11.7 Review Questions

Wanhua Su

11.7 Review Questions

An American roulette wheel contains 18 red numbers, 18 black numbers, and 2 green numbers. The following table shows the frequency with which the ball landed on each color in 300 trials.

Color Red Black Green

Frequency 140 120 40

At the 5% significance level, do the data suggest that the wheel is out of balance?
A gambler thinks a die may be loaded and that the six numbers are not equally likely. To test his suspicion, he rolled the die 150 times and obtained the data in the following table.

Number 1 2 3 4 5 6

Frequency 23 26 23 21 31 26

Do the data provide sufficient evidence to conclude that the die is loaded? Perform the hypothesis test at the 5% significance level.
The following table reported the survey results on how members would prefer to receive ballots in annual elections. At the 5% significance level, do the data provide sufficient evidence to conclude that gender (column) and preference (row) are associated?

Male Female Total

Mail 60 30 90

Email 150 90 240

Both 70 40 110

N/A 80 50 130

Total 360 210 570

Show/Hide Answer

If the wheel is balanced, the chance of landing on a red number is $p_{1} = \frac{18}{18 + 18 + 2}$ , on a black number is $p_{2} = \frac{18}{18 + 18 + 2}$ , and on a green number is $p_{3} = \frac{2}{18 + 18 + 2}$ . If the wheel is out of balance, at least one proportion differs from the specified value. We can use the chi-square goodness of fit test to test this. The assumptions of a goodness-of-fit test are:
1. All expected frequencies are at least 1.
2. At most 20% of the expected frequencies are less than 5.
3. Simple random sample (if you need to generalize the conclusion to a larger population)
The expected frequencies are $E_{1} = n p_{1} = 300 \times \frac{18}{38} = 142.1053$ , $E_{2} = n p_{2} = 300 \times \frac{18}{38} = 142.1053$ , $E_{3} = n p_{3} = 300 \times \frac{2}{38} = 15.78947$ . All the expected frequencies are greater than 5. We assume the trials were conducted randomly.

Steps:
1. Hypotheses. $H_{0} : p_{1} = \frac{18}{38}, p_{2} = \frac{18}{38}, p_{3} = \frac{2}{38}$ versus $H_{a} :$ at least one proportion is different from the specified value.
2. Significance level $α = 0.05$
3. Test statistic.
  
  Color Observed Expected Contribution
  
  red 140 142.10526 0.0311891
  
  black 120 142.10526 3.4385965
  
  green 40 15.78947 37.1228070
  
  $\begin{aligned} χ_{o}^{2} & = \sum_{i = 1}^{3} \frac{(O_{i} - E_{i})^{2}}{E_{i}} = \frac{(140 - 142.10526)^{2}}{142.10526} + \frac{(120 - 142.10526)^{2}}{142.10526} + \frac{(40 - 15.78947)^{2}}{15.78947} \\ = 0.03118908 + 3.43859649 + 37.12280702 \\ = 40.59259 . \end{aligned}$
  with $d f = k - 1 = 3 - 1 = 2$ .
4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value $χ_{o}^{2}$ under the chi-square density curve with degrees of freedom $d f = k - 1 = 3 - 1 = 2$ . $P-value = P (χ^{2} \geq χ_{o}^{2}) = P (χ^{2} \geq 40.593) < P (χ^{2} \geq 10.597) = 0.005 .$
  That is, P-value<0.005.
  If the critical value approach is used, we need to find the rejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is $α$ , so the critical value is $χ_{α}^{2} = χ_{0.05}^{2} = 5.991 .$
5. Decision. We reject $H_{0}$ since P-value<0.005<0.05 ( $α$ ). If the critical approach is used, we reject $H_{0}$ since the observed test statistic $χ_{o}^{2} = 40.593 > 5.991$ falls in the rejection region.
6. Conclusion. At the 5% significance level, we have sufficient evidence that the wheel is out of balance.
If the die is balanced, the proportion of landing on each of 1, 2, $\dots$ , 6 should be the same, i.e., $p_{i} = \frac{1}{6}, i = 1, 2, \dots, 6$ . All expected frequencies are $E_{i} = n p_{i} = 150 \times \frac{1}{6} = 25 > 5$ , the assumptions for a chi-square goodness-of-fit test are satisfied.
Steps:
1. Hypotheses. $H_{0} : p_{i} = \frac{1}{6}, i = 1, 2, \dots 6$ versus $H_{a}$ : at least one proportion is equal to $\frac{1}{6}$ .
2. Significance level $α = 0.01$
3. Test statistic. The following table is useful.
  $χ_{o}^{2} = \sum_{all cells} \frac{(O - E)^{2}}{E} = \frac{(23 - 25)^{2}}{25} + \frac{(26 - 25)^{2}}{25} + \frac{(23 - 25)^{2}}{25} + \frac{(21 - 25)^{2}}{25} + \frac{(31 - 25)^{2}}{25} + \frac{(26 - 25)^{2}}{25} = 2.48$
  with the degrees of freedom of the test $d f = k - 1 = 6 - 1 = 5.$
4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value $χ_{o}^{2}$ under the chi-square density curve with degrees of freedom $d f = k - 1 = 6 - 1 = 5$ . P-value= $P (χ^{2} \geq χ_{o}^{2}) = P (χ^{2} \geq 2.48) > P (χ^{2} \geq 9.236) = 0.1$ , i.e., P-value>0.1.
  Or
  Rejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is $α$ , so the critical value is $χ_{α}^{2} = χ_{0.01}^{2} = 15.086$ .
5. Decision: We cannot reject $H_{0}$ since P-value>0.1>0.01 ( $α)$ .
  Or
  We cannot reject $H_{0}$ since $χ_{o}^{2} = 2.48 < 15.086$ falls in the non-rejection region.
6. Conclusion: At the 5% significance level, we do not have sufficient evidence that the die is loaded.
We should use the chi-square independence test. The expected frequencies are given in the following table. All the expected frequencies are greater than 5, the assumptions for a chi-square independence test are satisfied.

