## Related questions with answers

Last year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was $1000 or more and to whether a physical injury resulted from the accident. The data follows:

Under$1000: Number of Accidents 32 and Number Involving injuries 10 $1000 or more: Number of Accidents 41 and Number Involving Injuries 23

a. Estimate the true proportion of accidents involving injuries when the damage was$1000 or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under $1000 and those with damage of$1000 or more. Use a 95% confidence interval.

Solution

Verified(a) Given:

$\begin{align*} n&=\text{Sample size}=41 \\ x=&\text{Number of successes}=23 \\ c&=\text{Confidence coefficient}=95\%=0.95 &\color{#4257b2}(\text{Assumption}) \end{align*}$

The sample proportion is the number of successes divided by the sample size:

$\hat{p}=\dfrac{x}{n}=\dfrac{23}{41}\approx 0.5610$

For confidence level $1-\alpha=0.95$, determine $z_{\alpha/2}=z_{0.025}$ using the normal probability table in the appendix (look up 0.05 in the table, the z-score is then the found z-score with opposite sign):

$z_{\alpha/2}=1.96$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}\hat{q}}{n}}=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=1.96\cdot \sqrt{\dfrac{0.5610(1-0.5610)}{41}}\approx 0.0312$

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