Question

out of a sample of 1,450 people, 1,000 said they prefer to drive themselves to work rather than carpool. construct a 95% confidence interval for the population mean of people who would rather drive themselves than carpool. ci = (65.84%, 72.09%) ci = (66.97%, 70.96%) ci = (65.27%, 72. 13%) ci = (66.58%, 71.35%)

Explanation:

Step1: Calculate the sample proportion ($\hat{p}$)

The sample proportion is the number of successes (people who prefer to drive themselves) divided by the sample size. Here, the number of successes $x = 1000$ and the sample size $n = 1450$. So, $\hat{p}=\frac{x}{n}=\frac{1000}{1450}\approx0.6897$ (or $68.97\%$).

Step2: Determine the z-score for a 95% confidence interval

For a 95% confidence interval, the z-score ($z_{\alpha/2}$) is 1.96 (this is a standard value for a 95% confidence level, corresponding to the middle 95% of the standard normal distribution).

Step3: Calculate the standard error (SE)

The formula for the standard error of a proportion is $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting the values, we get $SE=\sqrt{\frac{0.6897\times(1 - 0.6897)}{1450}}=\sqrt{\frac{0.6897\times0.3103}{1450}}\approx\sqrt{\frac{0.2140}{1450}}\approx\sqrt{0.0001476}\approx0.01215$.

Step4: Calculate the margin of error (ME)

The margin of error is $ME = z_{\alpha/2}\times SE$. Substituting the values, $ME = 1.96\times0.01215\approx0.02381$.

Step5: Calculate the confidence interval

The confidence interval is $\hat{p}\pm ME$. So, the lower bound is $0.6897 - 0.02381\approx0.6659$ (or $66.59\%$) and the upper bound is $0.6897 + 0.02381\approx0.7135$ (or $71.35\%$). Rounding to two decimal places for the percentage, the confidence interval is approximately $(66.58\%, 71.35\%)$ (the slight difference in the lower bound is due to rounding during intermediate steps).

Answer:

$\text{CI} = (66.58\%,\ 71.35\%)$ (corresponding to the last option)