Preference Male Female

Mail 56.84211 33.15789

Email 151.57895 88.42105

Both 69.47368 40.52632

N/A 82.10526 47.89474

Steps:
1. Hypotheses. $H_{0} :$ gender and preference are independent. versus $H_{a}$ : gender and preference are associated.
2. Significance level $α = 0.05$ .
3. Test statistic.
  $\begin{aligned} χ_{o}^{2} & = \sum_{all cells} \frac{(O - E)^{2}}{E} \\ = \frac{(60 - 56.842)^{2}}{56.842} + \frac{(30 - 33.158)^{2}}{33.158} + \frac{(150 - 151.579)^{2}}{151.579} + \frac{(90 - 88.421)^{2}}{88.421} \\ + \frac{(70 - 69.474)^{2}}{69.474} + \frac{(40 - 40.526)^{2}}{40.526} + \frac{(80 - 82.105)^{2}}{82.105} + \frac{(50 - 47.895)^{2}}{47.895} \\ = 0.678 \end{aligned}$
  with the degrees of freedom of the test $d f = (r - 1) \times (c - 1) = (4 - 1) \times (2 - 1) = 3.$
4. P-value. The chi-square test is always right-tailed. P-value is the area to the right of the observed value $χ_{o}^{2}$ under the chi-square density curve with degrees of freedom $d f = 3$ . P-value= $P (χ^{2} \geq χ_{o}^{2}) = P (χ^{2} \geq 0.678) > P (χ^{2} \geq 6.251) = 0.1$ , i.e., P-value >0.10.
  Or
  Rejection region. The chi-square test is always right-tailed. Make sure that the area of the rejection region is $α$ , so the critical value is $χ_{α}^{2} = χ_{0.05}^{2} = 7.815$ .
5. Decision: Don’t reject $H_{0}$ , since $P -value > 0.1 > 0.05 (α) .$
  Or
  We cannot reject $H_{0}$ since $χ_{o}^{2} = 0.678 < 7.815$ falls in the non-rejection region.
6. Conclusion: At the 5% significance level, we do not have sufficient evidence that gender and preference are associated.

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	Male	Female	Total
Mail	60	30	90
Email	150	90	240
Both	70	40	110
N/A	80	50	130
Total	360	210	570

Color	Observed	Expected	Contribution
red	140	142.10526	0.0311891
black	120	142.10526	3.4385965
green	40	15.78947	37.1228070

Preference	Male	Female
Mail	56.84211	33.15789
Email	151.57895	88.42105
Both	69.47368	40.52632
N/A	82.10526	47.89474

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