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In a town of 5,832 people, the mayor wants to determine if there is a difference in the proportion of voters aged 18-30 and 31-40 who support an increase in the food tax.

An assistant surveys:
- 85 voters aged 18-30, with 23 supporting the increase.
- 70 voters aged 31-40, with 56 supporting the increase.

Assuming the conditions for inference have been met, what is the 99% confidence interval for the difference in proportions of voters who support the increase in the food tax?

A. [tex]\((0.27-0.20) \pm 1.96 \sqrt{\frac{0.27(1-0.27)}{85}+\frac{0.20(1-0.20)}{70}}\)[/tex]

B. [tex]\((0.73-0.80) \pm 1.96 \sqrt{\frac{0.73(1-0.73)}{70}+\frac{0.80(1-0.80)}{85}}\)[/tex]

C. [tex]\((0.27-0.20) \pm 2.58 \sqrt{\frac{0.27(1-0.27)}{70}+\frac{0.20(1-0.20)}{85}}\)[/tex]

D. [tex]\((0.73-0.80) \pm 2.58 \sqrt{\frac{0.73(1-0.73)}{85}+\frac{0.80(1-0.80)}{70}}\)[/tex]

Sagot :

GinnyAnswer
To find the 99% confidence interval for the difference in proportions of voters who would support the increase in the food tax for the different age groups, follow these step-by-step calculations:

1. Sample Proportions:
- [tex]\( \hat{p}_1 \)[/tex] is the proportion of voters aged 18-30 who support the increase: [tex]\( \hat{p}_1 = \frac{62}{85} = 0.73 \)[/tex].
- [tex]\( \hat{p}_2 \)[/tex] is the proportion of voters aged 31-40 who support the increase: [tex]\( \hat{p}_2 = \frac{56}{70} = 0.80 \)[/tex].

2. Sample Sizes:
- [tex]\( n_1 = 85 \)[/tex] for voters aged 18-30.
- [tex]\( n_2 = 70 \)[/tex] for voters aged 31-40.

3. Standard Errors of the Sample Proportions:
- Standard error for [tex]\( \hat{p}_1 \)[/tex]: [tex]\( SE_1 = \frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} = \frac{0.73 \times (1 - 0.73)}{85} = 0.0028157142857142855 \)[/tex].
- Standard error for [tex]\( \hat{p}_2 \)[/tex]: [tex]\( SE_2 = \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2} = \frac{0.80 \times (1 - 0.80)}{70} = 0.0018823529411764702 \)[/tex].

4. Combined Standard Error:
- Combined standard error: [tex]\( SE_{\text{combined}} = \sqrt{SE_1 + SE_2} = \sqrt{0.0028157142857142855 + 0.0018823529411764702} = 0.0685424483578662 \)[/tex].

5. Difference in Sample Proportions:
- [tex]\( \hat{p}_1 - \hat{p}_2 = 0.73 - 0.80 = -0.07000000000000006 \)[/tex].

6. Z-Score for 99% Confidence Interval:
- [tex]\( z = 2.58 \)[/tex].

7. Margin of Error:
- Margin of error: [tex]\( ME = z \times SE_{\text{combined}} = 2.58 \times 0.0685424483578662 = 0.1768395167632948 \)[/tex].

8. Confidence Interval:
- Lower bound: [tex]\( (\hat{p}_1 - \hat{p}_2) - ME = -0.07000000000000006 - 0.1768395167632948 = -0.24683951676329485 \)[/tex].
- Upper bound: [tex]\( (\hat{p}_1 - \hat{p}_2) + ME = -0.07000000000000006 + 0.1768395167632948 = 0.10683951676329473 \)[/tex].

So, the 99% confidence interval for the difference in proportions of voters who would support the increase in the food tax for the different age groups is:

[tex]\[ (-0.24683951676329485, 0.10683951676329473) \][/tex]
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