Question

1a) A trend in urban development is to reduce the need for residents to have a car; city neighborhoods are often ranked for “walkability’.

• The US city with the lowest car ownership rate is New York City; a majority (56%) of households are “car-free” with only 44% of households owning any vehicles.
• San Jose has the highest car ownership rate of large US cities; only about 6% of households “car-free”.
• San Francisco’s percent of “car-free” households has changed rapidly in recent years.

Suppose a recent study of 1200 households in San Francisco showed that 372 households were “car-free”. Construct and interpret a 95% confidence interval for the true proportion of households in San Francisco that are “car-free”. Use a 95% confidence level.

Population parameter (p): =   

random variable (p') =

In the first part, the point estimate is ____ = _______ ; the confidence interval is _______________

(i) Can we conclude with 95% confidence that more than 25% of SF households are “car-free”? Explain.

(ii) Can we conclude with 95% confidence that more than 30% of SF households are “car-free”? Explain

(iii) What does it mean when we say the confidence level is 95% or that "we are 95% confident"?

Answer 1

Population parameter(P) : = 0.56

Random variable (p') = 372/1200 = 0.31

The point estimate is p' = 0.31

At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

P' +/- z_0.025 * sqrt(p'(1 - p')/n)

= 0.31 +/- 1.96 * sqrt(0.31(1 - 0.31)/1200)

= 0.31 +/- 0.0262

= 0.2838, 0.3362

I) Since 0.25 doesn't lie in the confidence interval, so we can conclude that with 95% confidence that more than 25% SF households are "car-free".

ii) Since 0.30 lies in the confidence interval, so we cannot conclude that more than 30% of SF households are " Car - free".

Iii) We are 95% confident that the true proportion of households in San Francisco that are "car - free" lies with the confidence limits 0.2838 and 0.3362.