a)
To state the null and alternative hypotheses.
a)

Answer to Problem 42E
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:
b)
To calculate expected counts.
b)

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:
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 calculate value of chi square test statistic, df and p-value.
c)

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 state the conclusion.
d)

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.
Chapter 11 Solutions
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