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To calculate the [tex]\(99\%\)[/tex] confidence interval for the true proportion of cars with damage from the storm, we need to follow these steps:
1. Determine the sample proportion [tex]\( \hat{p} \)[/tex]:
[tex]\[ \hat{p} = \frac{\text{number of damaged cars}}{\text{total sample size}} = \frac{18}{50} = 0.36 \][/tex]
2. Identify the z-value corresponding to the confidence level:
For a [tex]\(99\%\)[/tex] confidence level, the z-value is [tex]\(2.58\)[/tex].
3. Calculate the standard error [tex]\( SE \)[/tex]:
[tex]\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.36 \times (1 - 0.36)}{50}} \approx 0.06788225099390856 \][/tex]
4. Calculate the margin of error (MOE):
[tex]\[ MOE = z \times SE = 2.58 \times 0.06788225099390856 \approx 0.17513620756428408 \][/tex]
5. Construct the confidence interval:
[tex]\[ \text{Lower bound} = \hat{p} - MOE = 0.36 - 0.17513620756428408 \approx 0.1848637924357159 \][/tex]
[tex]\[ \text{Upper bound} = \hat{p} + MOE = 0.36 + 0.17513620756428408 \approx 0.535136207564284 \][/tex]
Therefore, the [tex]\(99\%\)[/tex] confidence interval for the true proportion of cars with damage is approximately [tex]\((0.1849, 0.5351)\)[/tex].
Among the given options, the one that matches our calculation for a [tex]\(99\%\)[/tex] confidence interval is:
[tex]\[ 0.36 \pm 2.58 \sqrt{\frac{0.36(1 - 0.36)}{50}} \][/tex]
This is the correct answer.
1. Determine the sample proportion [tex]\( \hat{p} \)[/tex]:
[tex]\[ \hat{p} = \frac{\text{number of damaged cars}}{\text{total sample size}} = \frac{18}{50} = 0.36 \][/tex]
2. Identify the z-value corresponding to the confidence level:
For a [tex]\(99\%\)[/tex] confidence level, the z-value is [tex]\(2.58\)[/tex].
3. Calculate the standard error [tex]\( SE \)[/tex]:
[tex]\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.36 \times (1 - 0.36)}{50}} \approx 0.06788225099390856 \][/tex]
4. Calculate the margin of error (MOE):
[tex]\[ MOE = z \times SE = 2.58 \times 0.06788225099390856 \approx 0.17513620756428408 \][/tex]
5. Construct the confidence interval:
[tex]\[ \text{Lower bound} = \hat{p} - MOE = 0.36 - 0.17513620756428408 \approx 0.1848637924357159 \][/tex]
[tex]\[ \text{Upper bound} = \hat{p} + MOE = 0.36 + 0.17513620756428408 \approx 0.535136207564284 \][/tex]
Therefore, the [tex]\(99\%\)[/tex] confidence interval for the true proportion of cars with damage is approximately [tex]\((0.1849, 0.5351)\)[/tex].
Among the given options, the one that matches our calculation for a [tex]\(99\%\)[/tex] confidence interval is:
[tex]\[ 0.36 \pm 2.58 \sqrt{\frac{0.36(1 - 0.36)}{50}} \][/tex]
This is the correct answer.
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