Chapter 11.2, Problem 42E

a)

To determine

To state the null and alternative hypotheses.

Answer to Problem 42E

  $H_0$ : There is no association between Opinion about the astrology and degree held.

  $H_a$ : There is association between Opinion about the astrology and degree held.

Explanation of Solution

Given:

		Degree held
		Associate's	Bachelor's	Master's
Opinion about astrology	Not at all scientific	169	256	114
Very or sort of scientific	65	65	18

Calculation:

The null and alternative hypotheses:

  $H_0$ : There is no association between Opinion about the astrology and degree held.

  $H_a$ : There is association between Opinion about the astrology and degree held

b)

To determine

To calculate expected counts.

Explanation of Solution

Given:

		Degree held
		Associate's	Bachelor's	Master's
Opinion about astrology	Not at all scientific	169	256	114
Very or sort of scientific	65	65	18

Formula:

  $E=\frac{r\times c}{n}$

Where, r = row total, c = column total and n = grand total.

Calculation:

		Associate's	Bachelor's	Master's	Total
Not at all scientific	Observed	169	256	114	539
	Expected	183.59	251.85	103.56	539.00
Very or sort of scientific	Observed	65	65	18	148
	Expected	50.41	69.15	28.44	148.00
Total	Observed	234	321	132	687
	Expected	234.00	321.00	132.00	687.00

c)

To determine

To calculate value of chi square test statistic, df and p-value.

Answer to Problem 42E

The chi-square test statistic is 10.58. The degrees of freedom = df = 2 and p-value = 0.0050

Explanation of Solution

Given:

		Degree held
		Associate's	Bachelor's	Master's
Opinion about astrology	Not at all scientific	169	256	114
Very or sort of scientific	65	65	18

Calculation:

		Associate's	Bachelor's	Master's	Total
Not at all scientific	Observed	169	256	114	539
	Expected	183.59	251.85	103.56	539.00
	(O - E)² / E	1.16	0.07	1.05	2.28
Very or sort of scientific	Observed	65	65	18	148
	Expected	50.41	69.15	28.44	148.00
	(O - E)² / E	4.22	0.25	3.83	8.30
Total	Observed	234	321	132	687
	Expected	234.00	321.00	132.00	687.00
	(O - E)² / E	5.38	0.32	4.88	10.58
		10.58	chi-square
		2	df
		.0050	p-value

Therefore, the chi-square test statistic is 10.58. The degrees of freedom = df = 2 and p-value = 0.0050.

d)

To determine

To state the conclusion.

Answer to Problem 42E

There is convincing evidence that there is association between Opinion about the astrology and degree held.

Explanation of Solution

Given:

The chi-square test statistic is 10.58. The degrees of freedom = df = 2 and p-value = 0.0050.

Calculation:

Decision: P-value < 0.05, reject H0.

Conclusion: There is convincing evidence that there is association between Opinion about the astrology and degree held.