Question

Question 8 options:

A person’s educational attainment and age group was collected by the U.S. Census Bureau in 1984 to see if age group and educational attainment are related. The counts in thousands are in following table ("Education by age," 2013):

Table: Educational Attainment and Age Group

Education	Age Group					Row Total
	25-34	35-44	45-54	55-64	>64
Did not complete HS	5416	5030	5777	7606	13746	37575
Competed HS	16431	1855	9435	8795	7558	44074
College 1-3 years	8555	5576	3124	2524	2503	22282
College 4 or more years	9771	7596	3904	3109	2483	26863
Column Total	40173	20057	22240	22034	26290	130794

Do the data show that educational attainment and age are independent? Test at the 5% level.

(i) Which of the following statements correctly defines the null hypothesis H_O?

A. level of educational attainment is not independent of age  

B. level of educational attainment is independent of age

C. neither of these statements is correct for H_O

Enter letter corresponding to correct answer

(ii) Which of the following statements correctly defines the alternate hypothesis H_A?

A. level of educational attainment is not independent of age  

B. level of educational attainment is independent of age

C. neither of these statements is correct for H_O

Enter letter corresponding to correct answer

(iii) Enter the level of significance ? used for this test:

Enter in decimal form. Examples of correctly entered answers:  0.01    0.02    0.05    0.10

  (iv) Are any of the expected frequency values (for which H_O is true) for any cell in your table less than 5? Calculate the expected frequencies using technology, the select most correct answer:

A. yes - this will not affect results of hypothesis test

B. yes - this may invalidate results of hypothesis test

C. no - this will not affect results of hypothesis test

D. no - this may invalidate results of hypothesis test

Enter letter corresponding to correct answer. Recommend using (Chi-Square Test of Independence Applet) to create contingency table of expected and observed frequency values in order to answer this question.

(v) Using calculator, spreadsheet, or online calculator, determine test statistic:

Enter value to nearest integer.

Recommended online calculator:  Chi-Square Test of Independence Applet

(vi) Using tables, calculator, or spreadsheet: Determine and enter p-value corresponding to test statistic.

Enter value in decimal form rounded to nearest ten-thousandth. Examples of correctly entered answers:

0.0001     0.0021     0.0305     0.6004      0.8143    1.0000

Recommended online calculator:  Chi-Square Test of Independence Applet

(vii) Comparing p-value and ? value, which is the correct decision to make for this hypothesis test?

A. Accept H_o

B. Accept H_A

C. Reject H_o

D. Fail to reject H_o

Enter letter corresponding to correct answer.

  (viii) Select the statement that most correctly interprets the result of this test:

A. The result is statistically significant at .05 level of significance. There is not enough evidence to support the claim that level of educational attainment and age are NOT independent.

B. The result is statistically significant at .05 level of significance. Sufficient evidence exists to support the claim that level of educational attainment and age are NOT independent.

C. The result is not statistically significant at .05 level of significance. Sufficient evidence exists to support the claim that level of educational attainment and age are NOT independent.

D. The result is not statistically significant at .05 level of significance. There is not enough evidence to support the claim that level of educational attainment and age are NOT independent.

Enter letter corresponding to most correct answer

Answer 1

i) Null Hypothesis H0: level of educational attainment is independent of age

ii) ALternative Hypothesis H1: level of educational attainment is not independent of age  

iii) Level of Significance

iv) D. no - this may invalidate results of hypothesis test

The Expected frequencies are

v) Under H0, the test statistic is

The Calculations for Chi square test statistic is

	Obs freq(O)	Exp Freq( E )	(O-E)^2/E
	5416	11541.05	3250.678015
	5030	5762.05	93.00460817
	5777	6389.19	58.65792003
	7606	6330.01	257.2113599
	13746	7552.69	5078.599645
	16431	13537.2	618.5975268
	1855	6758.66	3557.788289
	9435	7494.27	502.575025
	8795	7424.86	252.837578
	7558	8859.01	191.0627734
	8555	6843.85	427.8343801
	5576	3416.9	1364.310577
	3124	3788.8	116.6488176
	2524	3753.7	402.8457495
	2503	4478.75	871.5798074
	9771	8250.89	280.0588072
	7596	4119.39	2934.127891
	3904	4567.74	96.44830651
	3109	4525.43	443.3333285
	2483	5399.55	1575.365336
Total	130794	130794	22374

vi) Degrees of freedom = (r-1)*(c-1) = (4-1)*(5-1) = 3*4 = 12

The P-Value is < 0.00001

vii) Since p value is less than significance level, Reject H0

viii) B. The result is statistically significant at .05 level of significance. Sufficient evidence exists to support the claim that level of educational attainment and age are NOT independent.