given the data above :

Number of People Making Contribution

Ethnic Group $1-50 $51-100 $101-150 $151-200 Over $200 Row Total

A 77 63 54 39 18 251

B 90 48 76 26 24 264

C 73 67 59 37 30 266

D 108 82 70 50 30 340

Column Total 348 260 259 152 102 1121

1.) The level of significance, α/2 = 0.01/2

2.)

The hypothesis :

H0 : Contribution and Ethnic group are independent

H1 : Contribution and Ethnic group are not independent

The Chisquare statistic :

χ² = (Observed - Expected)²/ Expected

The expected value for each cell is calculated thus :

(Row total * column total) / sum total

The expected values :

77.9197 58.2159 57.992 34.0339 22.8385

81.9554 61.231 60.9955 35.7966 24.0214

82.5763 61.6949 61.4576 36.0678 24.2034

105.549 78.8582 78.5549 46.1017 30.9367

Using the χ² formula above ;

χ² values for each cell are :

0.010855 0.393154 0.274794 0.724635 1.02509

0.789645 2.85902 3.69099 2.68108 0.0000191

1.11055 0.456179 0.098278 0.0240936 1.38826

0.056933 0.125176 0.93165 0.32963 0.028359

Taking the sum :

0.010855+0.393154+0.274794+0.724635+1.02509+0.789645+2.85902+3.69099+2.68108+0.0000191+1.11055+0.456179+0.098278+0.0240936+1.38826+0.056933+0.125176+0.93165+0.3293+0.028359

χ² = 16.998

To obtain the Pvalue :

df = (Row - 1)(column - 1) = (5-1)*(4-1) = 4 * 3 = 12

Pvalue(16.998 ; 12) = 0.1497

If Pvalue < α ; Reject H0 ;

Since Pvalue > α ; WE fail to reject H0

where did you get them? if you picked from the grass, is is probably poisonous.

step-by-step explanation: